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 vapour pressure of $A$ (in torr) the solution if the total vapour pressure of the solution is $475torr.$

Answer 1

Answer: 32

Step I : Calculation of vapour pressure of pure liquid $A(p{°}_A)$ Number of moles of liquid $A$
$\left(n_A\right)=\frac{W_A}{M_A}=\frac{(100g)}{\left(140gmo{l}^{-1}\right)}=0.7143mol$
Number of moles of liquid $B$
$\left(n_B\right)=\frac{W_B}{M_B}=\frac{(1000g)}{\left(180gmo{l}^{-1}\right)}=5.5556mol$
Mole fraction of $A$
$\left(x_A\right)=\frac{n_A}{n_A+n_B}$
$=\frac{(0.7143mol)}{(0.7143+5.5556)mol}=\frac{0.7143}{6.2699}=0.1139$
Mole fraction of $B\left(x_B\right)=1-0.1139=0.8861$
Vapour pressure of pure liquid $B,B\left(p_B^0\right)=500$ torr
Total vapour pressure of solution $(p)=475$ torr
According to the Raoult's law
$p=p_A^o{x}_A+p_B^o{x}_B$
$475\text{ torr }=p_A^o\times (0.1139)+500\text{ torr }\times (0.8861)$
$475\text{ torr }=p_A^o\times (0.1139)+443.05\text{ torr }$
$p_A^o=\frac{(475-443.05)\text{ torr }}{(0.1139)}=\frac{31.95}{0.1139}\text{ torr }=280.5\text{ torr }$
Step II : Calculation of vapour pressure of $A$ in the solution $\left(p_A\right)$
According to Raoult's law,
$p_A=p_A^0{x}_A=(280.5\text{ torr })\times (0.1139)$
$p_A=32.0\text{ torr }$