Question

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

• A. $180$
• B. $160$
• C. $16$
• D. $100$

Answer 1

The correct option is B $160$

Number of moles of A= 3 mol
Number of moles of B = 2 mol
Mole fraction of A$x_A=\frac{n_A}{n_A+n_B}=\frac{3}{3+2}=\frac{3}{5}=0.6$
Molefraction of B $x_B=x_A-0.6=0.4p=p_A^{\circ }{x}_A+p_B^{\circ }{x}_{B}184=200\times 0.6+p_B^{\circ }\times 0.4184=120+p_B^{\circ }\times 0.4p_B^{\circ }=\frac{64}{0.4}=160\text{torr}$