Answer:
1) The value of the [tex]K_b[/tex] is 0.07742°C/m.
2) 0.261°C is the freezing-point depression of a solution.
Explanation:
1) [tex]\Delta T_b=T_b-T[/tex]
[tex]\Delta T_b=K_b\times m[/tex]
[tex]\Delta T_b=iK_b\times \frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Mass of solvent in Kg}}[/tex]
where,
[tex]\Delta T_b[/tex] =Elevation in boiling point
[tex]K_b[/tex] = boiling point constant of solvent= 3.63 °C/m
1 - van't Hoff factor (non-electrolyte solute)
m = molality
We have : [tex]\Delta T_b=2.4^oC[/tex]
m = 3.1 m
[tex]K_b=?[/tex]
[tex]\Delta T_b=K_b\times m[/tex]
[tex]2.4^oC=K_b\times 3.1 m[/tex]
[tex]K_b=\frac{2.4 ^oC}{3.1 m}=0.07742 ^oC/m[/tex]
The value of the [tex]K_b[/tex] is 0.07742°C/m.
2) [tex]\Delta T_f=T-T_f[/tex]
[tex]\Delta T_f=K_f\times m[/tex]
[tex]Delta T_f=iK_f\times \frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Mass of solvent in Kg}}[/tex]
where,
[tex]\Delta T_f[/tex] =depression in freezing point
[tex]K_f[/tex] = freezing point constant of solvent= 1.86°C/m
1 - van't Hoff factor (non-electrolyte solute)
m = molality
We have , Moles of solute = 0.705 mol
Mass of solvent = 5.02 kg
[tex]molality=\frac{\text{Moles of solute }}{\text{Mas of solvent(kg)}}[/tex]
m = [tex]\frac{0.705 mol}{5.02 kg}=0.1404 mol/kg[/tex]
[tex]K_f=1.86^oC/m[/tex]
[tex]\Delta T_f=iK_f\times m[/tex]
[tex]=1\times 1.86 ^oC/m\times 0.1404 m[/tex]
[tex]=0.261^oC[/tex]
0.261°C is the freezing-point depression of a solution.
