Question

Calculate the freezing point of the solution containing 0.105 m k2s. calculate the boiling point of the solution above. -g

Answer 1

The freezing point of a solution is
`-0.586^oC`

The boiling point of a solution is
`100.2^oC`

Explanation :

First we have to calculate the freezing point of solution.

Formula used :

`\Delta T_f=i* K_f* m\\\\T^o-T_s=i* K_f* m`

where,

`\Delta T_f` = change in freezing point

`\Delta T_s` = freezing point of solution = ?

`\Delta T^o` = freezing point of water =
`0^oC`

i = Van't Hoff factor = 3 (for K₂S electrolyte)

`K_f` = freezing point constant for water =
`1.86^oC/m`

m = molality = 0.105 m

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

`0^oC-T_s=3* (1.86^oC/m)* 0.105m`

`T_s=-0.586^oC`

Thus, the freezing point of a solution is
`-0.586^oC`

Now we have to calculate the boiling point of solution.

Formula used :

`\Delta T_b=i* K_f* m\\\\T_s-T^o=i* K_b* m`

where,

`\Delta T_b` = change in boiling point

`\Delta T_s` = boiling point of solution = ?

`\Delta T^o` = boiling point of water =
`100^oC`

i = Van't Hoff factor = 3 (for K₂S electrolyte)

`K_b` = boiling point constant for water =
`0.51^oC/m`

m = molality = 0.105 m

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

`T_s-100^oC=3* (0.51^oC/m)* 0.105m`

`T_s=100.2^oC`

Thus, the boiling point of a solution is
`100.2^oC`

Answer 2

Quantity of K2S m = 0.105 m
Number of ions i = 2(K) + 1(S) = 3
Freezing point depression constant of water Kf = 1.86
delta T = i x m x Kf = 3 x 0.105 x 1.86 = 0.586
Freezing point = 0 - 0.586 = 0.586 C
Boiling point constant of water Kb = 0.512
delta T = i x m x Kb = 3 x 0.105 x 0.512 = 0.161
Boiling point = 100 + 0.161 = 100.161 C