Question

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

Answer 1

Answer :

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 :  

$\Delta T_f=i\times K_f\times m\\\\T^o-T_s=i\times K_f\times m$

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

$0^oC-T_s=3\times (1.86^oC/m)\times 0.105m$

$T_s=-0.586^oC$

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

Now we have to calculate the boiling point of solution.

Formula used :  

$\Delta T_b=i\times K_f\times m\\\\T_s-T^o=i\times K_b\times m$

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

$T_s-100^oC=3\times (0.51^oC/m)\times 0.105m$

$T_s=100.2^oC$

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