Discussion Forum

A solution of 17.1 w% of surcose (molar mass =$342g mol^{-1}$)  is isotonic with a x w% solution of oxalic acid (molar mass =90g mol^-1)  . assume the degree of dissociation of  oxalic acid as 0.01 . what is x?

Answer:

Explanation:

Given,

 mass of sucrose (w₁) =1.71 g

 Molar mass of sucrose (M1)=342 g mol^-1

 Mass of oxalic  acid (w₂) =xg

 Molar mass of oxalic acid (M₂)  = 90 g mol^-1

 Degree of dissociation ($\alpha$)  of oxalic acid =0.01

  Step I To find the value of van't Hoff factor (i) 

$i= (1- \alpha)+n \alpha$

 where, n= number of parts in which one molecule of oxalic acid dissociates (=3)

 i=(1-0.01)+(3 x 0.01)

i= 0.99+0.03

   i=1.02

Step II  To find the value of (x)

 For isotonic  solutions:

 $\pi$ (sucrose) = $\pi $ ( oxalic acid)     [ $\because \pi =CRT$]

 or, Molarity (sucrose)  = i Molarity (oxalic acid)

 $\frac{w_{1}}{M_{1}} (sucrose) = i \times  ( \frac{x}{ M_{2}})$ (oxalic acid )

 $x= \frac{w_{1}}{M_{1}}\times\frac{M_{2}}{i}=\frac{17.1\times90}{1.02\times342}$

 $x= \frac{1539}{348.84}=4.41 g$

 hence, value of x=4.41 g

 Hence, option (c)  is correct.