Mole Concept

a. 20 H₂ + 20 O → 20H₂O

The reactant molecules of oxygen get consumed first. Because there is only atom of oxygen present.

b. If equal number of molecules of hydrogen and oxygen is taken, the reaction gets completed with none of the reactants remaining unreacted.

Carbon

The atomic mass of elements is expressed by considering 1/12 the mass of an atom of carbon-12 isotope as one unit. This mass is called unified mass (u)

Given:

Atomic mass of helium = 4

Atomic mass of oxygen = 16

Mass of oxygen = 40 g

Hence, we can write

By putting the values, we get:

By applying criss cross method:

We get:

Mass of helium × 16 = 4 × 40g

Mass of helium = 10g

Thus, 10 grams of helium is required.

Gram atomic mass: The mass of an element in grams is numerically equal to the atomic mass of the element is one gram atomic mass.

1 GAM (gram atomic mass) = Mass in grams equal to atomic mass

a. Given: Mass of helium = 100g

Atomic mass of helium = 4

1 GAM of helium = 4g

Apply the formula given below:

Number of gram atomic mass = 25g

Thus, number of gram atomic mass of helium is 25g

Given: Mass of oxygen = 200g

Atomic mass of oxygen = 16

1 GAM of oxygen = 16g

Apply the formula given below:

Number of gram atomic mass = 12.5

Thus, number of gram atomic mass of oxygen is 12.5.

Given: Mass of nitrogen = 70g

Atomic mass of nitrogen = 14

1 GAM of nitrogen = 14g

Apply the formula given below:

Number of gram atomic mass = 5

Thus, number of gram atomic mass of nitrogen is 5

Given: Mass of calcium = 1g

Atomic mass of calcium = 40

1 GAM of calcium = 40g

Apply the formula given below:

Number of gram atomic mass = 0.025g

Thus, number of gram atomic mass of calcium is 0.025

Gram molecular mass (GMM): The mass in grams equal to the molecular mass of an element or compound is called its gram molecular mass.

1 GMM = Mass in grams equal to atomic mass

HNO₃

Gram atomic mass of H=1g

Gram atomic mass of N=14g

Gram atomic mass of O=16g

GMM of HNO₃ = 1g + 14g + 3× 16g

GMM of HNO₃ = 63g

Thus, gram molecular mass of HNO₃ is 63g

CaCl₂

Gram atomic mass of Ca=40g

Gram atomic mass of Cl=35.5g

GMM of CaCl₂ = 40g + 2 × 35.5g

GMM of CaCl₂ = 111g

Thus, gram molecular mass of CaCl₂ is 111g

Na₂SO₄

Gram atomic mass of Na=11g

Gram atomic mass of S=32g

Gram atomic mass of O=16g

GMM of Na₂SO₄ = 2× 11g + 32g + 4× 16g

GMM of Na₂SO₄ = 22g + 32g + 64g

GMM of Na₂SO₄ = 118g

Thus, gram molecular mass of Na₂SO₄ is 118g

NH₄NO₃

Gram atomic mass of N=14g

Gram atomic mass of H=1g

Gram atomic mass of O=16g

GMM of NH₄NO₃ = 14g + 4×1g + 14g + 3× 16g

GMM of NH₄NO₃ = 14g + 4g + 14g + 48g

GMM of NH₄NO₃ = 80g

Thus, gram molecular mass of NH₄NO₃ is 80g

One mole: The amount of any substance containing 6.022 × 10²³ particles is called one mole.

a. Given: Mass of water = 400g

Gram molecular mass (GMM) of H₂O = 18g

Apply the formula:

Number of moles = 22.2

Thus, number of moles of the given sample is 22.2

b. Given: Mass of carbon = 400g

Gram molecular mass (GMM) of carbon = 12g

Apply the formula:

Number of moles = 33.3

Thus, number of moles of the given sample is 33.3

c. Given: Mass of helium = 400g

Gram molecular mass (GMM) of He = 4g

Apply the formula:

Number of moles = 100

Thus, number of moles of the given sample is 100

d. Given: Mass of hydrogen = 400g

Gram molecular mass (GMM) of H₂ = 2g

Apply the formula:

Number of moles = 200

Thus, number of moles of the given sample is 200

e. Given: Mass of glucose = 400g

Gram molecular mass (GMM) of C₆H₁₂O₆ = 180 g

Apply the formula:

Number of moles = 2.22

Thus, number of moles of the given sample is 2.22

ii) Arrangement of the samples in the increasing order of their number of moles is:

Glucose < Water <Carbon < Helium < Hydrogen

Given: Mass of water = 1kg =1000g

Gram molecular mass of H₂O = 18g

To calculate the number of moles, apply the formula given:

Number of moles = 55.5

Thus, number of moles in 1 kg of water is 55.5

Given: Mass of calcium carbonate = 500g

Gram molecular mass of CaCO₃ = 100g

To calculate the number of moles, apply the formula given:

