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

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* k_f* m`

or,

`T_f^o-T_f=i* k_f* 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* (1.86^oC/m)* m`

`m=0.806mole/Kg`

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

Answer 2

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