Question

in a solution of urea, 3g of it is dissolved in 100ml of water. find out the freezing point of the solution.

Answer 1

use this formula to calculate freezing point of the given solution
i hope it is helpful for you

Answer 2

Answer : The freezing point of a solution is 272.07 K.

Solution : Given,

$k_f=1.86^oC$    (for water)

Mass of solute (urea) = 3 g

Volume of solvent (water) = 100 ml

Molar mass of urea = 60 g/mole

Density of water = 1 g/ml

First we have to calculate the mass of solvent.

$Density=\frac{mass}{volume}$

Mass of solvent = Density of solvent × volume of solvent

Mass of solvent = 1 g/ml × 100 ml = 100 g

Now we have to calculate the molality of solution.

Formula used : $Molality=\frac{\text{ mass of solute}\times 1000}{\text{ Molar mass of solute}\times \text{ mass of solvent}}=\frac{3g\times 1000}{60g/mole\times 100kg}=0.5mole/kg$

The formula used for depression in freezing point is,

$\Delta T_f=k_f\times m\\T^o_f-T_f=k_f\times m$

where,

$T^o_f$ = freezing point of pure solvent = $0^oC$ = 273 K

$T_f$ = freezing point of solution

m = molality

$k_f$ = molal depression constant

Now put all the given values in this formula, we get

$0^oC-T_f=1.86^oC\times 0.5mole/kg$

$T_f=-0.93^oC=273-0.93=272.07K$

Therefore, the freezing point of a solution is 272.07 K.