Solutions

Mole fraction is useful in relating vapor pressure with the concentration of a solution.

According to Raoult’s law, the partial vapor pressure of each component in the solution is directly proportional to its mole fraction.

Thus, for a component C in a solution

Where P_C is the partial pressure of component C and x_c is a mole fraction of C in a solution.

Thus, option (i) is corerect and others are wrong.

An increase in the temperature increases the dissolution as it is easy to break the bonds of sugar at a higher temperature and dissolution is an endothermic process.

Powdered sugar dissolves faster in water, as the surface area of contact of the sugar molecules with water increases i.e. more molecules come in the contact with the molecules of water.

Thus, powdered sugar in hot water dissolves rapidly than other conditions.

Thus, Option (iv) is the correct.

At the state of Equilibrium, the rate of the dissolution of the solute is equal to the rate of the crystallization of the solute i.e. the solute crystal will remain at a constant size.

If the rate of the dissolution is greater than the rate of the crystallization, then the solute crystal in the solution will get smaller.

If the rate of crystallization is greater than the rate of dissolution, then the solute crystal will get larger in the solution.

This equality is the cause of the Equilibrium.

Thus, Option (iii) is the correct answer.

On the addition of the solute to a saturated solution, the solute settles down (precipitation).

For the concentrated and unsaturated solution, the excess solute gets dissolved.

A supersaturated solution is highly unstable and crystallization of solute takes place on a slight disturbance.

Thus, option (i) is the correct option.

Pressure has nearly no effect on the solubility of a solute in a solution as solute is nearly incompressible.

Thus, it’s solubility cannot be increased by pressure.

The solubility of solute depends on its nature and nature of the solvent.

The solubility of a solute increases with the increase in temperature as the molecules easily break at a higher temperature, due to the increase in the activation energy.

Thus, Option (iii) is the correct answer.

At higher altitudes, the layer of atmosphere is not as dense as it is on the sea level.

Thus, the atmospheric pressure decreases and the solubility of gases also decreases.

The presence of oxygen gas is less in the atmosphere at higher altitudes due to the above reason. Thus, people living in higher altitudes have a low concentration of oxygen in the blood.

Thus, Option 2 is the correct answer.

CH₃OH, methanol has hydrogen bonding between its molecules. When acetone is added the hydrogen bonding breaks and new bonds formation takes place, these bonds

are not so strong as intermolecular hydrogen bonding i.e. interaction between acetone -methanol is not as strong as the interaction between methanol-methanol and acetone-acetone molecules.

Since the boiling point has reduced the vapor pressure will increase above the liquid solution, which is a positive deviation from Raoult's law.

Thus, at a specific composition methanol-acetone mixture will form minimum boiling azeotrope and will show positive deviation from Raoult’s law.

Thus, Option (i) is the correct answer.

Colligative property depends upon the number of solute particles in the solution and is independent of its nature.

The van’t Hoff factor of a solute helps in determining it’s effect on the colligative property of the solution.

Thus, Option 2 is the correct answer.

An increase in the boiling point of the solution depends upon the van’t Hoff factor (i) of the solute when it is added to the solution.

In Option 1-NaOH dissociates into 2 ions,

In Option 2-Na₂SO₄ dissociates into 3 ions,

In Option 3-NH₄NO₃ dissociates into 2 ions and

In Option 4-KNO₃ dissociates into 2 ions.

Van’t Hoff factor is the maximum for option (ii).

Thus, the maximum increase in the boiling point will be on the addition of 1M Na₂SO₄ in solution.

Thus, Option (ii) is the correct answer.

In thermodynamics, Ebullioscopic constant relates molality to the boiling point elevation.

This relation is

Where is the elevation in the boiling point, i is the van’t Hoff factor, b is the molality and K_b is the Ebullioscopic constant.

Thus, the unit of an ebullioscopic constant is K (molality)^-1 or K Kg mol^-1, as molality= .

Thus, Option (i) is the correct answer.

Depression in freezing point = i × m × K​_f

Now, i = van’t Hoff factor, m is the molality and K_f is the freezing constant of the solvent.

Here, we see that molality is the same i.e. 0.01 M and K_f will be the same as the solvent is water.

Thus, depression in freezing point depends only on the van’t Hoff factor.

i = Total number of ions after dissociation/association/Total number of ions before dissociation/association.

i= 3 for MgCl2 as it dissociates into 3 ions

i = 1 for glucose as it does not undergo any association/dissociation.

Thus, option (iii) is the correct option.

Osmosis is the flow of the solvent from higher concentration to lower concentration through a semipermeable layer.

Unripe mango has a higher concentration of water and salt solution has a lower concentration of water.

Thus, water flows from mango to salt solution due to osmosis, as a result, mango shrivels.

Thus, Option (iv) is the correct answer.

At a given temperature osmotic pressure is given by a formula i.e.

Where is the osmotic pressure, is the van’t Hoff factor, M is the molar concentration, R is the universal gas constant and T is the temperature.

Thus, osmotic pressure directly proportional to the concentration.

For a concentrated solution, the osmotic pressure is greater than the osmotic pressure for a dilute solution.

Thus, Option (i) is the correct answer.

Depression in freezing point = i × m × K​_f

Now, i = van’t Hoff factor, m is the molality and K_f is the freezing constant of the solvent.

Now K_f is the property of the solvent and is different for different solvent.

Two different solutions of sucrose of the same molality prepared in different solvents will have different depression in freezing point.

Thus, statement (i) is false.

At a given temperature osmotic pressure is given by a formula i.e.

Where is the osmotic pressure, is the van’t Hoff factor, M is the molar concentration, R is the universal gas constant and T is the temperature.

