Question

A solution containing 8.6gL−1 of urea (molar mass 60gmol−1) is isotonic with a 5% solution of unknown solute.The molar mass of the solute will be

Answer 1

No of moles of urea =8.6/60
5% of unknown solution=5g/100ml
=50g/1000ml
=5g/litre
No of moles of unknown solute =50/m
8.6/60=50/m
m=348.9

Answer 2

Answer:  348.8 g/mol

Explanation: Isotonic solutions are those solutions which have the same osmotic pressure.

$\pi =CRT$ if osmotic pressures are equal at the same temperature, concentrations must also be equal.

$\pi$ = osmotic pressure

C= concentration

R= solution constant

T= temperature

Thus: $C_{urea}=C_s$ where concentration is in molarity.

For urea solution: 8.6 g of urea is dissolved in 1 L of solution.

For unknown solute s : 5 g of s is dissolved in 100 ml of solution.

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

$C_{urea}=\frac{8.6}{60}\times {1L}=0.14M$

$C_ s=\frac{5.6\times 1000}{M_s}\times 100$

As $C_{urea}=C_s$

$0.14=\frac{5.6\times 1000}{M_A}\times 100$

$M_s= 348.8 g/mol$