Question

A solution has mole fraction of a solute equal to 0.05 and vapour pressure
$2 \times 10 {}^{4}$
Nm
$- {}^{2}$
what id the vapour pressure of a pure solvent
​

Answer 1

Answer:

0.05 and vapour pressure

$2 \times 10 {}^{4}$

Nm

$- {}^{2}$

wA solution has mole fraction of a solute equal to 0.05 and vapour pressure

$2 \times 10 {}^{4}$

Nm

$- {}^{2}$

what id the vapour pressure of a pure solvent

hat id the vapour pressure of a pure solvent

Answer 2

Given info : A solution has mole fraction of a solute equals to 0.05 and vapor pressure of solution is 2 × 10⁴ N/m².

To find : The vapor pressure of pure solvent is...

solution : according to Raoult's law,

vapor pressure of pure solvent, P = P₀ × x

where P₀ vapor pressure of solution with solute and x is mole fraction of solvent.

here mole fraction of solute is 0.05 so mole fraction of solvent = 1 - 0.05 = 0.95

so x = 0.95

P₀ = 2 × 10⁴ N/m²

⇒P = 2 × 10⁴ × 0.95 = 1.9 × 10⁴ N/m²

Therefore the vapor pressure of pure solvent is 1.9 × 10⁴ N/m².