Thus, statement (ii) is correct.

When the concentration of the solute is same osmotic pressure depends on the van’t Hoff factor.

Van’t Hoff factor for BaCl₂ is 3, for KCl is 2, for CH₃COOH is 2 and sucrose it is 1.

Since KCl is ionic it will completely dissolve in the solution its osmotic pressure is greater than that of CH₃COOH.

Thus statement (iii) is also correct.

According to Raoult’s law, the partial vapor pressure of each component in the solution is directly proportional to its mole fraction.

Thus, for a component C in a solution

Where P_C is the partial pressure of component C and x_c is a mole fraction of C in a solution.

Thus statement (iv) is also correct.

The van’t Hoff factor is:

i = Total number of ions after dissociation/association/Total number of ions before dissociation/association.

Thus, for KCl, van’t Hoff factor = 2,

for NaCl van’t Hoff factor is 2 and

for K₂SO₄ van’t Hoff factor is 3.

Thus, Option 2 is the correct answer.

Statement (ii) is false, as in reverse osmosis, solvent molecules move through a semi-permeable membrane from a region of higher solute concentration to a lower solute concentration under the application of hydrostatic pressure.

The unit of atmospheric pressure and osmotic pressure is the same i.e. atm.

The value of molal depression constant depends upon the nature of the solvent i.e T_f =K_f * M, where K_f depends on the nature of the solvent.

The relative lowering of vapor pressure is a dimensionless quantity as it is equal to the mole fraction (which is dimensionless).

Thus, Option (ii) is the correct answer.

The value of K_H increases with temperature due to the increase in the molecular agitation.

Greater the molecular agitation (entopy) of gases it is difficult to dissolve it in solvent.

Thus, statement (i) is true.

The higher the K_h lower is the solubility of the gas in the solvent.

More pressure is required to dissolve a gas with higher K_h.

Thus, statement (ii) is true and others are false.

In reverse osmosis, solvent molecules move through a semi-permeable membrane from a region of higher solute concentration to a lower solute concentration under the application of hydrostatic pressure, vice versa is true for osmosis.

Thus if the pressure applied to piston B, is greater than osmotic pressure reverse osmosis will take place.

Thus, Option (ii) is correct.

The van’t Hoff factor for strong electrolyte is:

i = Total number of ions after dissociation/association/Total number of ions before dissociation/association.

For NaCl, the van’t Hoff factor is 2 as it dissociates into two ions i.e.

Van’t Hoff factor is independent of the molality of solute added.

Thus, i_A = i_B = i_C .

Thus, Option (iii) is the correct answer.

Solution A will follow Raoult's law as it is an ideal solution. The intermolecular interactions of A–A and B–B type are nearly same as A–B type interactions.

Solution B will form a minimum boiling azeotrope and show a positive deviation from Raoult’s law.

Solution C will form a maximum boiling azeotrope and show a negative deviation from Raoult’s law.

Thus, Option(ii) is the correct answer.

NaCl is a non-volatile substance when it is added to the water it increases the boiling point of the water simultaneously decreasing the vapor pressure of the water.

According to Raoult’s law, the partial vapor pressure of each component in the solution is directly proportional to its mole fraction.

Thus, for water as a solvent in a solution

Where P_w is the partial pressure of water and x_W is a mole fraction of water in the solution.

Thus, on adding NaCl the mole fraction of water reduces.

Thus, vapor pressure also reduces.

Thus, Option (i) is the correct answer.

If two liquids A and B form minimum boiling azeotrope at some specific composition then the bonds between A-B is weaker than A-A and B-B bonds because bonds are breaking at a lower temperature for a new interaction.

Thus, statement (iv) is true and others are false.

Molarity =

0.02= , n is the number of moles of NaCl

n=.08

The volume of the solution is changed to 5L (4+1)

Molarity=

Thus, molarity is 0.016.

Thus, Option (iv) is the correct answer.

CH₃OH, methanol has hydrogen bonding between its molecules. When acetone is added the hydrogen bonding breaks and new bonds formation takes place, these bond sare not so strong as intermolecular hydrogen bonding i.e. interaction between acetone -methanol is not as strong as the interaction between methanol-methanol and acetone-acetone molecules.

Since the boiling point has reduced the vapor pressure will increase above the liquid solution, which is a positive deviation from Raoult's law.

Thus, at a specific composition methanol-acetone mixture will form minimum boiling azeotrope and will show positive deviation from Raoult’s law.

Thus, Option(i) is the correct answer.

The higher the K_h lower is the solubility of the gas in the solvent.

More pressure is required to dissolve a gas with higher K_h.

Thus, the correct increasing order of the solubility is Ar < CO₂ < CH₄ < HCHO.

Thus, Option(iii) is the correct answer.

Many gases are soluble in liquids. It depends on the nature of the solute, hence (a) is a correct option. The solubility of gases decreases with increase in temperature as the process of dissolution can be considered similar to condensation and heat is evolved too. The higher the temperature the more molecules escape the liquid phase. Hence, (b) is the correct option. The solubility of gases increases on increase in the pressure in the system, which is due to the increase in the number of gaseous particles per unit volume striking the surface of the solution to enter it. Hence, (c) is a correct option.

Hence the correct factors and options are (i) and (ii).

