The vapour pressure of pure solvent is 0.8 mm of Hg at a particular temperature. On addition of a non - volatile solute 'A' the vapour pressure of solution becomes 0.6 mm of Hg. The mole fraction of component 'A' is:

Question

The vapour pressure of pure solvent is 0.8 mm of Hg at a particular temperature. On addition of a non - volatile solute 'A' the vapour pressure of solution becomes 0.6 mm of Hg. The mole fraction of component 'A' is: (A) 0.25 (B) 0.75 (C) 0.5 (D) 0.35

Tardigrade Admin · Accepted Answer

( P 0- P s / P 0)= x text solute ⇒ (0.8-0.6/0.8)= x text solute ⇒ 0.25

Q. The vapour pressure of pure solvent is $0.8 mm$ of $H g$ at a particular temperature. On addition of a non - volatile solute 'A' the vapour pressure of solution becomes 0.6 $mm$ of $H g$ . The mole fraction of component 'A' is:

0.25

0.75

0.5

0.35

$\frac{P ^{0} - P _{s}}{P ^{0}} = x_{solute}$

$\Rightarrow \frac{0.8 - 0.6}{0.8} = x_{solute} \Rightarrow 0.25$

Q. The vapour pressure of pure solvent is 0.8mm of Hg at a particular temperature. On addition of a non - volatile solute 'A' the vapour pressure of solution becomes 0.6 mm of Hg. The mole fraction of component 'A' is:

0.25

0.75

0.5

0.35

P0P0−Ps​​=xsolute ​ ⇒0.80.8−0.6​=xsolute ​⇒0.25

Q. The vapour pressure of pure solvent is $0.8 mm$ of $H g$ at a particular temperature. On addition of a non - volatile solute 'A' the vapour pressure of solution becomes 0.6 $mm$ of $H g$ . The mole fraction of component 'A' is:

$\frac{P ^{0} - P _{s}}{P ^{0}} = x_{solute}$

$\Rightarrow \frac{0.8 - 0.6}{0.8} = x_{solute} \Rightarrow 0.25$