Solutions Question 53

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}} $ .



NCERT Chapter Video Solution

Dual Pane