Question

At ${25}^{\circ }C$, the total pressure of an ideal solution obtained by mixing $3\text{mol}$ of ${}^{\prime }{A}^{\prime }$ and $2\text{mol}$ of ${}^{\prime }{B}^{\prime }$, is $184\text{torr}$. What is the vapour pressure (in $\text{torr}$) of pure ${}^{\prime }{B}^{\prime }$ at the same temperature (Vapour pressure of pure ${}^{\prime }{A}^{\prime }$ at ${25}^{\circ }C$ is $200\text{torr}$)?

• A. $180$
• B. $160$
• C. $140$
• D. $100$

Answer 1

The correct option is B $160$

$\text{Number of moles of A,}{n}_A=3\text{mol}\text{Number of moles of B,}{n}_B=2\text{mol}$

$\text{Mole fraction of A =}\frac{\text{No. of moles of A}}{\text{Total no. of moles of A and B}}$

$x_A=\frac{n_A}{n_A+n_B}=\frac{3}{3+2}=\frac{3}{5}=0.6x_B=1-0.6=0.4\text{By Raoult's law,}$
$p_A=x_A\times p_A^o$
$p_B=x_B\times p_B^o$
$P_{\text{total}}=x_A\times p_A^o+x_B\times p_B^{o}184=200\times 0.6+p_B^{\circ }\times 0.4184=120+p_B^{\circ }\times 0.4$
$p_B^{\circ }=160\text{torr}$