Question

3%solution of gluose is isotonic with 1% solution of a non volatile non electrolyte substance the molecular mass of the substance would be

Answer 1

Mglucose = Munknown
$\frac{3}{180} \times \frac{1000}{100} = \frac{1}{x} \times \frac{1000}{100}$
$\frac{3}{180} = \frac{1}{x}$
$x = \frac{180}{3}$
x=60

Answer 2

Answer: 62.5 g/mol

Explanation: Isotonic solutions are those solutions which have the same osmotic pressure.  If osmotic pressures are equal at the same temperature, concentrations must also be equal.

$\pi =CRT$   (for non electrolytes)

$\pi$ = osmotic pressure

C= concentration

R= solution constant

T= temperature

Thus: $C_{glucose}=C_A$ where concentration is in molarity.

For glucose solution: 3 g of glucose is dissolved in 100 ml of solution.

For solute A: 1 g of A is dissolved in 100 ml of solution.

$Molarity=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{volume of solution in ml}}$

$C_{glucose}=\frac{3.0\times 1000}{180\times 100}=0.16M$

$C_ A=\frac{1.0\times 1000}{M_A\times 100}$

As $C_{urea}=C_A$

$0.16=\frac{1.0\times 1000}{M_A\times 100}$

$M_A= 62.5g/mol$