Question

A solution contains 36 g of solute dissolved in 1 litre of water give am osmotic pressure 6.75 atm at 27 c the molal elevation constant of water is 52 calculate the boiling point of solution .

Answer 1

Answer: The boiling point of solution is $100.14^0C$

Explanation:

$\pi =CRT$ for non electrolytes

$\pi$ = osmotic pressure

C= concentration  in Molarity=$\frac{\text {given mass}}{\text {Molar mass}\times {\text {Volume of solution in L}}}=\frac{36}{M\times 1L}$

R= solution constant= 0.0821 Latm/Kmol

T= temperature  = $27^0C=27+273=300K$

$6.74atm=\frac{36}{M\times 1L}\times 0.0821\times 300$

$M=132g/mol$

2. Elevation in boiling point

$\Delta T_b=K_b\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}$

$\Delta T_b$ = change in boiling point

$K_b$ = boiling point constant = 0.52

$\Delta T_b=T_b-T_b^0=(T_b-100)^0C$

$T_b-100^0C=0.52\times \frac{36g}}{132}\times 1kg}$

$T_b=100.14^0C$