Some Basic Concepts Of Chemistry

Atomic Mass of Sodium = 23g

Atomic Mass of Sulphur = 32g

Atomic Mass of Oxygen = 16g

From the molecular formula of sodium sulphate, it is understood that in 1 mole of sodium sulphate, 1 mole of sulphur, 4 moles of oxygen and 2 moles of sodium is present.

So Molecular Mass of Sodium Sulphate = [23×2] + [32×1] + [16×4]

= 46 + 32 + 64

= 142g

Percentage of Sodium in sodium sulphate = [46×100]/142

= [4600/142]

= 32.394%

Percentage of Sulphur in sodium sulphate = [32×100]/142

= 3200/142

= 22.535%

Percentage of Oxygen in sodium sulphate = [64×100]/142

= 6400/142

=45.070%

Given:

Percentage of Iron in the oxide = 69.9%

Percentage of oxygen in the oxide = 30.1%

To determine the empirical formula of iron oxide, we need to find the simplest whole number ratio.

Simplest molar ratio of iron to oxygen:

= 1.25: 1.88

= 1: 1.5
for converting above ratio in the whole number ratio by 2,we will get

= 2: 3

The mole ratio of the element is represented by subscripts in empirical formula.

Therefore, the empirical formula for iron oxide is Fe₂O₃.

The reaction of carbon with oxygen is given as follows:

(i) By observing the above equation, it is understood that when 1 mole of carbon is burnt in air, then the amount of carbon dioxide produced is 44 g [Molecular Mass of carbon dioxide].

(ii) In the above equation 32 g of oxygen react with 1 mole of carbon, then 44 g of carbon dioxide is produced.

32 g of Oxygen → 44 g of Carbon Dioxide
16 g of Oxygen → X g of Carbon Dioxide

X = [44×16]/32

= 22 g

Therefore, 16 g of oxygen when reacted with 1 mole of carbon we get 22 g of CO₂.

(iii) Even though 2 moles of carbon is available for reaction but 16 g of oxygen can react with only 0.5 moles of carbon to give 22 g of CO₂. In this oxygen acts as limiting reagent and carbon acts as an excess reagent. So 1.5 moles carbon would be left un-reacted.

Given:

Molecular Mass of Sodium Acetate = 82.0245g/mol

Molarity of sodium acetate solution required = 0.375 M

0.375 molar solution of sodium acetate means 0.375g of sodium acetate dissolved in 1000ml of the solution.

But it is given in the question that the solution should be of 500ml.

So, 0.375g of CH₃COONa → 1000ml of solution

X g of CH₃COONa → 500ml of solution

X

= 0.1875 moles

Mass of Sodium acetate required to make 500ml solution = Number of moles of CH₃COONa × Molecular Mass of CH₃COONa

= 82.0245× 0.1875

= 15.379g

≈ 15.38g

Therefore 15.38g of CH₃COONa is required to make 500ml of solution.

Given:

Density of nitric acid = 1.41 g/ml

Percentage of Nitric acid = 69%

69% of nitric acid signifies 69g of nitric acid dissolved in 100g of solution.

Volume of nitric acid solution = [mass of solution]/ density of solution

= 100/1.41

= 70.922 ml

Number of moles of Nitric Acid

= 1.095 moles

Molarity, M_o

= 15.404 M

≈ 15.4 M

Therefore, concentration of nitric acid is 15.4 mole/L.

Given:

Mass of Copper Sulphate = 100g

Atomic Mass of Copper = 63.5g

Atomic Mass of Sulphur = 32g

Atomic Mass of Oxygen = 16g

Molecular Mass of CuSO₄ = 63.5 + 32 + [4×16]

= 159.5g

63.5g of Copper [Cu] → 159.5g of Copper Sulphate

X g of Copper [Cu] → 100 g of Copper Sulphate

X = [100×63.5]/159.5

= 6350/159.5

= 39.812

≈39.81g

Therefore 39.81g of copper can be obtained from 100g of copper sulphate.

