Question

At $300\text{K}$, the vapour pressure of an ideal solution containing $3$ moles of $A$ and $2$ moles of $B$ is $600$ torr. At the same temperature, if $1.5$ moles of $A$ and $0.5$ moles of $C$ (non volatile ) are added to this solution the vapour pressure of solution increases by $30$ torr. What is the value of $P_B^{\circ }$ (in Torr)?

Answer 1

According to Raoult's law,
$P_T=P_A^o{\chi }_A+P_B^o{\chi }_B$
$P_A^o$ and $P_B^o$ are the vapour pressure of pure $A$ and $B$ in liquid phase
Let, $n_A=\text{Moles of A}$
$n_B=\text{Moles of B}$
So, Mole fraction of $A$, ${\chi }_A=\frac{n_A}{n_A+n_B}$
Mole fraction of $B$, ${\chi }_B=\frac{n_B}{n_A+n_B}$
Now, putting the values,
$600=P_A^o\left(\frac{3}{3+2}\right)+P_B^o\left(\frac{2}{3+2}\right)$
$\Rightarrow 3000=3P_A^o+2P_b^o\cdot \cdot \cdot \left(1\right)$
Ig $1.5$ moles of $A$ and $0.5$ moles of $C$ are added, then, ${\chi }_A^{/}=\left(\frac{3+1.5}{3+1.5+2+0.5}\right)$
${\chi }_B^{/}=\left(\frac{2}{3+1.5+2+0.5}\right)$
Noe total pressue, $P=(600+30)\text{torr}=630\text{Torr}$
So, $630=P_a^o\left(\frac{4.5}{7}\right)+P_b^o\left(\frac{2}{7}\right)$
$\Rightarrow 4410=4.5P_A^o+2P_B^o\cdot \cdot \cdot \left(2\right)$
By solving $\left(1\right)$ and $\left(2\right)$ we get,
$P_A^o=940\text{Torr}$
By using equation $\left(1\right)$,
$P_B^0=\frac{3000-3P_A^o}{2}$
$\Rightarrow P_B^0=\frac{3000-3\times 940}{2}\text{Torr}=90\text{Torr}$