# A solution containing 10.0 g of an unknown liquid and 90.0 g water has a freezing point of -3.33° C. Given K_f = 1.86°C/m for water, what is the molar mass of the unknown liquid?

Jan 24, 2016

${\text{62.1 g mol}}^{- 1}$

#### Explanation:

The most important thing to realize here is that since the solute is a liquid, you must be dealing with a covalent compound, which as you know does not ionize in aqueous solution.

This implies that the van't Hoff factor, which tells you the ratio between the number of moles of particles produced in solution and the number of moles of solute dissolved, will be equal to $1$.

Simply put, every mole of your unknown compound will produce $1$ mole of particles in solution.

$\textcolor{b l u e}{\Delta {T}_{f} = i \cdot {K}_{f} \cdot b} \text{ }$, where

$\Delta {T}_{f}$ - the freezing-point depression;
$i$ - the van't Hoff factor
${K}_{f}$ - the cryoscopic constant of the solvent;
$b$ - the molality of the solution.

The freezing-point depression is defined as

$\textcolor{b l u e}{\Delta {T}_{f} = {T}_{f}^{\circ} - {T}_{f}} \text{ }$, where

${T}_{f}^{\circ}$ - the freezing point of the pure solvent
${T}_{f}$ - the freezin point of the solution

Use the given freezing point of the solution to find

$\Delta {T}_{f} = {0}^{\circ} \text{C" - (-3.33^@"C") = 3.33^@"C}$

Now, you strategy here will be to find the molaity of the solution, $b$. This will then get you the number of moles of solute present in this solution.

So, rearrange the above equation to solve for $b$. Keep in mind that the cryoscopic constant for water, ${K}_{f}$, is said to be equal to ${1.86}^{\circ} {\text{C kg mol}}^{- 1}$.

$\Delta {T}_{f} = i \cdot {K}_{f} \cdot b \text{ "implies" } b = \frac{\Delta {T}_{f}}{i \cdot {K}_{f}}$

b = (3.33 color(red)(cancel(color(black)(""^@"C"))))/(1 * 1.86 color(red)(cancel(color(black)(""^@"C"))) "kg mol"^(-1)) = "1.7903 mol kg"^(-1)

Now, a solution's molality is defined as the number of moles of solute divided by the mass of the solvent - expressed in kilograms.

$\textcolor{b l u e}{b = {n}_{\text{solute"/m_"solvent in kg}}}$

Since your solution is said to contain $\text{90.0 g}$ of water, you can say that

$b = {n}_{\text{solute"/m_"solvent in kg" " "implies" "n_"solute" = b * m_"solvent in kg}}$

You will have

${n}_{\text{solute" = "1.7903 mol" color(red)(cancel(color(black)("kg"^(-1)))) * 90.0 * 10^(-3)color(red)(cancel(color(black)("kg"))) = "0.16113 moles}}$

Finally, the molar mass of a substance can be found by dividing the mass of a given sample of that substance by the total number of moles it contains.

$\textcolor{b l u e}{{M}_{M} = \frac{m}{n}}$

In your case, the molar mass of the liquid will be

${M}_{M} = {\text{10.0 g"/"0.16113 moles" = "62.062 g mol}}^{- 1}$

Rounded to three sig figs, the answer will be

${M}_{M} = \textcolor{g r e e n}{{\text{62.1 g mol}}^{- 1}}$