Given:

Percentage of Iron in the oxide = 69.9%

Percentage of oxygen in the oxide = 30.1%

Finding Empirical Formula:

To determine the empirical formula of iron oxide, we need to find the simplest whole number ratio.

Therefore, the empirical formula for iron oxide is Fe₂O₃.

Finding Molecular Formula:

Empirical Formula Mass = [55.84×2] + [16×3]

= 111.68 + 48

= 159.68g

Molecular Mass [given] = 159.69g

n = [Molecular Mass]/ [Empirical Formula Mass]

= 159.69/159.68

= 1

Molecular Formula = n × Empirical Formula

= 1 × Fe₂O₃.

= Fe₂O₃

Therefore, the Molecular Formula is Fe₂O₃.

Given:

Atomic Mass of First Isotope of Chlorine = 34.9689g

Natural Abundance of First Isotope of Chlorine, P_a1 = 75.77%

= 0.7577

Total Atomic Mass of 1^st Isotope, M_a1 = 34.9689×0.7577

= 26.4959g

Atomic Mass of Second Isotope of Chlorine = 36.9659g

Natural Abundance of Second Isotope of Chlorine, P_a2 = 24.23%

= 0.2423

Total Atomic Mass of 2^nd Isotope, M_a2 = 36.9659×0.2423

= 8.9568g

Avg. Atomic Mass of Chlorine =
=(26.4959 gm + 8.9568 gm)

=35.4527 gm.

In 1 mole of ethane (C₂H₆), it is seen that 2 moles of carbon and 6 moles of hydrogen is present.

(i) Since three moles of ethane is present,

So, Number of moles of carbon present = 3×2

= 6

Therefore 6 moles of carbon is present in 3 moles of ethane.

(ii) Since three moles of ethane is present,

So, Number of moles of hydrogen present = 3×6

= 18

Therefore 18 moles of hydrogen is present in 3 moles of ethane.

(iii) 1 mole of ethane contains 6.023×10²³ molecules of ethane [By Avogadro’s Law]

Therefore, Number of molecules if ethane in 3 moles of ethane

= 3×6.023×10²³

= 18.069×10²³

Therefore, Number of molecules of ethane present in 3 moles of ethane is 18.069×10²³

Given:

Mass of sugar = 20g

Volume of solution = 2 L

Atomic Mass of Hydrogen = 1g

Atomic Mass of Oxygen = 16g

Atomic Mass of Carbon = 12g

So Molecular Mass of sugar = [12×12] + [22×1] + [16×11]

= 144 +22 + 176

= 342 g

Number of moles of sugar = [Mass of sugar]/ Molecular Mass of sugar

= 20/342

= 0.05848

Molarity, M_o

= 0.0292 mol/L

≈ 0.029 mol/L

Given:

Density of methanol = 0.793 kg/L

Molarity of solution = 0.25 M

Molecular Formula of Methanol = [CH₃OH]

Atomic Mass of Hydrogen = 1g

Atomic Mass of Oxygen = 16g

Atomic Mass of Carbon = 12g

So Molecular Mass of methanol= [12×1] + [4×1] + [16×1]

= 12 + 4 + 16

= 32g

= 0.032kg

Molarity =

0.25 =

No. of moles of methanol = 0.625 mole

Weight of methanol = ( 0.625 × 32 )gm

=20 gm

=0.02 kg

Volume of methanol =

=

=0.025 L

Therefore, volume needed for making 2.5 L of its 0.25 M solution is 0.025 L.

Given:

Pressure = Force/Area

=

= 1.01332×10⁵ kg/m/s²

We know that,

1N = 1km/ s²

1 Pa = 1 kg/m^-2/s²

= 1 kg/m^-1/s²

Therefore, Pressure in Pascal =1.01322×10⁵ Pa

The SI system was internationally accepted in 1960. SI is the abbreviation of “Systeme International” Units which is French equivalent of “International System of Units”.

The SI unit of Mass is kilogram.

1 kilogram is defined as the mass of a platinum-iridium cylinder kept in Paris.

In practice, the mass of 1litre of water at 4°C is 1 kilogram.

