Respuesta :

W0lf93
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

Answer :

The freezing point of a solution is [tex]-0.586^oC[/tex]

The boiling point of a solution is [tex]100.2^oC[/tex]

Explanation :

First we have to calculate the freezing point of solution.

Formula used :  

[tex]\Delta T_f=i\times K_f\times m\\\\T^o-T_s=i\times K_f\times m[/tex]

where,

[tex]\Delta T_f[/tex] = change in freezing point

[tex]\Delta T_s[/tex] = freezing point of solution = ?

[tex]\Delta T^o[/tex] = freezing point of water = [tex]0^oC[/tex]

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

[tex]K_f[/tex] = freezing point constant for water = [tex]1.86^oC/m[/tex]

m = molality  = 0.105 m

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

[tex]0^oC-T_s=3\times (1.86^oC/m)\times 0.105m[/tex]

[tex]T_s=-0.586^oC[/tex]

Thus, the freezing point of a solution is [tex]-0.586^oC[/tex]

Now we have to calculate the boiling point of solution.

Formula used :  

[tex]\Delta T_b=i\times K_f\times m\\\\T_s-T^o=i\times K_b\times m[/tex]

where,

[tex]\Delta T_b[/tex] = change in boiling point

[tex]\Delta T_s[/tex] = boiling point of solution = ?

[tex]\Delta T^o[/tex] = boiling point of water = [tex]100^oC[/tex]

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

[tex]K_b[/tex] = boiling point constant for water = [tex]0.51^oC/m[/tex]

m = molality  = 0.105 m

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

[tex]T_s-100^oC=3\times (0.51^oC/m)\times 0.105m[/tex]

[tex]T_s=100.2^oC[/tex]

Thus, the boiling point of a solution is [tex]100.2^oC[/tex]

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