Question

An aqueous solution was found to have an osmotic pressure of 2.51 atm at ${25}^{o}C$. The freezing point of solution if $K_f$ for water is $1.86Kmo{l}^{-1}kg$ will be:

Assume molarity and molality are same.

• A. $-{0.19}^{o}C$
• B. $-{0.55}^{o}C$
• C. $-{0.36}^{o}C$
• D. None of these

Answer 1

The correct option is A $-{0.19}^{o}C$

$\because \pi =\left(n/V\right)RT$

Given, $\pi =2.51atm,T=273+25=298K$

$\therefore 2.51=\left(n/V\right)\times 0.0821\times 298$

$\therefore \frac{n}{V}=\frac{2.51}{0.0821\times 298}=0.1026M$

or $Molarity=0.1026M=Molality$ (Given)

Now, $\Delta T=K_f\times molality$

$=1.86\times 0.1026=0.19$

$\therefore$ Freezing pt. of the solution $=0-0.19=-{0.19}^{o}C$