On atomic scale, 1 kilogram is equivalent to the mass of 5.0188× 10²⁵ atoms of ₆C¹² [isotope of carbon].

The total number of digits in a measured physical quantity is called the number of significant figures. The number of significant figures refers to the precision of a measured quantity and may be defined as follows:

The number of significant figures in a measured physical quantity refers to number of digits written, including the last one whose value is uncertain.

Rules for Determination of Significant Figures:

[i] All non-zero digits are significant.

For example: 146 cm has three significant figures.

[ii] The zeros placed to the left of the first non-zero digit in the given physical quantity are not significant.

For example: 0.56cm has two significant figures.

[iii] The zeros placed to the right of the decimal point are significant.

For example: 15,0cm has three significant figures.

[iv] The zeros placed between two non-zero digits are significant.

For example: 6.08cm has three significant figures.

Given:

Level of contamination = 15 ppm [by mass]

To find: Mass Percentage and Molality

Formula:

Molality

Mass Percentage of Solute

Calculation of Mass Percentage:

15 ppm means 15 parts of Chloroform in 1 million [10⁶] parts of drinking water

→Mass Percentage

= 1.5 × 10^-3 %

Calculation of Molality:

→Molecular Mass of Chloroform, CHCl₃ = [12] + [1] + [35.5×3]

= 119.5 g

→Number of Moles of Chloroform = [15 / 119.5]

= 0.1255 moles

Molality

= 1.255 × 10^-4

Therefore, the Mass Percentage is = 1.5 × 10^-3% and the Molality of the solution is = 1.255 × 10^-4 m.

(i) The scientific notation of 0.0048 is 4.8×10^-3

(ii) The scientific notation of 234,000 is 2.34×10⁵

(iii) The scientific notation of 8008 is 8.008×10³

(iv) The scientific notation of 500 is 5×10²

(v) The scientific notation of 6.0012 is 6.0012×10⁰

Rules for Determination of Significant Figures:

[i] All non-zero digits are significant.

For example: 146 cm has three significant figures.

[ii] The zeros placed to the left of the first non-zero digit in the given physical quantity are not significant.

For example: 0.56cm has two significant figures.

[iii] The zeros placed to the right of the decimal point are significant.

For example: 15,0cm has three significant figures.

[iv] The zeros placed between two non-zero digits are significant.

For example: 6.08cm has three significant figures.

(i) 0.0025 – 2 significant figures

(ii) 208 – 3 significant figures

(iii) 5005 – 4 significant figures

(iv) 126,000 – 3 significant figures

(v) 500.0 – 4 significant figures

(vi) 2.0034 – 5 significant figures

(i) 34.2

(ii) 10.4

(iii)0.0460

(iv)2810

The Law satisfied by the above data is named as Law of Multiple Proportions. This law was given by John Dalton in 1804.

Statement:

When two elements combine to form two or more compounds, then the different masses of one element, which combine with a fixed mass of the other element, bear a simple whole number ratio with another.

Nitrogen and oxygen combine together to form five compounds- Nitrous Oxide [N₂O], Nitric Oxide [NO], Nitrogen Trioxide [N₂O₃], Nitrogen Dioxide [N₂O₄] and Nitrogen pentoxide [N₂O₅]. In these compounds ratio of nitrogen and oxygen by mass is given in the above table.

Thus the ratio of the masses of oxygen which combine with fixed mass of nitrogen are 16:32:48:64:80 that is they are in simple whole number ratio of 1:2:3:4:5.