A solution of benzene and toluene constitutes an ideal solution. An ideal solution is one that follows Raoult’s Law over the entire range of all concentrations. Raoult’s law states that, “for a solution of volatile liquids the partial vapour pressure of each component of the solution is directly proportional to its mole fraction present in solution.” Ideal solutions have two major properties that the (1) enthalpy of mixing of the pure components to form the solution is zero and (2) the volume of mixing is also zero. Hence ΔmixH = zero and ΔmixV = zero. Hence, option (i) and (ii) are true. Also, since the solution formed is ideal in nature, it does not show positive and negative deviation from Raoult’s law, hence it does not form an azeotrope. Hence, (iii) and (iv) are not true about the solution.

The correct options are (iii) and (iv).

Colligative properties are those that depend on the number of solute particles irrespective of their nature relative to the total number of particles present in the solution. Relative lowering of vapour pressure is a colligative property as Raoult’s observations established that the lowering of vapour pressure depends only on the concentration of the solute particles and it is independent of their identity.

Relative lowering of vapour pressure is equal to the mole fraction of the solute.

The vapour pressure of a solution reduces on addition of a non-volatile solute, and does not depend on the nature of the solute molecules, which is due to the colligative property. Hence option (i) is correct.

When electrolytes are dissolved in solution they are separated into ions. The vapour pressure lowering does not depend on the electrolyte itself but on the number of ions. Hence (ii) is correct.

The options (iii) and (iv) are incorrect options as they go against colligative properties.

The correct options are (i) and (ii).

The change in the observed boiling point versus the expected boiling point is known to be a result of abnormal molar mass. The scientist van’t Hoff introduced a factor to understand the value of dissociation and association. The van’t Hoff factor “i” is the ratio of normal molar mass to abnormal molar mass. It is the ratio of observed colligative property to calculated colligative property. Hence, the correct options are (i) and (iii) and incorrect options are (ii) and (iv).

Two solutions which have the same osmotic pressure at a given temperature are said to be isotonic in nature. The osmotic pressure of a given solution is given by the formula

π = CRT = n/VRT

Where π is the osmotic pressure of the solution, C is the concentration, the moles of solute per volume of solvent and T is the temperature, with R being universal gas constant. Osmotic pressure is a colligative property, hence for two solutions to be isotonic, the nature of the solute does not matter, i.e. the solute in the two solutions do not have to be the same. Even the density, which does not have a place in the equation of osmotic pressure does not need to be the same. The isotonic solutions at a given temperature need to have the same volume and same molar concentration, hence they will also have the equal elevation in boiling point and depression in freezing point, two other colligative properties.

The correct answers are (iii) and (iv).

Binary mixtures that have the same composition in liquid and vapour phase are known as azeotropes. These solutions boil at a constant temperature. The two types of azeotropes are minimum boiling azeotropes and maximum boiling azeotropes. Water-nitric acid is a maximum boiling azeotrope and water-ethanol is a minimum boiling azeotrope. The options (i) and (iv) are examples of ideal solutions, hence they cannot be azeotropes.

Hence the correct answers are (ii) and (iii).

Two solutions which have the same osmotic pressure at a given temperature are said to be isotonic in nature. The osmotic pressure of a given solution is given by the formula

π = CRT = n/VRT

Where π is the osmotic pressure of the solution, C is the concentration, the moles of solute per volume of solvent and T is the temperature, with R being universal gas constant. Osmotic pressure is a colligative property, hence it depends on the concentration of the solute in the solution and not the identity of either two. Hence the solvent and solute may or may not be the same.

Hence the correct answers are (ii) and (iii).

Ideal solutions are binary solutions which follow the Raoult’s Law over the entire range of concentrations, which states that the partial vapour pressure of the components of a solution is directly proportional to its mole fraction in the solution. Also, total vapour pressure over the solution can be related to the mole fraction of any one component. The graphs between the partial vapour pressure for components of an ideal solution should be a straight line.

p1 ∝x1 and p2 ∝x2

From the given figures, the proportion is correctly given by options (i) and (iv). Hence the correct options are (i) and (iv).

Colligative properties are those that depend on the number of solute particles relative to the total particles present in the solution and not the nature of the solute. Colligative properties are observed when a non-volatile solute is added to a volatile solvent. Hence the correct options are (i) and (ii).

The solution in question is an example of an azeotrope. Azeotropic solutions are binary solutions which have the same composition in liquid and vapour phase and boil at a constant temperature. The components of an azeotropic solution cannot be separated by fractional distillation. An example of azeotropes is ethanol-water mixture formed after fermentation of sugars. Fractional distillation of this solution gives a solution of 95% of ethanol. Further fractional distillation does not give a higher concentration of ethanol. This composition is similar to the one in the question and is known as the azeotrope composition. In this composition, the liquid and vapour have the same composition, and no further separation occurs.

Addition of a non-volatile solute like NaCl to solvent like water increases the boiling point of the solution. On the other hand, methyl alcohol is more volatile than water, hence addition of methyl alcohol as solute to water decreases the boiling point.

“Like dissolves like” in solutions means that a solute dissolves in a solvent if their intermolecular interactions are similar. This means that if polar solutes can dissolve in polar solvents and non-polar solutes can dissolve in non-polar solvents.

The molarity of a solution is defined as the number of moles dissolved in one litre of solution. A change in temperature causes a change in volume, hence changes the molarity of a solution. On the other hand, the mass of a solution does not change with a change in temperature. Hence, terms such as mass percentage, ppm, mole fraction, and molality are independent of temperature as they are calculated on the basis of mass and do not involve volume.

Henry’s law states that in a solution, “the partial pressure of the gas in vapour phase (p) is proportional to the mole fraction of the gas (x) in the solution”

The equation is given as p = KHx, where KH is the Henry Law’s constant. The constant is a function of the nature of the gas.

KH = x/p, from this equation we can deduce that higher the value of KH at a given pressure, the lower is the solubility of the gas in the liquid.

