JEE Main On 8 April 2017 Question 10
Question: 5 g of $ Na _2SO _4 $ was dissolved in x g of $ H _2O. $ The change in freezing point was found to be $ 3.82{}^\circ C $ . If $ Na _2SO _4 $ is 81.5% ionised, the value of x ( $ K _{f} $ for $ water=1.86{}^\circ C $ kg $ mo{l^{\text{-1}}} $ ) is approximately: (molar mass of S = 32 g $ mo{l^{\text{-1}}} $ and that of Na = 23 g $ mo{l^{\text{-1}}} $ ) [JEE Online 08-04-2017]
Options:
A) 25 g
B) 65 g
C) 15 g
D) 45 g
Show Answer
Answer:
Correct Answer: D
Solution:
$$ \Delta T_f=i \cdot K_f \cdot m $$
where:
- $\Delta T_f=3.82^{\circ} \mathrm{C}$ is the change in freezing point.
- $i$ is the van ’t Hoff factor, which accounts for ionization.
- $K_f=1.86^{\circ} \mathrm{C} \mathrm{kg} \mathrm{mol}^{-1}$ is the cryoscopic constant for water.
- $m$ is the molality of the solution.
Van ’t Hoff Factor $i$ :
- Ideal $i$ for $\mathrm{Na}_2 \mathrm{SO}_4$ is 3 .
- Given $81.5 %$ ionization:
$$ i=1+0.815 \times(3-1)=2.63 $$
Molality $m$ :
- Molar mass of $\mathrm{Na}_2 \mathrm{SO}_4=142 \mathrm{~g} / \mathrm{mol}$.
- Moles of $\mathrm{Na}_2 \mathrm{SO}_4=\frac{5}{142} \approx 0.0352 \mathrm{~mol}$.
- Molality $m=\frac{0.0352 \times 1000}{x}$. Freezing Point Depression Equation:
$$ 3.82=2.63 \times 1.86 \times \frac{0.0352 \times 1000}{x} $$
$$ x=\frac{2.63 \times 1.86 \times 0.0352 \times 1000}{3.82} \approx 45 \mathrm{~g} $$
Therefore, the solution is 45 g of water. Hence, the answer is D) 45 g .
BETA
