Question

The osmotic pressure of an aqueoussolution of urea at 300 K is

Answer 1

Answer: The freezing point of solution is -0.09°C

Explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=imRT$

where,

$\pi$ = osmotic pressure of the solution = 120 kPa

i = Van't hoff factor = 1 (for non-electrolytes)

m = concentration of solute in terms of molality = ?

R = Gas constant = $8.31\text{L kPa }mol^{-1}K^{-1}$

T = temperature of the solution = 300 K

Putting values in above equation, we get:

$120kPa=1\times m\times 8.31\text{ L kPa }mol^{-1}K^{-1}\times 300K\\\\m=0.05m$

• To calculate the depression in freezing point, we use the equation:

$\Delta T=i\times K_f\times m$

where,

i = Vant hoff factor = 1 (for non-electrolytes)

$K_f$ = molal freezing point depression constant = 1.86°C/m

m = molality of solution = 0.05 m

Putting values in above equation, we get:

$\Delta T=1\times 1.86^oC/m\times 0.05m\\\\\Delta T=0.09^oC$

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.  

• The equation used to calculate freezing point of solution is:

$\Delta T=\text{freezing point of water}-\text{freezing point of solution}$

where,

$\Delta T$ = Depression in freezing point = 0.09 °C

Freezing point of water = 0°C

Freezing point of solution = ?

Putting values in above equation, we get:

$0.09^oC=0^oC-\text{Freezing point of solution}\\\\\text{Freezing point of solution}=-0.09^oC$

Hence, the freezing point of solution is -0.09°C