Question

At $25{}^\circ C,$ at 5% aqueous solution of glucose (molecular weight $=180\,g\,mo{{l}^{-1}}$ ) is isotonic with a 2% aqueous solution containing and unknown solute. What is the molecular weight of the unknown solute?

• A. 60
• B. 80
• C. 72
• D. 63

Answer 1

Answer: (C)

72

Answer 2

Since the two solutions are isotonic, they must have same concentrations in moles/litre. For glucose solution, concentration
$=5\text{ }g/100\text{ }c{{m}^{3}}$ (given)
$=50g/L$
$\therefore$ $\frac{50}{180}=\frac{20}{M}$
or $M=72$
( $\because$ Molar mass of glucose $=180\text{ }g\text{ }mo{{l}^{-1}}$ )
For unknown substance, concentration
$=2\text{ }g/100\text{ }c{{m}^{3}}$ (given)
$=20g/L=\frac{20}{M}mol/L$
$\therefore$ $\frac{50}{180}=\frac{20}{M}$ or $M=72$