Question

At 60 ∘C, compound X has a vapor pressure of 96 mmHg, benzene (C6H6) has a vapor pressure of 395 mmHg, and a 50:50 mixture by mass of benzene and X has a vapor pressure of 286 mmHg .
What is the molar mass of X?

Answer 1

According to Raoult's law n equation form, for a mixture of liquids A and B, this reads:

$$\begin{array}{l} p_{A} = x_{A} \times P_{A}^{0} \\ p_{B} = x_{B} \times P_{B}^{0} \\ \end{array}$$

In this equation, $P_A$ and $P_B$ are the partial vapour pressures of the components A and B.

And

The total vapour pressure of the mixture is equal to the sum of the individual partial pressures.

$$\text{Total vapour pressure} = p_{A} + p_{B} = P$$

Also $x_A$ and $x_B$ are the mole fractions of A and B. That is exactly what it says it is - the fraction of the total number of moles present which is A or B.

You calculate mole fraction using, for example:

$$x_{A} = \frac{\text{moles of A}}{\text{total number of moles}} =$$

From this

$$x_{A} + x_{B} = 1$$ $$P = P_{A} * x_{A} + P_{B} * x_{B} = P_{A} * x_{A} + P_{B} - P_{B} * x_{A}$$ $$x_{A} = (P - P_{B}) / (P_{A} - P_{B}) / (286 - 395) / (96 - 395) = 0.3645 = n_{A} / (n_{A} + n_{B})$$

n-number of moles

$$n_{A} = (n_{A} + n_{B}) * 0.3645$$

From the condition, we know that mixture we have equal masses C6H6 (we call it a component B) and compound X (we call it a component A). It follows that $M_A * n_A = M_B * n_B = > n_A = M_B * n_B / M_A$. Combining these two equations, we obtain $M_B * n_B / M_A = (M_B * n_B + M_A * n_B) / M_A * 0.3645 = > M_B * n_B = M_B * n_B * 0.3645 + M_A * n_B * 0.3645 = > M_B = M_B * 0.3645 + M_A * 0.3645$

$$=> M _ {A} = \left(M _ {B} - M _ {B} * 0.3645\right) / 0.3645 = \left(\left(6 * 12 + 6\right) - \left(6 * 12 + 6\right) * 0.3645\right) / 0.3645 = 1$$

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