Solutions Question 294
Question: The boiling point of water ( $ 100^{o}C $ ) becomes $ {{100.52}^{o}}C $ , if 3 grams of a nonvolatile solute is dissolved in $ 200ml $ of water. The molecular weight of solute is ( $ K_{b} $ for water is $ 0.6,K-m $ ) [AIIMS 1998]
Options:
A) $ 12.2,g,mo{l^{-1}} $
B) $ 15.4,g,mol $
C) $ 17.3,g,mo{l^{-1}} $
D) $ 20.4,g,mol $
Show Answer
Answer:
Correct Answer: C
Solution:
First boiling point of water = $ 100^{o}C $ Final boiling point of water = $ {{100.52}^{o}} $
$ w=3g $ , $ W=200,g $ , $ K_{b}=0.6,k{g^{-1}} $
$ \Delta T_{b}=100.52-100={{0.52}^{o}}C $
$ m=\frac{K_{b}\times w\times 1000}{\Delta T_{b}\times W} $
$ =\frac{0.6\times 3\times 1000}{0.52\times 200}=\frac{1800}{104}=17.3gmo{l^{-1}} $ .
BETA
