Question

What mass of cholesterol should be dissolved in `"295 g"` of benzene to lower the freezing point by `"0.450 K"`? (`M = "386.6 g/mol"` for cholesterol and `"78.11 g/mol"` for benzene.)

Answer 1

I got

`10.0_4` `"g"`

of cholesterol to three sig figs (the subscript indicates the first digit of which we are not certain) to reduce the normal freezing point of benzene to `5.05^@ "C"`.

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Given the freezing point depression equation:

`bb(DeltaT_f = T_f - T_f^"*" = -iK_fm)`,

where:

• `T_f` is the freezing point, and `"*"` indicates pure solvent.
• `i` is the van't Hoff factor, and is `1` for nonelectrolytes.
• `K_f` (not given) is the freezing point depression constant of benzene, the solvent.
• `m` is the molality of the solution, `"mol solute/kg solvent"`.

Since we don't have `K_f`, we'll have to get it from (Physical Chemistry: A Molecular Approach, McQuarrie, Ch. 25-3):

`\mathbf(K_f = (M_"solvent")/("1000 g/kg")(R(T_f^"*")^2)/(DeltabarH_f))`

where `M` is the molar mass in `"g/mol"`, `R` is the universal gas constant, and `DeltabarH_f` is the molar enthalpy of fusion in `"J/mol"`.

From NIST, `DeltabarH_f = "9.87 kJ/mol"`, so the `K_f` is:

`K_f = ("78.11 g/mol")/("1000 g/kg") ("8.314472 J/mol"cdot"K" cdot (5.50 + 273.1"5 K")^2)/("9870 J/mol")`

`= "5.11 K"cdot"kg/mol"`

This could be verified here... This allows us to get the molality:

`color(green)(m) = (DeltaT_f)/(-iK_f)`

`= (-"0.450 K")/(-(1)("5.11 K"cdot"kg/mol"))`

`=` `color(green)"0.0881 mol solute/kg solvent"`

Since we know that `"295 g benzene"`, or `"0.295 kg"`, were used, and we know that cholesterol has a molar mass of `"386.6 g/mol"`:

`"0.0881 mol solute"/cancel"kg solvent" xx 0.295 cancel"kg solvent"`

`=` `"0.0260 mol cholesterol"`

As a result, we have this mass of solute dissolved in benzene:

`0.0260 cancel"mols cholesterol" xx "386.6 g"/cancel"mols cholesterol"`

`=` `color(blue)("10.0 g cholesterol")` to three sig figs.