Question

What is the molality of a solution of water and KCl if the freezing point of the solution is –3°C? (Kf = 1.86°C/m; molar mass of water = 18
g.

Answer 1

Some of the solutions exhibit colligative properties. These properties depend on the amount of solute dissolved in a solvent. These properties include freezing point depression, boiling point elevation, osmotic pressure and vapor pressure lowering. Calculations are as follows:

ΔT(freezing point)  = (Kf)mi
3  = 1.86 °C kg / mol (m)(2)
3 =3.72m
m = 0.81 mol/kg

Answer 2

Answer : The molality of a solution is, 0.806 mole/Kg

Explanation :

First we have to calculate the Van't Hoff factor (i) for KCl.

The dissociation of $KCl$ will be,

$KCl\rightarrow K^++Cl^-$

So, Van't Hoff factor = Number of solute particles = $K^++Cl^-$ = 1 + 1 = 2

Now we have to calculate the molality of solution.

Formula used for lowering in freezing point :

$\Delta T_f=i\times k_f\times m$

or,

$T_f^o-T_f=i\times k_f\times m$

where,

$\Delta T_f$ = change in freezing point

$T_f$ = temperature of solution = $-3^oC$

$T^o_f$ = temperature of pure water = $0^oC$

$k_f$ = freezing point constant  = $1.86^oC/m$

m = molality  = ?

i = Van't Hoff factor = 2

Now put all the given values in this formula, we get the molality of the solution.

$0^oC-(-3^oC)=2\times (1.86^oC/m)\times m$

$m=0.806mole/Kg$

Therefore, the molality of a solution is, 0.806 mole/Kg