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# Calculate the mass of the compound $(\text{molar mass} = 256 g/mol)$ to be dissolved in $75g$ of benzene to lower its freezing point by $0.48K$ $\left ( K_{f}= 5.12 K kg/mol \right )$

Last updated date: 20th Jun 2024
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Hint: We can use Henry’s law to formulate the formula required for the given question which is given as follows
$\Delta {{T}_{f}}={{K}_{f}}m$
Where, $\Delta {{T}_{f}}$ is the change in the freezing point, ${{K}_{f}}$ is the freezing constant and $m$is the molality of the solution in consideration.

Let W g of compound is to be dissolved in the benzene.
Number of moles of compound will be given by the following formula
$n=\dfrac{mass}{\text{molar mass}}$

$n=\dfrac{W}{256}$​

Mass of benzene will be = $75 g$ $=\dfrac{75}{1000}​kg$ = $0.075 kg$
Now we can calculate molality of the solution, $m=\dfrac{moles}{\text{mass of solvent}}$

$m=\dfrac{W}{256\times 0.075}$​= $19.2W ​molal$

The depression in the freezing point, $\Delta {{T}_{f}}$ ​ = $0.48K$
The depression in the freezing point constant = ${{K}_{f}}$​= $5.12 Kg/mol$
$\Delta {{T}_{f}}={{K}_{f}}m$

Now, substituting these values in the given formula we get,
$0.48=5.12\times \dfrac{W}{19.6}$
$W=\dfrac{0.48\times 19.2}{5.12}=1.8$

Hence the correct answer is that $1.8 g$ of solute is to be dissolved.

$C$ = $k[P_{gas}]$
Where, $C$ is the solubility of a gas at a fixed temperature in a particular solvent, $k$ is Henry's law constant and $[P_{gas}]$ is the partial pressure of the gas