Solutions Question 323
Question: A dry air is passed through the solution, containing the 10 gm of solute and 90 gm of water and then it pass through pure water. There is the depression in weight of solution wt by 2.5 gm and in weight of pure solvent by 0.05 gm. Calculate the molecular weight of solute [Kerala CET 2005]
Options:
50
180
100
25
51
Show Answer
Answer:
Correct Answer: C
Solution:
Lowering in weight of solution $ \propto $ solution pressure Lowering in weight of solvent $ \propto $ osmotic pressure
$ P^{0}-P_{s} $ ( $ \because $
$ p^{0}= $ vapour pressure of pure solvent) $ \frac{p^{0}-p_{s}}{p_{s}}=\frac{\text{Lowering in vapour pressure of solvent}}{\text{Lowering in vapour pressure of solution}} $
$ \frac{p^{0}-p_{s}}{p_{s}}=\frac{w\times M}{m\times W} $
$ \frac{0.05}{2.5}=\frac{10\times 18}{90\times m} $
$ \Rightarrow $ $ m=\frac{2\times 2.5}{0.05}=\frac{2\times 250}{5}=100 $
BETA
