# What is the "molality" of a 5.0*g mass of toluene dissolved in 255*g of benzene?

Sep 29, 2015

Molalilty is (moles of solute)/(kg solvent). You have the mass of solute, and can determine the amount of moles, and you also have the mass of solvent. The exercise is arithmetic.

#### Explanation:

Here, you could equally have a solution of benzene in toluene or a solution of toluene in benzene. The advantage of using molality, or opposed to molarity, is that because we have used mass as the measure of solvent, molality is independent of volume. Temperature would affect the volume of both toluene and benzene, therefore, a measurement of molality is independent of temperature.

Sep 29, 2015

$m = \frac{5.4 \times {10}^{- 2} m o l}{0.255 k g} = 0.21 \frac{\text{mol}}{g}$

#### Explanation:

The molality is calculated by: $m = \left({n}_{\text{solute")/(mass_"solvent}}\right)$

Where, $\left({n}_{\text{solute}}\right)$ is the number of mole of the solute (Toluene), and $m a s {s}_{\text{solvent}}$ is the mass of the solvent in kg.

For Toluene:
$M {M}_{\text{Toluene"=92 g/"mol}}$
$m a s {s}_{\text{Toluene}} = 5.0 g$

Then, n_"Toluene"=(5.0cancel(g))/(92 cancel(g)/"mol")=5.4xx10^(-2) mol

The molality then: $m = \frac{5.4 \times {10}^{- 2} m o l}{0.255 k g} = 0.21 \text{mol"/"kg}$