Solutions Question 253
Question: A current of dry air is passed through a bulb containing 5 g of a solute in 100 g of water and then through water alone. The losses in weight of the solution and pure water were respectively 0.78 g and 0.02 g. Calculate relative lowering of vapour pressure.
Options:
A) 2.04
B) 1.05
C) 0.03
D) 0.09
Show Answer
Answer:
Correct Answer: C
Solution:
Loss in mass of solution =0.78 g Loss in mass of solvent =0.02 g
$ \Rightarrow \frac{P_{o}-P_{S}}{P_{S}}=\frac{0.02}{0.78}=\frac{Lossinmassofsolvent}{Lossinmassofsolution} $ Let $ A\to solventB\to Solute $
$ \therefore \frac{P_{o}-P_{S}}{P_{S}}=\frac{n_{B}}{n_{A}}=\frac{w_{B}}{M_{B}}\times \frac{M_{A}}{W_{A}} $
$ \frac{0.02}{0.78}=\frac{5}{M_{B}}\times \frac{18}{100} $
$ \Rightarrow M_{B}=35.1g\approx 35g $ Relative lowering of $ VP=\frac{P_{o}-P_{S}}{P_{o}}=X_{B} $
$ \therefore n_{B}=\frac{5}{35}=\frac{1}{7},n_{A}=\frac{100}{18} $
$ X_{B}=\frac{\frac{1}{7}}{\frac{1}{7}+\frac{100}{18}}=\frac{18}{(18+700)}=0.0251\cong 0.03 $