(i) We know the basic relation that

1km = 1000m

1 m = 1000 mm

Therefore 1 km = 1000×1000

= 10⁶ mm

We also know that

1 m = 10¹²pm

Therefore 1 km = 1000× 10¹² pm

= 10¹⁵pm

(ii) We know the basic relation that

1 kg = 1000 g

1 g = 1000 mg

So, 1 mg = 10^-3g

Therefore 1 mg = 10^-3 × 10^-3 kg

= 10^-6 kg

1 g = 10⁹ng

1000 mg = 10⁹ ng

1 mg = [10⁹]/10³ ng

= 10⁶ ng

(iii) We know the basic relation that

1 L = 1000 ml

∴ 1 ml = 10^-3 L

1 ml = 10^-3 dm³

Given:

Speed of light = 3.0 × 10⁸ m s^–1

Time in which distance is to be covered = 2.00 ns

Speed = [Distance]/ Time

Distance = [3.0 × 10⁸] × [2×10^-9]

= 0.6 m

While solving the problems relating to chemical equation, the reactant reacts according to the balanced chemical equation. Quite often, one of the reactant is present in lesser amount while the other may be present in higher amount. The reactant which is present in lesser amount is known as limiting reactant or reagent.

[i] According to the above given reaction, 1 atom of A reacts with 1 molecule of B. Thus, 200 molecules of B will react with 200 atoms of A, thereby leaving 100 atoms of A unused. Hence, B is the limiting reagent as atom B is in lesser amount [200].

[ii] According to the reaction given in the question, 1 mol of A reacts with 1 mol of B. Thus, 2 mol of A will react with only 2 mol of B. As a result, 1 mol of A is not being consumed in the reaction process. Hence, A is the limiting reagent.

[iii] According to the given reaction given in the question, 1 atom of A combines with 1 molecule of B. Thus, all 100 atoms of A will combine with all 100 molecules of B. Hence, the mixture is stoichiometric where no limiting reagent is present.

[iv] 1 mol of atom A combines with 1 mol of molecule B. Thus, 2.5 mol of B will combine with only 2.5 mol of A. As a result, 2.5 mol of A will be left as such as it is not consumed in the reaction. Hence, B is the limiting reagent because B is less as compared to A.

[v] According to the reaction in the question, 1 mol of atom A combines with 1 mol of molecule B. Thus, 2.5 mol of A will combine with only 2.5 mol of B and the remaining 2.5 mol of B will be left as such. Hence, A is the limiting reagent as it is present in lesser amount.

The reaction given in the question is not balanced.

Balancing the above chemical reaction:

N₂ + 3H₂→ 2NH₃

(i) Finding mass of NH₃ produced:

Atomic Mass of Hydrogen = 1g

Atomic Mass of Nitrogen = 14g

Molecular Mass of NH₃ = 14 + 3

= 17 g

N₂ + 3H₂→ 2NH₃
28gm 6gm 34gm

From the chemical equation we observe that 28g of nitrogen react with 6g of hydrogen to give 34g of ammonia

Therefore 1 mole of dinitrogen reacts with 3 moles dihydrogen to give 2 moles of ammonia.

Therefore, amount of dihydrogen required to react with 2000g of dinitrogen = [6×2000]/28

= 12000/28

= 428.57g

Therefore, 2000g of dinitrogen will produce = [34×2000]/28

= 68000/28

= 2428.57g of Ammonia

Therefore 2428.57g of ammonia is produce when 2000g of dinitrogen is used.

(ii) Since dinitrogen is present in lesser amount(428.57 gm), it is the limiting reagent and dihydrogen is the excess reagent and hence H₂ will be left unreacted.

(iii) Mass of dihydrogen left unreacted = 1000 – 428.57

= 571.43g

Atomic Mass of Oxygen = 16g

Atomic Mass of Carbon = 12g

Atomic Mass of Sodium = 23g

Molecular Mass of Sodium Carbonate = [23×2] + [12×1]+[16×3]

= 46 + 12 + 48

= 106g

1 mole of Na₂CO₃ = 106 g

0.5 mole of Na₂CO₃ = X g

X = [0.5×106]/1

= 53g of Na₂CO₃

0.5M of Na₂CO₃ signifies 0.5 moles of Na₂CO₃ dissolved in 1 litre of water or rather 53g of Na₂CO₃ dissolved in 1 litre of water.

The reaction of hydrogen with oxygen is given as below:

2H₂ + O₂→ 2H₂O

By referring to above equation, 2 volumes of hydrogen and 1 volume of oxygen react together to form 2 volumes of water.

