Question

$161\times {10}^{-2}\text{kg}$ of Glauber's salt is dissolved in water to obtain $1{\text{dm}}^3$ of a solution of density $1520{\text{kg m}}^{-3}$. Which of the following is true?

• A. Molarity of the solution $=5\text{M}$
• B. Molality of the solution $=6.17\text{m}$
• C. Mole fraction of the solute $=0.1$
• D. Mole fraction of the solute $=0.0043$

Answer 1

Answer: (A), (B), (C)

The correct options are
A Molarity of the solution $=5\text{M}$
B Molality of the solution $=6.17\text{m}$
C Mole fraction of the solute $=0.1$

The formula of Glauber's salt is $N{a}_{2}S{O}_4.10H_{2}O$
Molecular mass of $N{a}_{2}S{O}_4.10H_{2}O$
$=[2\times 23+32+4\times 16]+10(1.0\times 2+16)=322{\text{g mol}}^{-1}$
Weight of the Glauber's salt taken $=1610\text{g}$
Out of $1610\text{g}$ of salt, weight of anhydrous $N{a}_{2}S{O}_4$
$=\frac{142}{322}\times 1610=710\text{g}$
Number of moles of $N{a}_{2}S{O}_4$ per $d{m}^3$ of the solution
$=\frac{710}{142}=5$
Molarity of the solution $=5\text{M}$
Density of solution $=1520kg$ $m^{-3}$
$=\frac{1.520\times {10}^3}{{10}^6}$ $gc{m}^{-3}$ = $1.520{\text{g cm}}^{-3}$
Total weight of solution $=V\times d=1{\text{dm}}^3\times d$
$=1000{\text{cm}}^3\times 1.520gc{m}^{-3}=1520\text{g}$
Weight of water $=1520-710=810\text{g}$
Molality of solution. $=\frac{5}{810\text{g}}\times 1000\text{g}=6.17\text{m}$
Number of moles of water in the solution $=\frac{810}{18}=45$
Mole fraction of $N{a}_{2}S{O}_4$
$=\frac{\text{No. of moles of}N{a}_{2}S{O}_4}{\text{Total number of moles}}=\frac{5}{5+45}$
$=0.1$