Question

An aqueous solution of urea has a freezing point of $-{{0.52}^{o}}C$. Assume molarity and molality by same. ${{K}_{f}}$ for ${{H}_{2}}O$ is $1.86Kmo{{l}^{-1}}Kg$. The osmotic pressure of solution at ${{27}^{o}}C$ in atm is (nearest integer)

Answer 1

Hint: You can find the solution of the given question by simply using the van’t Hoff equation that is $\pi =cRT$. First try to find out the change in freezing temperature and then calculate the concentration of the aqueous solution and then by using the equation, you will get the osmotic pressure.

Complete step by step solution:
The freezing point of an aqueous solution of urea is $-{{0.52}^{o}}C$.
The molal freezing point depression constant that is ${{K}_{f}}$ of water is $1.86\text{ Kmo}{{\text{l}}^{-1}}\text{Kg}$.
We have to find out the osmotic pressure of the aqueous solution at ${{27}^{o}}C$ in the atmosphere. So, we can find the answer by applying the van’t Hoff equation, which describes that, the substances in dilute solution obey the ideal gas laws and the osmotic pressure (which is denoted by $\pi $) will be directly proportional to the molar concentration of the solution (i.e. denoted by $c$) and its temperature (denoted by the letter $T$). So, the equation results to a formula as shown below,
$\pi =cT$
First of all, let us find the change in the freezing point.
So, the change in the freezing point i.e. difference between final freezing point and freezing point at ${{0}^{o}}C$.
Therefore, $\Delta T=0-{{(-0.52)}^{o}}C={{0.52}^{o}}C$
Then by using the formula,
$\Delta T={{K}_{f}}\times \text{ molality}$, we will get the value of molality.
So, $Molality=\dfrac{\Delta T}{{{K}_{f}}}$.
As per the given question molar concentration i.e. molarity is the same that of molality. So, $c=\dfrac{\Delta T}{{{K}_{f}}}$.
We have got the value of $c$ and $T$ i.e. ${{27}^{o}}C$. Now we can find the osmotic pressure by using the equation.
So, $\pi =cT$
Then, $\pi =\dfrac{\Delta T}{{{K}_{f}}}\times T$
Then, $\pi =\dfrac{0.52\times 27}{1.86}=7.54\text{ atm }\cong \text{ }8\text{ atm}$

Hence, the osmotic pressure of the solution at ${{27}^{o}}C$ will be $8atm$.

Note: In freezing point depression, the freezing point of the solvents decreases as there is an increase in the addition of the solutes and this is a colligative property of solutions that is generally proportional to the molality of the added solute.