Answered: An aqueous solution of 2 percent…

Chemistry: Principles and Practice

3rd Edition

ISBN:9780534420123

Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer

Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer

Chapter12: Solutions

Section: Chapter Questions

Problem 12.72QE

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Question

An aqueous solution of 2 percent non-volatile solute exerts a pressure of 1.004 bar at the boiling point of the solvent. What is the molecular mass of the solute?
[Vapour pressure of water = 1.013 bar]

Expert Solution

Step 1

Given,

$V a p o r p r e s s u r e o f w a t e r = 1.013 b a r V a p o r p r e s s u r e o f s o l u t i o n s = 1.004 b a r P e r c e n t a g e o f s o l u t e = 2 %$

To find -

The molar mass of the solute

Step 2

According to Raoult's Law,

$\begin{array}{rcl} \frac{(p_{1} ° - p_{1})}{p_{1} °} & = & \frac{(w_{2} \times M_{1})}{(M_{2} \times w_{1})} \\ w h e r e, \\ p_{1} ° & = & v a p o r p r e s s u r e p u r e s o l v e n t \\ p_{1} & = & v a p o r p r e s s u r e o f s o l u t i o n \\ w_{2} & = & M a s s o f s o l u t e \\ w_{1} & = & m a s s o f s o l v e n t \\ M_{1} & = & m o l a r m a s s o f s o l v e n t \\ M_{2} & = & m o l a r m a s s o f s o l u t e \end{array}$

Step 3

Here,

Assume-

$M a s s o f s o l u t e t o b e 2 g T h e n, M a s s o f s o l v e n t = (100 - 2) g = 98 g$

Thus,

$\begin{array}{rcl} \frac{(p_{1} ° - p_{1})}{p_{1} °} & = & \frac{(w_{2} \times M_{1})}{(M_{2} \times w_{1})} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1) \\ w h e r e, \\ p_{1} ° & = & v a p o r p r e s s u r e p u r e s o l v e n t i . e w a t e r = 1.013 b a r \\ p_{1} & = & v a p o r p r e s s u r e o f s o l u t i o n = 1.004 b a r \\ w_{2} & = & M a s s o f s o l u t e = 2 g \\ w_{1} & = & m a s s o f s o l v e n t = 98 g \\ M_{1} & = & m o l a r m a s s o f s o l v e n t i . e w a t e r = 18 g m o l^{- 1} \\ M_{2} & = & m o l a r m a s s o f s o l u t e t o b e c a l c u l a t e d \end{array}$

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