Number of moles = 5

Thus, number of moles in 500g of CaCO₃ is 5

Given: Mass of CO₂ = 88g

Gram molecular mass of CO₂ = 44g

To calculate the number of molecules, first we apply the formula:

Number of gram molecules = 2

Now, number of molecules = 2 × 6.022 × 10²³

= 1.2 × 10²⁴

Thus, number of molecules of CO₂ is1.2 × 10²⁴

There are three atoms in CO₂ (1atom of carbon and two atoms of oxygen)

∴ Total number of atoms = 3 × Number of molecules

= 3 × 1.2 × 10²⁴

= 3.6 × 10²⁴

Thus, total number of atoms of CO₂ is3.6 × 10²⁴

Given: Mass of ammonia at STP = 170g

Molecular mass of NH₃ = 17g

First, we will calculate the number of moles by applying the formula given:

Number of moles = 10

Thus, number of moles of CaCO₃ is 10

Now 1 mole of any gas occupies 22.4 litres of volume.

∴ 1 mole of NH₃ = 22.4 litre

10 moles of NH₃ = 22.4 × 10 = 224 litres

Thus, the volume occupied by 170g of NH₃ is 224 litres.

Given: Volume of CO₂ gas at STP = 112L

Gram molecular mass of CO₂ = 44g

First, we will calculate the number of moles so that we can find the mass of CO₂ gas

As we know 1 mole of any gas occupies 22.4 litres of volume.

∴ 1 mole of NH₃ = 22.4 litre

22.4 × Number of moles = 112 litres

Number of moles = 5

Now, we can calculate the mass of CO₂ gas by applying the formula given:

We can write,

Mass of CO₂ gas = Number of moles × GMM of CO₂

Mass of CO₂ gas = 5 × 44g

Mass of CO₂ gas = 220g

Thus, mass of 112 L CO₂ gas at STP is 220g

Now, number of molecules = Number of moles × 6.022 × 10²³

= 5 × 6.022 × 10²³

= 3.011 × 10²⁴

Thus, number of molecules of CO₂ is3.011 × 10²⁴

Atomic mass of carbon = 12

Atomic mass of oxygen = 16

Atomic mass of helium = 4

Mass of helium = 1g

For carbon:

By putting the values, we get:

By applying criss cross method:

We get:

Mass of carbon × 4 = 12 × 1g

Mass of carbon = 3g

Thus, 3 grams of carbon is required.

For oxygen:

By putting the values, we get:

By applying criss cross method:

We get:

Mass of oxygen × 4 = 16 × 1g

Mass of oxygen = 4g

Thus, 4 grams of oxygen is required.

i) Calculation of number of molecules in each:

a. Given: Mass of He = 20g

Gram molecular mass of He = 4g

To calculate the number of molecules, first we apply the formula:

Number of gram molecules = 5

Now, number of molecules = 5 × 6.022 × 10²³

= 3.011 × 10²⁴

Thus, number of molecules of He is3.011 × 10²⁴

b. Given: Volume of NH₃ at STP = 44.8L

Gram molecular mass of NH₃ = 17g

First we will calculate the number of moles so that we can find the number of molecules.

As we know that 1 mole of any gas occupies 22.4 litres of volume.

∴ 1 mole of NH₃ = 22.4 litre

22.4 × Number of moles = 44.8 litres

Number of moles = 2

Now, number of molecules = No. of moles × 6.022 × 10²³

= 2 × 6.022 × 10²³

= 1.204 × 10²⁴

Thus, number of molecules of NH₃ is1.204 × 10²⁴

c. Given: Volume of N₂ at STP = 67.2L

Gram molecular mass of N₂ = 28g

First we will calculate the number of moles so that we can find the number of molecules.

As we know that 1 mole of any gas occupies 22.4 litres of volume.

∴ 1 mole of NH₃ = 22.4 litre

22.4 × Number of moles = 67.2 litres

Number of moles = 3

Now, number of molecules = No. of moles × 6.022 × 10²³

= 3 × 6.022 × 10²³

= 1.806 × 10²⁴

Thus, number of molecules of N₂ is1.806 × 10²⁴

d. 1 mol of H₂SO₄

Given: Number of moles = 1mol

Number of molecules = No. of moles × 6.022 × 10²³

= 1 × 6.022 × 10²³

= 6.022 × 10²³

Thus, number of molecules of H₂SO₄ is6.022 × 10²³

e. Given: Mass of water = 180g

Gram molecular mass of H₂O = 18g

To calculate the number of molecules, first we apply the formula:

Number of gram molecules = 10

Now, number of molecules = 10 × 6.022 × 10²³

= 6.022 × 10²⁴

Thus, number of molecules of H₂O is6.022 × 10²⁴

The arrangement of the samples in increasing order of the number of molecules in each is:

44.8 L of NH₃ at STP < 67.2 L of N₂ at STP < 20 g of He < 1 mol of H₂SO₄ < 180 g of water

Calculation of number of atoms in each:

a. Number of molecules = 3.011 × 10²⁴ (calculated)

