The vapour pressure of pure liquid solvent is 0.80 atm. When a non-volatile substance B is added to the solvent, its vapour pressure drops to 0.6 atm. What is the mole fraction of component B in this solution?

Question

The vapour pressure of pure liquid solvent is 0.80 atm. When a non-volatile substance B is added to the solvent, its vapour pressure drops to 0.6 atm. What is the mole fraction of component B in this solution? (A) 0.75 (B) 0.25 (C) 0.48 (D) 0.3

Tardigrade Admin · Accepted Answer

According to Raoult's law, Relative lowering of vapour pressure propto mole fraction of solute or (po-ps/p0B)= chiB or (0.80-0.60/0.80)= chiB ∴ chiB=(1/4)=0.25

Q. The vapour pressure of pure liquid solvent is $0.80 a t m$ . When a non-volatile substance $B$ is added to the solvent, its vapour pressure drops to $0.6 a t m$ . What is the mole fraction of component $B$ in this solution?

0.75

0.25

0.48

0.3

According to Raoult's law, Relative lowering of vapour pressure $\propto$ mole fraction of solute or
$\frac{p ^{o} - p _{s}}{p ^{0 B}} = χ_{B}$
or $\frac{0.80 - 0.60}{0.80} = χ_{B}$
$∴ χ_{B} = \frac{1}{4} = 0.25$

Q. The vapour pressure of pure liquid solvent is 0.80atm. When a non-volatile substance B is added to the solvent, its vapour pressure drops to 0.6atm. What is the mole fraction of component B in this solution?

0.75

0.25

0.48

0.3

According to Raoult's law, Relative lowering of vapour pressure ∝ mole fraction of solute or p0Bpo−ps​​=χB​ or 0.800.80−0.60​=χB​ ∴χB​=41​=0.25

Q. The vapour pressure of pure liquid solvent is $0.80 a t m$ . When a non-volatile substance $B$ is added to the solvent, its vapour pressure drops to $0.6 a t m$ . What is the mole fraction of component $B$ in this solution?

According to Raoult's law, Relative lowering of vapour pressure $\propto$ mole fraction of solute or
$\frac{p ^{o} - p _{s}}{p ^{0 B}} = χ_{B}$
or $\frac{0.80 - 0.60}{0.80} = χ_{B}$
$∴ χ_{B} = \frac{1}{4} = 0.25$