Question

If 12 g urea is dissolved in 400 g of water then freezing point of the solution will
be (Kof water = 1.86 K kg mol)
-1.93°C
-0.25°C
-0.93°C
-1.25°C​

Answer 1

Given:

12 g urea is dissolved in 400 g of water.

To find:

Freezing point of solution?

Calculation:

• Van't Hoff Factor of urea is 1.

Molality of urea is :

$\rm \: m = \dfrac{moles \: of \: solute}{mass \: of \: solvent}$

$\rm \implies\: m = \dfrac{( \frac{12}{60} )}{( \frac{400}{1000} )}$

$\rm \implies\: m = \dfrac{0.2}{0.4}$

$\rm \implies\: m = 0.5$

Now, applying expression for DEPRESSION OF FREEZING POINT:

$\rm\Delta T = i \times k_{f} \times m$

$\rm \implies\Delta T = 1 \times 1.86 \times 0.5$

$\rm \implies\Delta T = {0.93}^{ \circ} C$

So, freezing point will be :

$\rm \implies T = {0}^{ \circ}C - {0.93}^{ \circ} C$

$\rm \implies T = - {0.93}^{ \circ} C$

So, freezing point of solution is -0.93°C.