Question

If `"6.5 g"` of solute `X` is dissolved in `"50 g"` of ether solvent (`"74 g/mol"`) whose vapor pressure dropped from `"445 mm Hg"` to `"410 mm Hg"`, what is the molar mass of `X`?

Answer 1

I got `"112.88 g/mol"`, though you only have two sig figs. If you use `"74 g/mol"` for the ether molecular weight, you'd get `"112.78 g/mol"`.

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This is asking you to use Raoult's law to look at vapor pressure reduction from `P_A^"*"` to `P_A`:

`P_A = chi_(A(l))P_A^"*"`

where:

• `P_A` is the vapor pressure of solvent `A`.
• `"*"` indicates pure solvent.
• `chi_(A(l)) = (n_A)/(n_A + n_X)` is the mol fraction of `A` in the solution phase. `n_A` is mols of solvent `A`.

Ether is also called diethyl ether, or `"H"_3"C""H"_2"C"-"O"-"CH"_2"CH"_3`, indeed with a molecular weight of `"74.122 g/mol"`. I'll use that, since it's more accurate. Given the mass of solute, we need its mols to find its molar mass.

Those mols can be found from the mol fraction.

`chi_(A(l)) = P_A/P_A^"*"`

`= ("410 mm Hg")/("445 mm Hg")`

`= n_A/(n_A + n_X) = 0.9213`

We know the mols of the solvent to be:

`n_A = m_A/M_A = 50 cancel"g" xx "1 mol ether"/(74.122 cancel"g")`,

(`m` being the mass in `"g"` and `M` being the molar mass in `"g/mol"`.)

`=> n_A =` `"0.6746 mols ether"`

So, we can solve for the mols of the solute `X`.

`0.9213 = ("0.6746 mols ether")/("0.6746 mols ether" + n_X)`

`=> 0.9213 (0.6746 + n_X) = 0.6746`

`=> 0.6215 + 0.9213n_X = 0.6746`

`=> n_X = "0.05758 mols"`

This means the molar mass of `X` is:

`color(blue)(M_X) = m_X/n_X = "6.5 g X"/"0.05758 mols X" = color(blue)("112.88 g/mol")`

though you only have two sig figs, apparently.