Question

On dissolving ${\bf{0}}.{\bf{25}}{{ }}{\bf{g}}$ of a non-volatile substance in ${\bf{30}}{{ }}{\bf{mL}}$ benzene (density ${\bf{0}}.{\bf{8}}{{ }}{\bf{g}}{{ }}/{{ }}{\bf{mL}}$), its freezing point decreases by ${\bf{0}}.{\bf{4}}{{\bf{0}}^o}{\bf{C}}$. Calculate the molecular mass of non-volatile substance (${{\bf{K}}_f} = {\bf{5}}.{\bf{12Kkgmo}}{{\bf{l}}^{ - 1}}$).

Answer 1

Hint:
The freezing point of a Solution will decrease, that is it will freeze at a lower temperature when a non volatile solute is added to the Solution. Therefore, in this Solution the freezing point decreases from $273K \to 273.40K$. Higher the value of ${K_f}$ more is the ability of the non volatile solute to decrease the freezing point.

Formula Used:
$D = \dfrac{M}{V}$
where $D$ is the density, $M$ is the mass, $V$ is the volume
 $\Delta T = \dfrac{{{K_f} \times {W_{solute}} \times 1000}}{{{M_{solute}} \times {W_{solvent}}}}$
Here, $\Delta T$ is the difference in temperature,
${K_f}$ is the cryoscopic constant, ${W_{solute}}$ is the weight of the solute, ${W_{solvent}}$ is the weight of the solvent, ${M_{solute}}$ is the molar mass of the solute.

Complete step by step answer:
This question is based on the colligative property of a Solution known as the Depression of freezing point. This occurs due to addition of a non volatile solute into a pure solvent which leads to the decrease in the freezing point.

The decrease in the freezing point is expressed in terms of difference in the temperatures. But in this question, we know the difference in the temperature. It is required to find the mass of the solute. This can be done by altering the formula as demonstrated below:
 $\Delta T = \dfrac{{{K_f} \times {W_{solute}} \times 1000}}{{{M_{solute}} \times {W_{solvent}}}}$
Taking the molar mass of the solute from the denominator and sending it to the numerator, we get,
${M_{solute}} = \dfrac{{{K_f} \times {W_{solute}} \times 1000}}{{\Delta T \times {W_{solvent}}}}$
We know that, ${K_f}$ is the cryoscopic constant = ${\bf{5}}.{\bf{12Kkgmo}}{{\bf{l}}^{ - 1}}$
${W_{solute}}$ = ${\bf{0}}.{\bf{25}}{{ }}{\bf{g}}$
The mass of the solvent can be found from the density using the values mentioned in the question. This is done in the steps shown below:
Density = ${\bf{0}}.{\bf{8}}{{ }}{\bf{g}}{{ }}/{{ }}{\bf{mL}}$
Volume = ${\bf{30}}{{ }}{\bf{mL}}$
Density = $\dfrac{{Mass}}{{Volume}}$

Substituting these values in the equation given above:
$0.8 = \dfrac{{mass}}{{30}}$
$0.8 \times 30 = mass$
Multiplying the two
$24g = Mass$
Therefore, the weight of the solvent will be $24g$.
That is, ${W_{solvent}}$ = $24g$

Now we can find the molar mass of the solute through the steps demonstrated below:
${M_{solute}} = \dfrac{{{K_f} \times {W_{solute}} \times 1000}}{{\Delta T \times {W_{solvent}}}}$
On substituting the values,we get
$ \Rightarrow \dfrac{{5.12 \times 0.25 \times 1000}}{{0.40 \times 24}}$
Calculating the numerator and denominator, we get,
$ \Rightarrow \dfrac{{1280}}{{9.6}}$
$ \Rightarrow 133.33g$

Therefore the answer to the question will be $133.33g$. That is the molar mass of the non volatile solute will be $133.33g$.

Note: The freezing point of a solvent decreases when a solute is added because the solute and solvent form bonds and this will contribute to a lower freezing point. ${K_f}$ is the constant that can determine the depression of freezing point of a Solution upon addition of one mole of solute to one kilogram of solvent.