NEET Solved Paper 2016 Question 41
Question: At $ 100^{o}C $ the vapour pressure of a solution of 6.5g of a solute in 100 g water is 732 mm. If $ K _{b}=0.52, $ the boiling point of this solution will be :-
Options:
A) $ 101^{o}C $
B) $ 100^{o}C $
C) $ 102^{o}C $
D) $ 103^{o}C $
Show Answer
Answer:
Correct Answer: A
Solution:
$ ( \frac{P^{0}-P _{s}}{P^{0}} )=\frac{n}{N}=\frac{w _{solute}}{M _{solute}}\times \frac{M _{solvent}}{W _{solvent}} $ at $ 100{{\ }^{o}}C P^{o}=760mm $ $ \frac{760-732}{760}=\frac{6.5\times 18}{M _{solute}\times 100} $ $ M _{solute}=31.75\ g\ mo{l^{-1}} $ $ \Delta T _{b}=m\times K _{b}=\frac{w _{solute}\times 1000}{M _{solute}\times w _{solvent}}\times K _{b} $ $ \Delta T _{b}=\frac{0.52\times 6.5\times 1000}{31.75\times 100}=1.06{{\ }^{o}}C $
$ \therefore $ boiling point of solution $ =100^{o}C+{{1.06}^{o}}C\simeq 101{{\ }^{o}}C $
BETA