Henry’s Law states that “the partial pressure of the gas in vapour phase (p) is proportional to the mole fraction of the gas (x) in the solution” and is given by the equation p = KHx, where KH is the Henry Law’s constant. From the equation, it is deduced that higher the KH at a given pressure, lower is the solubility of the gas in the liquid. For O2, the value of KH also increases on increase of temperature, thus an increase in temperature will reduce the solubility of the gas. Therefore, a reduction in temperature will increase the solubility. Hence, aquatic species are more comfortable in cold water due to the increased solubility and availability of O2.

(a) Henry’s Law states that, “the partial pressure of the gas in vapour phase (p) is proportional to the mole fraction of the gas (x) in the solution” and is given by the equation p = KHx, where KH is the Henry Law’s constant.

(i) The bends is a condition which occurs with scuba divers due to changes in pressure underwater. An increase in pressure increases solubility of gases in blood. Scuba divers have to breathe air under high pressures underwater in the presence of high concentrations of dissolved gases. The major gases present in the atmosphere are N2 and O2. These gases are not very soluble at normal pressures, however, underwater at high pressure, they dissolve in the blood. Once the divers come up to the surface, the gases escape the blood by forming bubbles of N2 in blood. These bubbles can block capillaries and obstruct the flow of O2, which forms the painful condition of bends.

(ii) At higher altitudes, atmospheric pressure is low and it reduces the solubility of oxygen in blood and tissues of people traveling or living at high altitudes. The lack of oxygen causes weakness and unable to think properly, a condition known as anoxia.

(b) Soda water and soft drinks are produced by adding CO2 to the liquid and sealing the bottles under high pressure, in order to increase the solubility of CO2. When a soda water bottle is opened at room temperature and normal pressure conditions, the pressure inside the bottle reduces and solubility of CO2 reduces, thus causing the gas bubbles to escape and soda water fizzing.

In the pure water solvent, the molecules present on the surface are only of water and the vapour pressure exerted is from the solvent molecules. When a non-volatile solute is dissolved in water such as glucose, the vapour pressure exerted on the solution is only from solvent molecules. Due to the glucose dissolving in water, a fraction of molecules covering the surface of water are occupied by glucose molecules. The presence of the solute molecules prevents the escape of solvent molecules from the surface into the vapour phase, hence the vapour pressure of an aqueous solution of glucose is lower than water.

Addition of salt to snow reduces the freezing point of ice. The lowering of the freezing point helps in melting the ice. The freezing point of water is 0°C. Lowering of the freezing point causes the point to go below 0°C and the ice melts. Hence salt is a de-icing agent and can be used to melt ice covered roads in hilly areas.

A semipermeable membrane is a membrane that allows only the flow of solvent molecules but not solute molecules. During the process of osmosis and reverse osmosis, the semipermeable membranes allow the movement of solvent molecules. Examples of semipermeable membranes are cellulose acetate and phospholipid bilayer in biological cells.

Reverse osmosis is the process where the direction of osmosis can be reversed if a pressure larger than the osmotic pressure is applied to the solution side. That is, the pure solvent flows out of the solution through the semi permeable membrane. Examples of materials used for semipermeable membrane include cellulose acetate and polyamides.

(i) Saturated solution – (d) A solution which contains the maximum amount of solute which can be dissolved in a given amount of solvent at a particular temperature is called as saturated solution

Explanation: When a maximum amount of solute has been added to a solvent at a given temperature, then this solution is said to be saturated.

For example, let’s say we keep dissolving salt in water at a given temperature. After a maximum amount has been reached, we can no longer dissolve more salt. Then this solution is said to be saturated.

When the amount exceeds this maximum amount of solute, the solution is said to supersaturate.

(ii) Binary solution- (c) Solution with two components

Explanation: Binary means two. A binary solution contains two components, each of which might be a solid, liquid or a gas.

(iii) Isotonic solution –(a)Solution having the same osmotic pressure at a given temperature as that of a given solution.

Explanation: Isotonic solutions are two solutions having the same osmotic pressure. There is no pressure difference, so the solutions can move across and mix with each other easily.

(iv) Hypotonic solution – (b) A solution whose osmotic pressure is less than that of another.

Explanation:Hypo means less. A hypotonic solution has less osmotic pressure than that of another solution. It is very dilute, i.e. it has less solute dissolved compared to another solution.

(v) Solid solution – (f) A solution in solid phase

Explanation: When two or more components form a homogenous solution in the solid state then they form a solid solution. For example, metal alloys are an amalgamation of two solid components.

(vi) Hypertonic solution – (e) A solution whose osmotic pressure is more than that of another.

Explanation:Hypertonic solutions have more solutes dissolved in them which makes them more concentrated. This gives rise to higher osmotic pressure.

(i) Soda water- (e) A solution of the gas in liquid

Explanation: Carbonated drinks are a system of gas in a liquid. The solubility of CO2 in water is increased by applying high pressure. This is also known as Henry’s law. Henry’s law states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas present above the surface of liquid or solution.

(ii) Sugar solution – (c) A solution of solid in liquid

Explanation:Sugar crystals(solute) are solids dissolved in a liquid(solvent). Therefore, it is a solution of solid in a liquid.

(iii) German silver – (d) A solution of solid in solid

Explanation:German silver is an alloy of nickel, zinc and copper. Alloys are amalgamations of two or more solids. Therefore, it is a solution of solid in solid.

(iv) Air – (b) A solution of the gas in gas

Explanation: Air is a homogenous mixture of different gases. Different gases mix together uniformly, and therefore it is a solution of the gas in gas.

