Question

The molar mass of a solute X in ${\text{gm mol}}^{-1}$, if its 1% solution is isotonic with a 5% solution of cane sugar (molar mass = 342 ${\text{gm mol}}^{-1}$), is

• A. 68.4
• B. 34.2
• C. 136.2
• D. 171.2

Answer 1

Answer: (A)

68.4

Let M g/mol be the molar mass of solute X.

The wt percent of solute X is converted to number of moles in 100 g of solution.

$\frac{1}{100\times M}$ --------- (1)

The wt percent of sucrose is converted to number of moles in 100 g of solution.

$\frac{5}{100\times 342}$ ------- (2)

1% solution of solute X is isotonic with a 5% solution of cane sugar.

Hence, (1) $=$ (2)

$\frac{1}{100\times M}=\frac{5}{100\times 342}$

$\Rightarrow M=68.4g/mol$