Given below are the Mole Concepts Questions with Detailed Solutions

a. Concepts questions

b. Calculation problems

c. percentage composition

d. Mole fraction & molarity

a. Concepts questions

b. Calculation problems

c. percentage composition

d. Mole fraction & molarity

How many molecules of water are there in 54 g of $H_2O$ ?

Molar Mass of $H_2O$ = 2 + 16 = 18 g/moles

So ,number of moles of $H_2O$ = Mass/Molar Mass = 54/18 =3 moles

Now 1 moles = $6.022 \times 10^{23}$ molecule

So 3 moles will have $18.066 \times 10^{23}$ molecule

Calculate the mass of $6.022 \times 10^{23}$ molecule of $NH_4Cl$ ?

Molar mass (Molecular mass in gram) of $NH_4Cl$= 14+4+35.5= 53.5g

No. of moles of $NH_4Cl$

$=\frac { 6.022 \times 10^{23}}{ 6.022 \times 10^{23}}$

= 1 mole

Mass of $NH_4Cl$

$= \text {No. of moles} \times \text {molar mass}$

$= 1 \times 53.5g =53.5g$.

Calculate the mass of $12.044 \times 10^{23}$ Oxygen atoms.

No. of moles of Oxygen atoms

$= \frac {12.044 \times 10^{23}}{6.022 \times 10^{23}}$

= 2 mole

Mass of Oxygen atoms

$= \text {No. of moles} \times \text{atomic mass}$

$= 2 \times 16$

= 32 g

How many atoms of hydrogen are there in 36 g of $NH_4$?

Molar mass (Molecular mass in gram) of $NH_4$= 14+4= 18 g

No. of moles of $NH_4$?

= 36/18 = 4 moles

Now Total Moles of Hydrogen Atoms

= 16 moles

$= 16 x 6.022 \times 10^{23}=96.352 \times 10^{23}$

Calculate the number of hydrogen atoms in 1 mole of H

1 molecule of H_{2}= 2 hydrogen atoms

So, 1 mole of H_{2}= 2 mole hydrogen atoms

$= 2 \times 6.022 \times 10^{23}=12.044 \times 10^{23}$ hydrogen atoms.

Calculate the number of Cu atoms in 0.3175 g of Cu.

No. of moles of Cu

$= \frac {\text {Mass of Cu}}{\text {Atomic mass}}$

= 0.3175/63.5

=0.005 mole

No. of Cu atoms

$= \text {No. of moles} \times \text {Avogadro constant}$

$= 0.005 \times 6.022 \times 10^{23}$

$= 30.11 \times 10^{20}$ Cu atoms.

Calculate the number of molecules in 22.4 litres of CH

1 mole of CH_{4}= 22.4 L (at NTP)

Therefore

22.4 L of CH_{4}gas_{}= 1 mole CH_{4}gas_{}= $6.022 \times 10^{23}$ CH_{4} gas molecules.

a. Find the volume of 1 g of H

b.Find the volume of 20g H

c.What is the volume occupied by $6.022 \times 10^{23}$ molecules of any gas at NTP?

a. Molar Mass of H_{2}gas = 2 gm

$\text {Number of moles} = \frac {Mass}{\text {Molar Mass}} = \frac {1}{2} = .5$

1 mole of H_{2}= 22.4 L (at NTP)

Therefore

.5 mole of H_{2}= .5 x 22.4 = 11.2 litre

b.

No. of moles of H_{2}= 20/2 =10

1 mole of H_{2}= 22.4 L (at NTP)

Therefore

10 moles = $10 \times 22.4$

=224 L

c.

$6.022 \times 10^{23}$ molecules = 1 mole molecules, and

1 mole molecules of any ideal gas occupies 22.4 L at NTP.

An atom of some element Y weighs $6.644 \times 10^{-23}$ g. Calculate the number of gram-atoms in 40 kg of it.

