Question

2) Calculate the decrease in
vapour pressure when 1.81x 102
kg of a solute is dissolved in 0.1
kg of water.
[V.p. of water 1.223 x 10^pa.
molecular wt. of solute=57.]​

Answer 1

Answer:

$\boxed{P-Po=699Pa}$

GIVEN:

Molar mass of solute : 57

Given mass of solute : 18.1 gms

VP of water : 1.223 X $10^4Pa$

Explanation:

Using the formula, for relative lowering of vapour pressure

$\frac{P-Po}{Po} =\frac{n2}{n1}$

where P₀= VP of pure solvent

           P= VP of mixture

           n₂= number of moles of solvent

           n₁= number of moles of solute

Calculating n₂ and n₁:-

∵ $n2=\frac{18.1}{57}=0.317$

∵ $n1=\frac{100}{18}=5.55$

⇒ $\frac{P-Po}{1.22Xa0^{4}} =\frac{0.317}{5.55}$

⇒ $\boxed{P-Po=699Pa}$

Hope this answer helped you :)