So if 10 volumes of hydrogen react with 5 volumes of oxygen then 10 volumes of water is produced.

(i) 1 pm = 10^-12 m

28.7 pm = 28.7×10^-12m

= 2.87×10^-11m

(ii) 1 pm = 10^-12 m

15.15 pm = 15.15×10^-12m

= 1.515×10^-11m

(iii) 1 mg = 10^-3 g

1 mg = 10^-6 kg

∴ 25365 mg = 25365×10^-6 kg

= 2.5365×10^-2 kg

(i) Gram atomic mass of Au= 197 g

Or

197g of Au contains = 6.022 x 10²³

Therefore 1gm of Au contains = 6.022 x 10²³/197X1 = 3.06 x 10²¹atoms

(ii) Gram atomic mass of Na = 23 g

Or

23 g of Na contains atoms = 6.022x 10²³

Or

1gm of Na contains atoms = 6.022x10²³/23 X1 = 26.2 x10²¹atoms

(iii) Gram atomic mass of Li = 7

Or

7g of Li contains atoms = 6.022 x 10²³

Or

1g of Li contains atoms = 6.022 x 10²³/7 X1= 86.0 x 10²¹atoms

(iv) Gram atomic mass of Cl = 71 Or 71g of Cl contains atoms = 6.022x10²³

Or

1 g of Cl contains atoms = 6.022x10²³/71 X 1= 8.48 x 10²¹atoms

Therefore, 1 g of Li will have the largest number of atoms.

Given:

Mole Fraction of Ethanol = 0.040

Density of water = 1g/ml(given)

So, Mole fraction of water = 1- 0.04 = 0.96

So, Moles of Ethanol = 0.04 moles.

Therefore, mass of 0.04 moles of ethanol = 0.04 mole × 46 gm/mole = 1.86 gm.

Density of ethanol = 0.789 g/ml.

Now, moles of water= 1-0.04 = 0.96 moles.

Volume of water = =17.28 ml = 0.017 L.

Molarity of Solution = ==2.043 mol/L.

By Avogadro’s Law

1 mole of carbon contain 6.023×10²³ carbon atoms = 12g of Carbon

6.023×10²³ atoms of carbon → 12g of Carbon

1 atom of Carbon → X g of Carbon

X = [12]/ 6.023×10²³

= 1.992×10^-23g of Carbon

Therefore, mass of 1 atom of Carbon is 1.992×10^-23g.

(i) = 1.648

Least precise number of calculation = 298.15

Number of significant numbers in result = Number of significant figures in least precise number = 2

Correct Answer = 1.65

(ii) Calculated Answer = 26.82

Least precise number of calculation = 5.364

Number of significant numbers in result = Number of significant figures in least precise number = 3

Correct Answer = 26.820

(iii) Calculated Answer = 0.8204

Number of significant numbers in result = Number of significant figures in least precise number = 4

Correct Answer = 0.8204

Atomic Mass of First Isotope of Argon = 35.96755g

Natural Abundance of First Isotope of Argon, P_a1= 0.337% = 0.00337

Total Atomic Mass of 1^st Isotope, M_a1 = 35.96755×0.00337

= 0.12121g

Atomic Mass of Second Isotope of Argon = 37.96272g

Natural Abundance of Second Isotope of Argon, P_a2 = 0.063%

=0.00063

Total Atomic Mass of 2^nd Isotope, M_a2 = 37.96272×0.00063

= 0.02392g

Atomic Mass of Third Isotope of Argon = 39.9624g

Natural Abundance of Third Isotope of Argon, P_a3 = 99.6%

Total Atomic Mass of 3^rd Isotope, M_a3 = 39.9624×0.996

= 39.8026g

So Average Atomic Mass of Argon =

→

→ 39.9477g

Therefore, the average atomic mass of Argon is 39.95g

(i) According to Avogadro’s Law,

1 mole of Argon → 6.023×10²³ atoms of Argon

So, 52 moles of Argon → X atoms of Argon

X =52×6.023×10²³

=3.13196×10²⁵

Therefore in 52 moles of Argon 3.13196×10²⁵ atoms are present.

