When 8.05 g of an unknown compound X was dissolved in 100. g of benzene (C6H6), the vapor pressure of the benzene decreased from 100.0 torr to 94.8 torr at 26 degrees C. What is the mole fraction and molar mass of X?

I got `"114.51 g/mol"`.

The reduction of vapor pressure is a colligative property. From Raoult's law, we have:

`P_j = chi_jP_j^"*"`,

where `P_j` is the partial pressure above the solution from component `j`, `P_j^"*"` is the pure vapor pressure of `j` by itself, and `chi_j` is the mol fraction of `j` in the solution.

(We consider `bbj` to be the solvent and `bbi` to be the solute.)

But mol fractions are always `<= 1`. Therefore, for a two-component solution, `chi_j = 1 - chi_i`. Thus:

`P_j = (1 - chi_i) P_j^"*"`

Since `chi_j <=1`, it follows that `1 - chi_i <= 1` as well, or that `chi_i >= 0`. Therefore, adding any solute into a solvent decreases its vapor pressure.

This reduction is calculated as follows. Define `DeltaP_j = P_j - P_j^"*"` to get:

`color(green)(DeltaP_j) = chi_jP_j^"*" - P_j^"*"`

`= (chi_j - 1)P_j^"*"`

`= (1 - chi_i - 1)P_j^"*"`

`= color(green)(-chi_iP_j^"*")`

Since `P_j^"*"` and `chi_i` are positive, the reduction in vapor pressure is of course negatively-signed. From this equation then, we can determine the mol fraction of what was dissolved in benzene.

Plug `P_j = "94.8 torr"` and `P_j^"*" = "100.00 torr"` in to get:

`(94.8 - 100.00) "torr" = -chi_i ("100.00 torr")`

`=> color(blue)(chi_i = 0.052)`

Now that we have the mol fraction of `X`, we can find the molar mass of `X` simply by solving for the mols of `X`, and dividing the given mass by the mols of `X`.

`0.052 = (n_X)/(n_X + n_"Ben")`

`=> 0.052n_X + 0.052n_"Ben" = n_X`

`=> (0.052 - 1)n_X = -0.052n_"Ben"`

`=>n_X = 0.0549n_"Ben"`

But we know the molar mass of benzene; it's `6xx12.011 + 6xx1.0079 = "78.1134 g/mol"`. So, we get the mols of benzene by its molar mass:

`"100.0 g Benzene" xx "1 mol"/"78.1134 g" = "1.2802 mols Benzene solvent"`

Therefore, the mols of `X` are:

`n_X = 0.0549("1.2802 mols") = "0.0703 mols X"`

This gives a molar mass for `X` as:

`color(blue)(M_(m,X)) = "8.05 g X"/"0.0703 mols X" = color(blue)("114.51 g/mol")`