(v) Hydrogen gas in palladium- (a) A solution of the gas in solid

Explanation:Hydrogen(a gas) gets adsorbed onto the surface of palladium( a solid metal). Therefore, it is a solution of the gas in solid.

(i) Raoult’s law- (c) p=x1p1° + x2p2°

Explanation: Raoult’s law states that for a solution of volatile liquids, the partial pressure of each component of the solution is directly proportional to its mole fraction present in solution.

p=x1p1° + x2p2°

where p1° is the vapour pressure of pure component 1,

x1 is the mole fraction of component 1.

p2° is the vapour pressure of pure component 2,

x2 is the mole fraction of component 2

and p is the total pressure over the solution phase

(ii) Henry’s law – (e) p=KH.x

Explanation: Henry’s law states that at a constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas present above the surface of liquid or solution.

p=KH.x

where KH is Henry’s law constant,

x is the mole fraction of gas in solution and

p is the partial pressure of the gas in the vapour phase.

(iii)Elevation of the boiling point – (d) ∆Tb=Kbm

Explanation:

In the presence of non-volatile solute, the vapour pressure of the solution decreases. To reach the boiling point, vapour pressure should be increased to equal with atmospheric pressure, by increasing the temperature. This leads to an increase in the boiling point. Elevation in boiling point is measured as,

∆Tb=Kbm

Where ∆Tb is the elevation in boiling point

Kb is the boiling point elevation constant,

m is the molality of the solution

(iv)Depression in freezing point- (a) ∆Tf=Kfm

Explanation:

When the vapour pressure of solution equals vapour pressure of pure solid, then the solution will freeze. When a non- volatile solid is added to the solution, it lowers the vapour pressure which brings down the freezing point. Depression in freezing point can be calculated as,

∆Tf=Kfm

Where ∆Tf is the depression in freezing point

Kf is the freezing point depression constant

m is the molality of the solution

(v)Osmotic pressure-(b)π = CRT

Explanation: The excess pressure applied to the solution side to stop the flow of solvent molecules through the semi-permeable membrane is called osmotic pressure. Osmotic pressure can be represented as,

π = CRT

where, π is the osmotic pressure

C is the molarity of the solution

T is the temperature

R is the gas constant

(i) Mass percentage - (d)

Explanation: Mass percentage of a component in the solution is the mass of that solute component present in solution divided by the total mass of the solution multiplied by 100

_{Mass percentage=}

(ii) Volume percentage – (c)

Explanation: Volume percentage is the volume of the solute component present in solution divided by the total volume of solution, multiplied by 100

_{Volume percentage=}

(iii) Mole fraction –(b)

Explanation: Mole fraction is the ratio of the number of moles of a particular component present in the system by the total number of moles of all the components present.

Mole fraction=

(iv) Molality- (e)

Explanation: Molality is defined as the ratio of number of moles of solute in a given mass of solvent( expressed in kg)

Molality=

(v) Molarity- (a)

Explanation: Molarity is defined as the ratio of number of moles of solute present in a given volume of solution(expressed in L)

Molarity=

Molality is temperature independent, but molarity is temperature dependent.

Molality= (Number of moles of the solute components)/(Mass solvent in ki log⁡r ams)

Molality depends on the mass of solvent which remains constant even on increasing temperature. Mass never increases or decreases with temperature.

Molarity=(Number of moles of solute component)/(Volume of solution( in litres))

But molarity, on the other hand, depends on the volume of the solution volume changes when the temperature changes. Therefore, molarity is temperature dependent.

In the presence of non-volatile solute, the vapour pressure of the solution decreases. Vapour pressure should be increased to equal with atmospheric pressure, by increasing the temperature. This leads to an elevation in the boiling point. Elevation in boiling point is measured as,

∆Tb=Kbm

Where ∆Tb is the elevation in boiling point

Kb is the boiling point elevation constant,

m is the molality of the solution

But on the addition of a volatile solute, vapour pressure increases. In this case, to equal vapour pressure to atmospheric pressure temperature needs to be brought down which decreases the boiling point.

When the vapour pressure of solution equals vapour pressure of pure solid, then the solution will freeze. When a non- volatile solid is added to the solution(here NaCl), it disturbs the dynamic equilibrium between solid and liquid and lowers the vapour pressure. Lowering of vapour pressure means that the temperature at which vapour pressure of frozen solid and liquid equals will also fall. Therefore, there is a depression in freezing point. This depression can be calculated as,

∆Tf=Kfm

Where ∆Tf is the depression in freezing point

Kf is the freezing point depression constant

m is the molality of the solution

Osmosis is a process in which solvent molecules pass through a semi-permeable membrane from lower concentration to a higher concentration. Here, the reason statement is false because the movement is not from higher concentration to a lower concentration but the other way around.

(i) w/w or mass percentage of a component of a solution is defined as the mass of the individual component in the solution in the total mass of the solution in percent. The formula is given as

w/w = (Mass of the component of the solution)/(Total Mass of the solution) x 100

(ii) V/V or volume percentage is defined as the percentage of the volume occupied by the component in the total volume of the solution. The formula is given as

V/V = (Volume of the component)/(Total volume of the solution) x 100

(iii) Mass by volume percentage, w/V, is defined as mass of solute dissolved in 100mL of the solution. This can be calculated by measuring the mass of the solute in grams dissolved in milliliters of solute and converting it into the percent form.

(iv) Ppm or parts per million, it is a convenient method to express concentration when a solute is present in trace quantities. It is defined as:

Parts per million = (Number of parts of the component× 10^6)/(Total number of parts of all the components in the solution)

(v) Mole fraction or x is defined as the ratio of the number of moles of the component in a solution to the total number of moles of all components in the solution.

x = (Number of moles of a component)/(Total number of moles of all the components of the solution)

Mole fraction with a subscript on its right-hand side denotes its component.

