Question

1.
The vapour pressure of pure liquid A at 310°C is
0.158 bar. The vapour pressure of this liquid in
solution with liquid B is 0.095 bar. Calculate the
mole fraction of Ain the solution if the mixture
obeys Raoult's law.

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Answer 1

Answer : The mole fraction of A in the solution will be, 0.6

Explanation :

According to the Raoult's law,

$p_A=X_A\times p^o_A$

where,

$p_A$ = vapor pressure of solution = 0.095 bar

$p^o_A$ = vapor pressure of solution = 0.158 bar

$X_A$ = mole fraction of liquid A = ?

Now put all the given values in this formula, we get the mole fraction of liquid A.

$0.095\text{ bar}=X_A\times 0.158\text{ bar}$

$X_A=0.6$

Therefore, the mole fraction of A in the solution will be, 0.6