In He, there is only one atom

Thus, total number of atoms of He = 1 × Number of molecules

= 3.011 × 10²⁴

b. Number of molecules = 1.204 × 10²⁴ (calculated)

In NH₃, there are four atoms

Thus, total number of atoms of NH₃ = 4 × Number of molecules

= 4 × 1.204 × 10²⁴

= 4.8 × 10²⁴

c. Number of molecules = 1.806 × 10²⁴ (calculated)

In N₂, there are two atoms

Thus, total number of atoms of N₂ = 2 × Number of molecules

= 2 × 1.806 × 10²⁴

= 3.6 × 10²⁴

d. Number of molecules = 6.022 × 10²³ (calculated)

In H₂SO₄, there are seven atoms

Thus, total number of atoms of H₂SO₄ = 7 × No. of molecules

= 7 × 6.022 × 10²³

= 4.21 × 10²⁴

e. Number of molecules = 6.022 × 10²⁴ (calculated)

In H₂O, there are three atoms

Thus, total number of atoms of H₂O = 3 × Number of molecules

= 3 × 6.022 × 10²⁴

= 1.8 × 10²⁵

The arrangement of the samples in increasing order of the number of atoms in each is:

20 g of He < 67.2 L of N₂ at STP < 1 mol of H₂SO₄ < 44.8 L of NH₃ at STP < < 180 g of water

Calculation of mass:

For 44.8 L of NH₃ at STP (b)

Number of moles = 2 (calculated in part (i)-b)

Now, we can calculate the mass of NH₃ by applying the formula given:

We can write,

Mass of NH₃ = Number of moles × GMM of NH₃

Mass of NH₃ = 2 × 17g

Mass of NH₃ = 34g

Thus, mass of 44.8L of NH₃ at STP is 34g

For 67.2 L of N₂ at STP (c)

Number of moles = 3 (calculated in part (i)-c)

Now, we can calculate the mass of N₂ by applying the formula given:

We can write,

Mass of N₂ = Number of moles × GMM of N₂

Mass of N₂ = 3 × 28g

Mass of N₂ = 84g

Thus, mass of 67.2 L of N₂ at STP is 84g

1 mol of H₂SO₄

Number of moles = 1

Now, we can calculate the mass of H₂SO₄ by applying the formula given:

We can write,

Mass of H₂SO₄ = Number of moles × GMM of H₂SO₄

Mass of H₂SO₄ = 1 × 98g

Mass of H₂SO₄ = 98g

Thus, mass of 1 mol of H₂SO₄ is 98g

Given: Mass of water = 90g

Gram molecular mass of H₂O = 18g

To calculate the number of molecules, first we apply the formula:

Number of gram molecules = 5

Now, number of molecules = 5 × 6.022 × 10²³

= 3.01 × 10²⁴

Thus, number of molecules of 90 gm of water is3.011 × 10²⁴

Number of molecules = 3.011 × 10²⁴ (calculated)

In H₂O, there are three atoms

Thus, total number of atoms of H₂O = 3 × Number of molecules

= 3 × 3.011 × 10²⁴

= 9.033 × 10²⁴

Number of gram molecules = 5 (calculated in part a)

Now, 1 molecule of water contains:

1 × 8 electron of oxygen

2x1 electron of Hydrogen

Total 10 electrons per molecule.

So total number of electrons = Number of gram molecules × number of electron per molecule

Total number of electrons = 5 × 10

Total number of electrons = 50 moles of electrons

As we know 1 mole = 6.022 × 10²³

∴ 50 moles of electrons = 50 × 6.022 × 10²³

= 3.01 × 10²⁵ number of electrons

Thus, the total number of electrons = 3.01 × 10²⁵

Preparation of 1M solution of NaOH

As we know that 1 mole of NaOH has 40gm of mass.

Hence, if we want to prepare 1M NaOH solution:

i. First we need to weigh 40gm of NaOH in weighing balance.

ii. Pipette out 1L water from 200mL water and put in a beaker.

iii. Now, dissolve 40gm of NaOH in the beaker.

That’s how 1M solution of NaOH is prepared.

Given: Volume of solution of NaCl (common salt) = 500mL

= 0.5L

Molarity of solution = 1M

We can write,

Number of moles = Molarity × volume in litres

Number of moles = 1M × 0.5L

Number of moles = 0.5

Now, to find the mass in grams, apply the formula given:

Gram molecular mass of NaCl = GMM of Na + GMM of Cl

= 23g + 35.5g

= 58.5g

Therefore,

Mass of NaCl in grams = 0.5 × 58.5g

Mass of NaCl in grams = 29.25g

Thus, 29.25 grams ofcommon salt are dissolved.

Note: If one litre of the solution contains 1mol of solute, it is called 1M (molar) solution.

If the solution is diluted with water to a volume of 2 L, the molarity will be:

As number of moles = 0.5 (calculated in part a)

Volume in litres = 2L (given)

∴

Molarity = 0.25M

Thus, the molarity will be 0.25M