Question

5g of $N{{a}_{2}}S{{O}_{4}}$ was dissolved $xg$ of ${{H}_{2}}O$ . The change in freezing point was found to be ${{3.82}^{\circ }}C$ . If $N{{a}_{2}}S{{O}_{4}}$ is $81.5%$ ionized , the value of x is \[\left( {{k}_{f}}for\text{ }water={{1.86}^{\circ }}Kgmo{{l}^{-1}} \right)\]is approximately:
A. 45g
B. 65g
C. 15g
D. 25g

Answer 1

Hint: When the solute is added to solvent there is a decrease in freezing point of solvent and that decrease in freezing point is known as depression in freezing point. It is calculated by formula ΔT f​= iKf​m.

Complete answer:
Molar mass can be calculated by adding the sum of each element present in the compound.
Molar mass of $N{{a}_{2}}S{{O}_{4}}$= 142g/mol
Number of moles in any solution is calculated by dividing the given mass to molecular mass.
\[Number\text{ }of\text{ }moles=\text{ }\dfrac{given\text{ }moles}{molecular\text{ }mass}\]
Mass of $N{{a}_{2}}S{{O}_{4}}$ in solution = 5g Number of moles in $N{{a}_{2}}S{{O}_{4}}$= $\dfrac{ 5 }{ 142 }$ = 0.03 moles \[Mass\text{ }of\text{ }water=x g=\dfrac{x} {1000} kg=0.001 x\] \[Molality~(m)=~\dfrac{0.0352 ~ mol~ Na2SO4}{0.001~x~mol~H2O}=\dfrac{35.2}{x}mol/kg\] Now as we know, 81.5 % of $N{{a}_{2}}S{{O}_{4}}$ is ionized which means if there is 100 g of solute out of that 81.5g of solute is ionized. So the degree of dissociation tells how much of The solute is ionized in solution. The degree of dissociation α=$\dfrac{81.5}{100} = 0.815$
The van't Hoff factor is the dimension that is unitless; it indicates how the molecule is dissociated in solution. The Van't Hoff factor is written as the symbol “i”. formula to calculate is :
\[i=[1+(n-1)\alpha \] i=[1+(3−1) × 0.815]=2.63
The solution freezes at a temperature lower than the freezing point of solvent. This property of solutions is called depression in freezing point.
The expression for depression in freezing point is shown as: ΔT f​= iKf​m
Where i is Van't hoff factor,ΔT f is change in freezing point ,m is molality and Kf is cryoscopic constant. \[{{3.82}^{\circ }}C=2.63\times {{1.86}^{o}}CKg/mol\times \dfrac{35.2}{x}mol/kg\] $x = 45g$ So the 45g of water is required for 5g of $N{{a}_{2}}S{{O}_{4}}$

The correct option is A.

Note: Depression in freezing point is calculated by formula ΔT f​= iKf​m
For which we should know van't hoff factor and molality. Molality is found by calculation the number of moles of solute and then dividing them by mass of solvent that is taken in kg.