# Question 655b9

May 16, 2016

Answer will be (2) 140 mm of Hg .

#### Explanation:

We have , A + B -------> AB

Mole fraction (X) of A = 0.8 ; X of B will be = 1 - 0.8 = 0.2

Now , we are given , V.P and of pure A and V.P of solution AB as 70 and 84 mm of Hg respectively .

As it forms an ideal solution , therefore ;

V.P of AB = Partial V.P of A + Partial V.P of B

Partial V.P of any component = X of that component * V.P of that component in pure state

Therefore , substituting the values in
V.P of AB = Partial V.P of A + Partial V.P of B ; we get ,

84 = 70 * 0.8 + V.P of pure B * 0.2

Solving this we get 140 mm of Hg as the V.P of pure B .

May 16, 2016

I get $\text{140 mm Hg}$.

The temperature is just for context.

This question is basically asking you to use Raoult's law for ideal binary mixtures:

$\setminus m a t h b f \left({P}_{j} = {\chi}_{j} {P}_{j}^{\text{*}}\right)$

where:

• ${P}_{j}$ is the "non-pure" vapor pressure of the solution (i.e. when there is component $i$ in it).
• ${P}_{j}^{\text{*}}$ is the vapor pressure of the pure component j (i.e. the solution with no component $i$).
• ${\chi}_{j}$ is the $\setminus m a t h b f \left(\text{mol}\right)$ fraction of component $j$ in solution. As ${\chi}_{j} \to 1$, ${P}_{j} \to {P}_{j}^{\text{*}}$.

You have:

• ${P}_{A}^{\text{*" = "70 mm Hg}}$
• ${\chi}_{A} = 0.8$
• ${P}_{\text{tot" = "84 mm Hg}}$
• P_B^"*" = ?

in an ideal binary mixture of $A$ and $B$. But this information is really all you need. Recall Dalton's law of partial pressures:

$\setminus m a t h b f \left({P}_{\text{tot}} = {P}_{A} + {P}_{B}\right)$

Now, combine this with Raoult's law (best for ideal binary mixtures!) to get:

${P}_{\text{tot" = chi_AP_A^"*" + chi_BP_B^"*}}$

Here we can use the fact that ${\chi}_{A} + {\chi}_{B} = 1$ to get:

color(blue)(P_"tot") = chi_AP_A^"*" + (1 - chi_A)P_B^"*"#

$= {\chi}_{A} {P}_{A}^{\text{*" + P_B^"*" - chi_AP_B^"*}}$

Now you can solve for ${P}_{B}^{\text{*}}$.

${P}_{\text{tot" - chi_AP_A^"*" = P_B^"*}} \left(1 - {\chi}_{A}\right)$

$\frac{\textcolor{b l u e}{{P}_{B}^{\text{*") = (P_"tot" - chi_AP_A^"*}}}}{1 - {\chi}_{A}}$

$= \frac{84 - 0.8 \cdot \left(70\right)}{1 - 0.8}$

$= \frac{84 - 56}{0.2}$

$= 28 \cdot 5 = \textcolor{b l u e}{\text{140 mm Hg}}$