Question

On dissolving $0.5g$ of a non-volatile, non-ionic solute to $39g$ of benzene, its vapour pressure decreases from $650mmHg$ to $640mmHg$. The depression is the freezing point of benzene (in K) upon the addition of the solute is _________.
Given data for benzene: Molar mass = $78gmo{{l}^{-1}}$, Molal freezing point depression constant = $5.12\text{ }K\text{ }kg\text{ }mo{{l}^{-1}}$.

Answer 1

Think about the formula that is used to relate the change in vapour pressure of the solvent before and after the addition of the solute and the number of moles of the solute and solvent. The Van’t Hoff factor is also used in this formula.

Complete step by step solution:
The formula that is used to calculate the depression in freezing point requires us to know the values of the cryoscopic constant (molal freezing point depression constant), the Van’t Hoff factor and the molality of the solution. We need to find the molality of the solution using the information that is given to us. The formula for the molality of a solution is:
$\text{molality = }\dfrac{\text{no}\text{. of moles of the solvent}}{\text{weight of solvent in kgs}}$
We already know the weight of the solvent as it is given in the question itself, we need to find the number of moles of the solute. We can find this by using the following formula that relates the pressure and the number of moles.
$\dfrac{{{P}_{0}}-{{P}_{s}}}{{{P}_{s}}}=i\left( \dfrac{{{n}_{solute}}}{{{n}_{solvent}}} \right)$
Here, ${{P}_{0}}$ is the vapour pressure of the solvent before the addition of the solute, ${{P}_{s}}$ is the vapour pressure of the solution after the addition of the solute. $i$ is the Van’t Hoff factor, and $n$ denotes the number of moles of the solute and the solvent.
The values of ${{P}_{0}}$ and ${{P}_{s}}$ are given as $650mmHg$ and $640mmHg$ respectively. The Van’t Hoff factor depends solely on the degree of dissociation and the number of ions formed, since it is given in the question that the solute is non-ionic, the Van’t Hoff factor will be 1. We can calculate the value of ${{n}_{solvent}}$ using the formula:
$n=\dfrac{\text{given weight}}{\text{molar mass}}$
For benzene, the given weight is $39g$ and the molar mass is $78gmo{{l}^{-1}}$ so the number of moles of benzene will be:

& {{n}_{solvent}}=\dfrac{39g}{78gmo{{l}^{-1}}} \\\ & {{n}_{solvent}}=0.5mol \\\ \end{aligned}$$ Now we can put all the values in the formula and obtain the value of the number of moles of the solute. The expression will be solved as follows: $$\begin{aligned} & \dfrac{650-640}{640}=1\times \left( \dfrac{{{n}_{solute}}}{0.5} \right) \\\ & \dfrac{10}{640}\times 0.5={{n}_{solute}} \\\ & {{n}_{solute}}=\dfrac{0.5}{64} \\\ \end{aligned}$$ The formula that we use to calculate the depression in freezing point is as follows: $$\begin{aligned} & \Delta {{T}_{f}}=i\times {{K}_{f}}\times m \\\ & \Delta {{T}_{f}}=i\times {{K}_{f}}\times \dfrac{{{n}_{solute}}}{weight\text{ }of\text{ }solvent} \\\ \end{aligned}$$ Where, $\Delta {{T}_{f}}$ is the depression in freezing point, $i$ is the Van’t Hoff factor, ${{K}_{f}}$ is the cryoscopic constant, and $m$ is the molality as is defined above. We will now put all the values that we have in the equation and solve for the depression in freezing point. $$\begin{aligned} & \Delta {{T}_{f}}=1\times 5.12K\text{ }kg\text{ }mo{{l}^{-1}}\times \dfrac{\dfrac{0.5}{64}mol}{39\times {{10}^{-3}}kg} \\\ & \Delta {{T}_{f}}=5.12\times \dfrac{0.00781\times {{10}^{3}}}{39}K \\\ & \Delta {{T}_{f}}=\dfrac{5.12\times 7.81}{39}K \\\ & \Delta {{T}_{f}}=1.043K \\\ \end{aligned}$$ **Hence, the answer to this question is that the depression in freezing point after the addition of solute is $1.043K$.** **Note:** The information about the weight of the solute given is extra information, but you can also use it to find the molar mass of the solute separately and then find the total number of moles. We were going to use the value of the number of moles anyway so you can keep it as it is too.