Question

$100g$ of liquid A (molar mass $140gmo{l}^{-1})$ was dissolved in $1000g$ of liquid B (molar mass $180gmo{l}^{-1})$. The vapour pressure of pure liquid B was found to be $500torr$. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is $475torr$.

Answer 1

Number of mole is the ratio of mass to molar mass.

For liquid A, number of moles $=\frac{100}{140}=0.714$

For liquid B, number of moles $=\frac{1000}{180}=5.556$

Mole fraction of A $=\frac{0.714}{0.714+5.556}=0.114$

Mole fraction of B $=1-0.114=0.886$

$P_{total}=P_A+P_B=P_A^o{X}_A+P_B^o{X}_B$

$475=P_A^o\times 0.114+500\times 0.886$

$475=0.114P_A^o+443$

Vapour pressure of pure A, $P_A^o=280.7$ torr

Vapour pressure of A in solution $=280.7\times 0.114=32$ torr

Thus vapour pressure of pure solute A is 280.7 torr and its vapour pressure in solution is 32 torr