Question #ab304

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Question #ab304

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Answer 1 · 2017-05-31T19:42:41

I got $253.47 g/mol$ $"253.47 g/mol"$ for the molecular mass, although it asks for the molecular formula of the solute. Maybe I'm missing something here?

My best guess is that the gram atomic mass is supposed to be $32.065 g/mol$ $"32.065 g/mol"$ so that $32.065 \times 8 = 256.52$ $32.065 xx 8 = 256.52$ , which is somewhat close to $253.47$ $253.47$ , indicating $S_{8}$ $S_8$ ...?

This seems to be combining the idea of vapor pressures via Raoult's law and utilizing the mol fraction, but the question is conflicting... not sure why it asks for the molecular formula if it gives a gram atomic mass.

Define the solute as $X$ $X$ and the solvent as $Y$ $Y$ . Then:

$P_{Y}^{*} = 854 torr$ $P_Y^"*" = "854 torr"$
$P_{Y} = 848.9 torr$ $P_Y = "848.9 torr"$

where $*$ $"*"$ indicates for the pure solvent.

From Raoult's law for ideal solutions, we have:

$P_{Y} = χ_{Y (l)} P_{Y}^{*}$ $P_Y = chi_(Y(l))P_Y^"*"$

$= (1 - χ_{X (l)}) P_{Y}^{*}$ $= (1 - chi_(X(l)))P_Y^"*"$ ,

where $χ_{a (l)} = \frac{n_{a (l)}}{n_{a (l)} + n_{b (l)}}$ $chi_(a(l)) = (n_(a(l)))/(n_(a(l)) + n_(b(l)))$ is the mol fraction of $a$ $a$ in the solution phase.

Since we know the drop in vapor pressure of the solvent $Y$ $Y$ due to adding the solute $X$ $X$ , we can calculate the mol fraction of $X$ $X$ .

$χ_{X (l)} = 1 - \frac{P_{Y}}{P_{Y}^{*}}$ $chi_(X(l)) = 1 - P_Y/P_Y^"*"$

$= 1 - \frac{848.9 torr}{854 torr}$ $= 1 - ("848.9 torr")/("854 torr")$

$= 0.005972$ $= 0.005972$

We can obtain the mols of $X$ $X$ in solution by the definition of the mol fraction.

$χ_{X (l)} = 0.005972 = \frac{n_{X (l)}}{n_{X (l)} + n_{Y (l)}}$ $chi_(X(l)) = 0.005972 = n_(X(l))/(n_(X(l)) + n_(Y(l)))$

We know the molar mass of the solvent, so we can get the mols of solvent:

$100 cancel("g CS"_2) xx "1 mol"/(76.141 cancel("g CS"_2)) = "1.313 mols CS"_2$ $"in solution"$

Therefore, we have:

$0.005972 = n_(X(l))/(n_(X(l)) + "1.313 mols")$

$=> 0.005972n_(X(l)) + 0.005972(1.313) = n_(X(l))$

$=> 0.9940n_(X(l)) = 0.007843$

$=> n_(X(l)) = "0.007890 mols"$

So, the molecular mass should be:

$"2.0 g"/"0.007890 mols" = color(blue)("253.47 g/mol")$ .

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