Mass of 1 mole Y atoms

$= \text {mass of 1 atom} \times \text {Avogadro constant}$

$= 6.644 \times 10^{-23} \times 6.022 \times 10^{23}$

= 40 g

So, the atomic mass of Y = 40

No. of gram-atoms (or moles) of X

$ = \frac {\text {mass of Y}}{\text {atomic mass}}$

$= \frac {40 \times 1000}{40} =1000$

Find the number of moles and number of atoms of H and S in 10 mole of $H_2S$.

1 mole of H_{2}S contains 2 mole of H, 1 mole of S

Therefore

10 mole of H_{2}S contains

20 mole of H = $20 \times 6.022 \times 10^{23}= 12.044 \times 10^{24}$ H atoms

10 mole of S = $10 \times 6.022 \times 10^{23}= 6.022 \times 10^{24}$ S atoms

Calculate the number of atoms of each element in 245 g of $KClO_3$.

Molecular mass of KClO_{3}= $39 +35.5+3 \times 16 = 122.5$

No. of mole of KClO_{3}= 245 g/122.5g = 2 mole

2 mole of KClO_{3}contains

2 mole of K = $2 \times 6.022 \times 10^{23} = 12.044 \times 10^{23}$ K atoms

2 mole of Cl = $2 \times 6.022 \times 10^{23} = 12.044 \times 10^{23}$ Cl atoms

6 mole of O = $6 \times 6.022 \times 10^{23}= 1.806 \times 10^{24}$ O atoms.

If 4 g of NaOH dissolves in 36 g of $H_2O$, calculate the Mole fraction of each component in the solution. Also, determine the Molarity of solution (specific gravity of solution is 1 gm/L)

Mass of NaOH = 4 g

Number of moles of = 4 g NaOH/40 g = 0.1 mol

Mass of $H_2O$ = 36 g

Number of moles of = 36 g/18= 2 mol

$\text {Mole fraction of water} =\frac {\text{Number of moles of water}} {\text{ No. of moles of water} + \text {No. of moles of NaOH}}$
$ = \frac {2}{2 +.1}= 0.95$

$\text {Mole fraction of NaOH} = \frac {\text {Number of moles of NaOH}}{ \text {No. of moles of NaOH} +\text { No. of moles of water}}$
$ = \frac {0.1}{ 2 + 0.1} = 0.047$

Mass of solution = mass of water + mass of NaOH = 36 g + 4 g = 40 g

Volume of solution = $40 \times 1 = 40$ mL (Since specific gravity of solution is = 1 g m/L)

$\text {Molarity of solution}= \frac {\text {Number of moles of solute }}{\text {Volume of solution in litre}} =\frac {0.1}{0.04} =2.5$ M

Calculate the number of atoms in 12.2 litres of the below gas at NTP

(i) mono-atomic

(ii) diatomic gas

No. of moles in 12.2 L gas at NTP = 12.2/22.4 = 0.50

No. of molecules in 12.2 L gas

$= 0.50 \times 6.022 \times 10^{23}$

$= 3.011 \times 10^{23}$ molecules

In mono-atomic gases, No. of atoms = No. of molecules =$3.011 \times 10^{23}$.

In diatomic gases, No. of atoms = $2 \times \text{No. of molecules} =2 \times 3.011 \times 10^{23}=6.022 \times 10^{23}$.

If a mole were to contain $1 \times 10^{24}$ particles, what would be the mass of

(i) one mole of oxygen

(ii) a single oxygen molecule?

Mass of one mole of oxygen molecule (O_{2})

= molecular mass of oxygen molecule (O_{2}) in gram

= 32 g.

Mass of a single oxygen molecule

$= \frac {32}{1 \times 10^{24}}$

$=3.2 \times 10^{-23}$ g

Calculate the standard molar volume of oxygen gas. The density of $O_2$ gas at NTP is 1.429 g/L.

Standard molar volume is Volume of 1 mole of gas i.e 32 gm O_{2} gas

Now

$Density = \frac {Mass}{volume}$

$Volume= \frac {Mass}{Density}$

= 32/1.429

=22.39 Litres.

if 2 mol of Calcium Carbonate (Formula Weight =100) occupies a volume of 67.0 ml, Find the density ?

Solution