(vi) Molarity (M) of a solution is defined as the number of moles of solute dissolved in 1 litre [or 1000mL or 10 cubics decimetre] of solution.

Molarity = (Moles of solute(n))/(Volume of solution in litre(V))

(vii) Molality (m) of a solution is defined the number of moles of the solute per kilogram (kg) of solvent. It is expressed as:

Molality = (Moles of solute(n))/(Mass of solvent in kg)

Out of these values, (i) mass percentage, (iv) ppm, (v) mole fraction and (vii) molality are independent of temperature, whereas (ii) volume percentage, (iii) mass by volume percentage, (vi) molarity is a function of temperature. This is because volume depends on temperature and the mass does not. For instance, heating a liquid substance will cause expansion of the constituent molecules and cause an increase in the volume. The mass is still the sum of the individual masses of the particles.

1. Raoult’s law states that ‘for a solution of volatile liquids, the partial vapour pressure of each component of the solution is directly proportional to its mole fraction present in solution.’ This means that the vapour pressure of the solvent above the solution is equal to the vapour pressure of the pure solvent at the same temperature regulated by the mole fraction of the solvent present. Thus for one of the two components of a binary solution,

p1 ∝ x1

and, p1 = p10 x1

Where p1 is the partial vapour pressure of component 1 and p10 is the vapour pressure of the pure solvent at the same temperature, and x1 is the mole fraction of component 1.

Similarly, for component 2, its partial vapour pressure is denoted as

p2 = p20 x2

2. Consider CHCl3 and CH2Cl2, chloroform and dichloromethane, both being volatile liquids, chloroform being non-polar and dichloromethane being lightly polar yet miscible with chloroform forming a binary solution, the total pressure will be the sum of the partial vapour pressures of the two volatile components,

ptotal = p1 + p2

= p10 x1 + p20 x2

= (1-x2)p10 + x2 p20

= p10 + (p20 – p10)x2

And similarly, p20 + (p10 – p20)x1.

(i) This equation explains the Dalton’s law of partial pressures - the total pressure (ptotal) exerted by the gases over the solution phase in a container will be the sum of the partial pressures of the components of the gas. In this case, it will be the sum of the volatile components exerting vapour pressure on the binary solution. This equation also explains the relation between the vapour pressure of the components and the mole fraction.

(ii) The vapour pressure over the solution can be related to the mole fraction of either of the component, CHCl3 or CH2Cl2. The total vapour pressure of the solution depends on the mole fraction of both the components in their pure state and the mole fraction of the second component in solution.

(iii) If the equation were plotted for an ideal solution, it would be a straight line in the equation form of y = mx + c [ptotal = (p20 – p10)x2], where the line is between ptotal and x2 whose slope is given by (p2o-p1o) and the y-intercept is equal to p1o.

3. For NaCl(s) and H2O(l), a solution contains a non-volatile solute in the form of NaCl. Both are polar molecules and with the principle “like dissolves like”, NaCl dissolves in water and exists in ionic form. The fraction of the liquid solvent molecules on the surface of the liquid gets replaced by dissolved solute molecules, reducing the solvent molecules escaping the surface, reducing the vapour pressure of the solution. Raoult’s law is applicable only to vapourisable component, H2O (component 1), and total vapour pressure is written as

ptotal = p1 = x1p10. The plot between the vapour pressure and the mole fraction is linear in this case.

1. Raoult’s Law can be generally defined as, ‘For any solution, the partial vapour pressure of each volatile component in the solution is directly proportional to its mole fraction.’ However, this law has its limitations, especially in the case of a large number of solutions with high concentrations and complexity of components whose intermolecular forces vary considerably from this law.

2. Hence, liquid-liquid solutions are classified into ideal and non-ideal solutions on the basis of Raoult’s Law.

3. Ideal solutions are mixtures which obey the law over the entire range of their concentrations. Two important properties of these solutions are (i) Enthalpy of mixing the pure components to form the solution is zero where no heat is absorbed or evolved when the two components are mixed [ΔmixH = 0] and (ii) Volume of mixing of the solution is zero, where the volume of the solution would be equal to the sum of volumes of the two components without an increase or decrease in volume [ΔmixV = 0].

4. Ideal mixtures do not exist practically but some binary solutions behave ideally to an extent. Examples include n-hexane and n-heptane, benzene and toluene, propan-1-ol and propan-2-ol, bromoethane and chloroethane and so on.

5. A simpler definition of Raoult’s law states ‘The partial vapor pressure of a component in a mixture is equal to the product of the vapour pressure of the pure component at that temperature and its mole fraction in the mixture’. A non-ideal solution does not obey the law and does not follow the properties of ideal solutions. The vapour pressure of these type of solutions are experimentally higher or lower than that predicted by the equation derived from the law.

6. The ideal and non-ideal behaviour of solutions can be explained by understanding the intermolecular interactions of these solutions. The molecules of the two components of a binary solution can be named A and B. In pure components, the intermolecular interactions will be between the molecules themselves i.e. A-A and B-B. In binary solutions, the interactions will be between the two sets of the molecules i.e. A-B in addition to the two before. In a pure solvent, some of the molecules will gain enough energy to overcome the intermolecular attractions and escape from the liquid surface to vapour form. This process of escape is dependent on intermolecular forces. If the forces are of smaller magnitude, more molecules will escape the surface at a constant temperature. In the case of an ideal mixture, this tendency of molecules A as well as B will be the same. If ideal solutions existed, the intermolecular forces between A-A, B-B and A-B will be identical, which is impossible in most cases. It happens to an extent in the examples mentioned previously because the sizes of the constituent molecules are similar and so are the Van der Waals forces. Ideal solution behaviour is also assumed for very dilute solutions.

