Question

A $5\%$ w/v solution of cane sugar is isotonic with $1\%$ w/v solution of a substance 'x'. The molecular weight of 'x' is ( in g/mol):

• A. 34.2
• B. 171.2
• C. 68.4
• D. 136.8

Answer 1

The correct option is C 68.4

Since it is given that the solutions are isotonic,
So ${\pi }_1={\pi }_2$
Here ${\pi }_1=$ for cane sugar solution
${\pi }_2=$ for substance $x$
$\therefore C_{1}RT=C_{2}RTWhereC=Concentration$
$\Rightarrow \left(\frac{wt.ingr.}{M.wt.ofcanesugar}\right)=\left(\frac{wt.ingr.}{M.wt.ofx}\right)$
So
$\frac{5}{342}=\frac{1}{M.wt.ofx}$
So M.wt.of $x=\frac{1\times 342}{5}=68.4$