Question

The vapour pressure of pure liquid solvent A is 0.80 atm. When a nonvolatile substance B is added to the solvent its vapour pressure drops to 0.60 atm. What is the mole fraction of component B in the solution?

• A. $0.25$
• B. $0.28$
• C. $0.30$
• D. $0.31$

Answer 1

The correct option is A $0.25$

According to raoult's law, $p=p^0\times x$

p=partial vapour pressure of pure component A

$p^0=$ vapour pressure of component A

$x=$ mole fraction of component A.

Substitute values in the above expression.

$0.60=0.80\times x$

Hence, the mole fraction of component A is 0.75.

The mole fraction of component B $=1-0.25=0.75$

Hence, the correct option is $A$