(ii) Atomic Mass of Helium [He] = 4 u

This means that mass of one atom of Helium is = 4 u

So, Number of atom of Helium contained in 52 u = 52/4= 13 atoms

Therefore, in 52 u of Helium 13 atoms are present.

(iii) Atomic Mass of Helium [He] = 4 u

Number of Moles of Helium in 52 g = 52/4= 13 moles

So, Number of atoms in 52g of Helium [He] = 13×6.023×10²³= 7.8299×10²⁴

Therefore, in 52 g of Helium 7.83×10²⁴ atoms are present.

Given:

Mass of Carbon = 3.38g

Volume of welding gas = 10 L

Mass of 10 L of welding gas = 11.6g

Finding Percentage of Carbon:

44 parts of CO₂→12 parts of C

OR

44g of CO₂→12 g of C

Therefore, according to the question

3.38 g of CO₂→ X g of C

X = [12× 3.38] /44

=0.921g

18 g of water → 2g of hydrogen

So, 0.690 g of water → X g of hydrogen

X = [2× 0.690]/18

=0.0767g

Since carbon and hydrogen are the only constituents of the compound, the total mass of the compound is obtained as follows:

=0.9217 g + 0.0767 g = 0.9984 g

Percentage of carbon = [100×weight of carbon]/weight of compound

= [0.921× 100]/0.998

= 92.32%

Also percentage of hydrogen =[100×weight of hydrogen]/weight of compound

=[0.0766× 100]/0.998

=7.68%

(i) Finding Empirical Formula:

Empirical formula of the compound = CH

Therefore, the Empirical Formula of the compound is CH

(ii) Weight of 22.4 L of gas of STP = [11.6g×22.4L]/10.0L

= 25.985g

≈ 26 g

Therefore, the molecular mass of the gas is 26 g.

(iii) Finding Molecular formula:

Empirical formula mass = 12+1 = 13g

Also molecular mass =26 g [as obtained in step (ii)]

n = [Molecular Mass] / [Empirical Formula Mass]

=26/13

=2

Now molecular formula = n ×Empirical Formula = 2×CH = C₂H₂

Therefore, the molecular formula of the compound is C₂H₂.

Given:

Volume of HCl = 25 ml

Molarity of HCl = 0.75 M

Atomic Mass of Hydrogen = 1g

Atomic Mass of Chlorine = 35.5g

Molecular Mass of HCl = 1 + 35.5

= 36.5g

By Avogadro’s Law,

1 mole of HCl→ 36.5g of HCl

So 0.75 moles of HCl → X g of HCl

X = 0.75×36.5

X = 27.35 g

Thus, 1000 mL of solution contains HCl = 27.375g

1000ml of HCl → 27.357g of HCl

So 25 ml of HCl → X g of HCl

X = [25×27.375]/1000

= 684.375/1000

=0.68438 g

From the chemical equation given in the question,

CaCO₃+2 HCL →CaCL₂+H₂O

We observe that

2 moles of HCl reacts with 1 mole of CaCO₃

73 g of HCl reacts with 100g of CaCO₃

0.68438g of HCl reacts with X g of CaCO₃

X = [0.68483×100]/73

= 68.483/73

= 0.93812 g

So amount of CaCO₃ required is 0.94g.

Given:

Mass of Manganese Dioxide = 5 g

Atomic Mass of Manganese = 55g

Atomic Mass of Oxygen = 16g

Atomic Mass of Hydrogen = 1g

Molecular Mass of MnO₂ = 55 + [16×2]

= 55 + 32

= 87g

Molecular Mass of HCl = 1 + 35.5

= 36.5g

So by observing the chemical reaction we see that,

1moles of MnO₂ reacts with 4moles of HCl

87g of MnO₂→ 4×36.5g of HCl

So 5g of MnO₂→ X g of HCl

X = [146×5]/87

X = 730/87

X = 8.39g

So, 8.4g of HCl will react completely with 5g of MnO₂.