Solutions Question 315
Question: Equal masses of a solute are dissolved in equal amount of two solvents A and B, respective molecular masses being $ M_{A} $ and $ M_{B} $ . The relative lowering of vapour pressure of solution in solvent A is twice that of the solution in solvent B. If the solutions are dilute, $ M_{A} $ and $ M_{B} $ are related as
Options:
A) $ M_{A}=M_{B} $
B) $ 2M_{A}=M_{B} $
C) $ M_{A}=2M_{B} $
D) $ M_{A}=4M_{B} $
Show Answer
Answer:
Correct Answer: C
Solution:
For dilute solution, $ \frac{\Delta P}{P{}^\circ }\Rightarrow \frac{n_{solute}}{n_{solvent}} $ For solution in A, $ \frac{\Delta P_{A}}{P_A^{{}^\circ }}=\frac{W/M}{W_{A}/M_{A}}=\frac{W}{M}\times \frac{M_{A}}{W_{A}} $ …(i) For solution in B, $ \frac{\Delta P_{B}}{P_A^{{}^\circ }}=\frac{W}{M}\times \frac{M_{B}}{W_{B}} $ ….(ii) From (i) and (ii),
$ \frac{\Delta P _{A}/{P^{\circ} } _{A}}{\Delta P _{B}/{P^{\circ} } _{B}}=$
$2=\frac{M_{A}W_{B}}{M_{B}W_{A}}=\frac{M_{A}}{W_{B}}(W_{A}=W_{B}) $
$ M_{A}=2M_{B} $