7. Considering non-ideal solutions, the enthalpy of the mixture as well as the volume of mixing is not equal to zero [ΔmixH ≠ 0; ΔmixV ≠ 0]. Heat is either evolved [ΔmixH < 0] or absorbed [ΔmixH > 0] on mixing the components to form a solution, and the volume of the solution is usually not equal to the sum of volumes of the components.

8. Non-ideal solutions are formed in two situations, one where there is positive deviation from Raoult’s law, and the other where there is negative deviation from Raoult’s law.

9. In case of positive deviation from Raoult’s law, the intermolecular forces between solute-solvent molecules are weaker than solute-solute and solvent-solvent particles. That is, A-B interactions are weaker than A-A and B-B interactions. In these solutions, it will be easier for A or B molecules to escape than be in a pure state. This increases vapour pressure and results in a positive deviation. Here, p1 > p10 x1 and p2 > p02 x2, as the total vapour pressure (p10 x1 + p02 x2) is higher than expected from Raoult’s law. The resulting solution has larger enthalpy than its constituents, causing the reaction to be endothermic and heat is absorbed. The volume of mixing is positive as the volume increases on mixing the constituents. ΔmixH > 0 and ΔmixV > 0.

10. In case of negative deviation of Raoult’s law, intermolecular forces between A-A and B-B are weaker than A-B, that is, solute-solvent interactions are stronger than solute-solute and solvent-solvent interactions, which leads to molecules staying in the pure state than escaping, leading to a reduction in vapour pressure and thus negative deviation. Here, p1 < p10 x1 and p2 < p02 x2, as the total vapour pressure (p10 x1 + p02 x2) is lower than expected from Raoult’s law. The resulting solution has negative enthalpy than its constituents, causing the reaction to be exothermic and heat is released. The volume of mixing is negative as the volume decreases on mixing the constituents. ΔmixH < 0 and ΔmixV < 0.

Figure 1: Solution showing positive deviation from Raoult's Law

1. Various liquids on mixing form azeotropes, which are binary mixtures which have the same composition in liquid as well as vapour phase and boil at a constant temperature. These mixtures cannot be separated into their individual components by fractional distillation.

2. Let us consider mixtures of ethanol and water. The lowest boiling point value of this mixture occurs at 95.6% concentration by mass of ethanol in this solution. The boiling point of this solution is 78.2°C, while that of pure ethanol is 78.5°C and of pure water is 100°C. The 0.3°C difference is important when it comes to separation of ethanol water mixtures. A graph showing a vapour composition curve of ethanol and water explains this further.

3. Boiling of any ethanol water mixture above 0% and below 95.6% mass will give a vapour of higher concentration of ethanol. Condensing and reboiling that mixture will finally lead to 95.6% ethanol. Reboiling further gives vapour and condensate of the same concentration, as graphically, this is where the liquid curve and the vapour curve meet. It is impossible to obtain pure ethanol from this mixture. It boils as a pure liquid with a constant boiling point. This is an azeotropic mixture or a constant boiling mixture.

4. Azeotropes can either have a lower boiling point or a higher boiling point than their constituents. The various types of azeotropes are as follows.

5. (i) Heterogeneous and homogeneous azeotropes: If the constituents of the azeotrope are not completely miscible, they are called heterogeneous azeotropes. Homogeneous azeotropes are solutions with completely miscible components. (ii) Positive and Negative azeotropes: An azeotropic mixture which has a lower boiling point than its components is termed as a positive azeotrope. These mixtures show positive deviation from the vapour pressure calculated by Raoult’s Law. They form minimum boiling azeotropes. The ethanol water mixture mentioned above is an example of this type. Negative azeotropes are mixtures with higher boiling points than their constituent boiling points. They show large negative deviation from vapour pressure calculated by Raoult’s law. They form maximum boiling azeotropes. An example of negative azeotrope is nitric acid and water. This azeotrope has the approximate composition, 68% nitric acid and 32% water by mass, with a boiling point of 393.5 K.

1. Osmosis is the diffusion of water molecules across a semipermeable membrane from an area of lower concentration solution to an area of higher concentration. Certain biological and synthetic membranes appear as a continuous film or sheet, yet they contain a network of submicroscopic pores or holes. Small solvent molecules like water pass through these pores yet large solute molecules are hindered. Membranes with such properties are known as semipermeable membranes. The flow of solvent molecules from a pure substance to a concentrated solution is osmosis.

2. For comparing two solutions with unequal concentrations, the solution with higher concentration of solute is hypertonic and the one with the lower concentration is hypotonic. Once the solutions have equal concentrations through osmosis, state of equilibrium is reached and the solutions are called isotonic.

3. There are two types of osmosis which also explains the experiment below, (i) Endosmosis, which occurs when a substance with membrane or a cell is placed in a relatively hypotonic solution, the solvent molecules move through its membrane and the cell becomes turgid, and (ii) Exosmosis, which occurs when a cell is placed in a hypertonic solution, the solvent molecules move out of the substance into the surrounding solution, making the cell flaccid.

4. Raisins are grapes from which the water content has been evaporated. The dried fruit is made up cells covering its surface as well internally, which have a membrane surrounding the cytoplasm. When raisins are placed in water, the water acts as a hypotonic solution to the content of the raisin and its molecules will flow through the cell membrane which acts semipermeable and partakes in osmosis. The type of osmosis in this case is endosmosis, making the cells turgid and causing the raisin to swell.

