Solutions Question 411
Question: The vapour pressure lowering of 0.2 molal urea solution at 40°C is (assuming latent heat of vaporization $ (\Delta H) $ is 10 Kcal/mol.)
Options:
A) 0.2 torr
B) 0.1 torr
C) 0.5 torr
D) 0.3 torr
Show Answer
Answer:
Correct Answer: A
Solution:
Using Clapeyron - Clausius equation first calculate vapour pressure of water of $ \Delta H=\frac{2.303,RT_1T_2}{T_2-T_1}\log \frac{P_2}{P_1} $
$ T_1=273+100^{o}=373,K $
$ T_2=273+40^{o}=313,K $
$ 10=\frac{2.303\times 0.002\times 313\times 373}{(313-373)}\log \frac{P_2}{760} $ This gives vapour pressure $ P_2 $ of water at $ 40^{o}C=58.2,Torr $
By Raoult law $ \frac{\Delta P}{P^{o}}=\frac{w_1}{m_1}\times \frac{m_2}{w_2} $
$ \frac{w_1}{m_1}=0.2 $ mole urea $ w_2=1000,g,H_2O $
$ m=18,g/mol $
$ \Delta P=P^{o}\times \frac{w_1}{m_1}\times \frac{w_2}{m_2}=\frac{58.2\times 0.2\times 18}{1000}=0.20,Torr $
