 # Mole Concepts Problems with Solutions

Given below are the Mole Concepts Questions with Detailed Solutions
a. Concepts questions
b. Calculation problems
c. percentage composition
d. Mole fraction & molarity

Known Values Question 1
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

Question 2
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 $NH_4Cl$
$= \text {No. of moles} \times \text {molar mass}$
$= 1 \times 53.5g =53.5g$.

Question 3
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

Question 4
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
$= 16 x 6.022 \times 10^{23}=96.352 \times 10^{23}$

Question 5
Calculate the number of hydrogen atoms in 1 mole of H2.

1 molecule of H2= 2 hydrogen atoms
So, 1 mole of H2= 2 mole hydrogen atoms
$= 2 \times 6.022 \times 10^{23}=12.044 \times 10^{23}$ hydrogen atoms.

Question 6
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.

Question 7
Calculate the number of molecules in 22.4 litres of CH4 gas at NTP.

1 mole of CH4= 22.4 L (at NTP)
Therefore
22.4 L of CH4gas= 1 mole CH4gas= $6.022 \times 10^{23}$ CH4 gas molecules.

Question 8
a. Find the volume of 1 g of H2gas in litres at N.T.P
b.Find the volume of 20g H2at NTP.
c.What is the volume occupied by $6.022 \times 10^{23}$ molecules of any gas at NTP?

a. Molar Mass of H2gas = 2 gm
$\text {Number of moles} = \frac {Mass}{\text {Molar Mass}} = \frac {1}{2} = .5$
1 mole of H2= 22.4 L (at NTP)
Therefore
.5 mole of H2= .5 x 22.4 = 11.2 litre
b.
No. of moles of H2= 20/2 =10
1 mole of H2= 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.

Question 9
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
$= \frac {\text {mass of Y}}{\text {atomic mass}}$
$= \frac {40 \times 1000}{40} =1000$

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

1 mole of H2S contains 2 mole of H, 1 mole of S
Therefore
10 mole of H2S 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

Question 11
Calculate the number of atoms of each element in 245 g of $KClO_3$.

Molecular mass of KClO3= $39 +35.5+3 \times 16 = 122.5$
No. of mole of KClO3= 245 g/122.5g = 2 mole
2 mole of KClO3contains
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.

Question 12
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

Question 13
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 = $2 \times \text{No. of molecules} =2 \times 3.011 \times 10^{23}=6.022 \times 10^{23}$.

Question 14
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 (O2)
= molecular mass of oxygen molecule (O2) in gram
= 32 g.
Mass of a single oxygen molecule
$= \frac {32}{1 \times 10^{24}}$
$=3.2 \times 10^{-23}$ g

Question 15
Calculate the standard molar volume of oxygen gas. The density of $O_2$ gas at NTP is 1.429 g/L.

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