5. There are various home as well as industrial applications of osmosis. Raw mango is placed in a highly concentrated salt solution, so it loses its water by osmosis and shrivels into pickles. Preservation of meat by salting and fruit by adding a high concentration of sugars cause bacteria in these foods to shrivel and die, thus helping to preserve the food. Blood cells or animal cells need to be stored in 0.9% saline solution as it is isotonic to physiological fluids and if the concentration is lower, water will flow inside the cells causing them to swell.

The process of flow of a solvent from a lower concentration to a higher concentration of solution through a semipermeable membrane is a colligative property and is known as osmosis. Osmosis has a large importance in our daily life as well as industrial applications.

1. Raw mangoes can be pickled when placed in concentrated salt-water which is a hypertonic solution and causes the mangoes to shrivel.

2. Many areas in animal and human bodies are known to transport water in and out of cells is through osmosis and diffusion.

3. Wilted flowers when placed in pure water revive to become fresh again. Even carrots which shrink and wilt due to loss of water to the atmosphere can be placed in fresh water to regain the shape.

4. Water flows through the plant roots and towards all the parts of the plant partially through osmosis.

5. Preservation of meats and fruit is carried out by salting, pickling and adding high amounts of sugar. The hypertonic salt outside the fruits and meats causes the water to flow out of the cell and reduce bacterial growth and increase shelf life of the products. Also, the hypertonic solution causes bacterial cells to lose their water content, causing them to shrivel and die.

6. Industrial applications of osmosis are numerous. Processes include power generation, desalination, wastewater treatment, food processing, medical product enrichment, drug release and others.

7. One of the most important applications of osmosis is reverse osmosis. In the case of osmosis, the lower concentrated solution or pure solvent flows through a semipermeable membrane to the higher concentrated solution. Osmotic pressure is the pressure required to stop this flow of solvent molecules through the semipermeable membrane. When enough pressure is applied on the hypertonic solution compartment, the process is reversed and pure solvent molecules flow from the hypertonic side to the hypotonic side through the semipermeable membrane.

8. Reverse osmosis is used for desalination of seawater and technology of the same exists in water purifiers.

Figure 2: A simplistic image of reverse osmosis. A: Piston and input B: Hypertonic solution C: Solute molecules D: Semipermeable membrane E: Pure solvent F: Output. Image obtained under Creative Commons license. By Colby Fisher - Own work, CC BY-SA 3.0.

9. In this process, when pressure higher than the osmotic pressure of the saline solution is applied, pure water is pushed through the SPM into the pure water (solvent) side. The pressure applied is very high for this to be successful.

10. The SPM used is cellulose acetate film which allows molecules of water to pass through but not impurities or ions.

1. The given experiment can be successfully done by using laboratory chemicals as well as household items. The process of osmosis is used in this experiment.

2. The hard calcium carbonate membrane can be removed without damaging the SPM by dipping it in a dilute mineral acid or vinegar for a few hours or overnight. On treatment with acetic acid/vinegar, the acids eat away the calcium carbonate and release calcium acetate and carbon dioxide.

3. After the shell has degraded, a soft, translucent egg is seen. It is placed in a hypertonic aqueous solution of sodium chloride, causing water from the egg to escape into sodium chloride and causing the egg to shrivel in size. The egg can easily be inserted in a narrow-necked bottle.

4. After inserting in a bottle, pure water is added to the bottle, which acts hypotonic to the contents inside the egg, causing water to flow through the SPM and returning the egg to the original shape. The shrivelling of the egg due to concentrated sodium chloride and swelling to original shape due to water is due to osmosis.

Use the diagram below to understand the experiment.

1. We know that molecules of certain substances when dissolved in solvents either associate or dissociate. This behaviour is shown by ionic substances dissociating or associating in water.

2. If we take potassium chloride or KCl, dissolving it in water will cause it to dissociate into K+ and Cl- ions. This leads to the increase in the chemical particles in the solution compared to the particles of the pure solute added. Properties of solutions which depend on the number of solute particles in it are known as colligative properties. Osmotic pressure, lowering of vapour pressure, elevation of boiling point and depression of freezing point are colligative properties. So calculation of molar mass on the basis of colligative properties would be higher than experimental value in this case.

3. In the case of association of particles, ethanoic acid (acetic acid) molecules dissolved in benzene dimerize due to hydrogen bonding. The number of particles in the solution thus reduces. Here the molar mass calculated from colligative properties, will be lower than the experimental value.

4. A molar mass that is either lower or higher than the expected value deduced theoretically is called as abnormal molar mass.

5. The van’t Hoff factor, i, was introduced to account for the extent of association or dissociation. It can be defined as the ratio of the experimental colligative property to the calculated colligative property.

i = (Normal Molar Mass)/(Abnormal Molar Mass)

i = (Observed Colligative Property)/(Calculated Colligative Property)

i = (Total number of moles of particles after association/dissociation)/(Number of moles of particles before association/dissociation)

6. In the case of an association where the observed molar mass is more than normal, the value of i is less than unity. For dissociation where the observed molar mass is lesser than expected, it is greater than unity.

7. Inclusion of van’t Hoff factor in colligative properties modifies their equations as follows:

i. Relative lowering of the vapour pressure of solvent,

(p_1^0- p_1^o)/(p_1^0 )=i.n_2/n_1

ii. Elevation of Boiling point, ∆Tb = i.Kb.m

iii. Depression of Freezing point, ∆Tf = i.Kf.m

iv. Osmotic pressure of solution, π = i.n2.R.T/V