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:
- $ Na _2SO _4\to 2N{a^{+}}+SO_4^{2-} $ $ x=1+(3-1)0.815=2.63 $ $ 3.82=1.86\times 2.63\times \frac{5\times 1000}{142\times X} $
$ \therefore $ $ X=\frac{1.86\times 2.63\times 5000}{142\times 3.82} $ =45gm
