Respuesta :
Answer:
1.67 m is the molality of a solution .
The boiling point of a solution is 100.85°C.
The freezing point of a solution is -3.1°C.
Explanation:
[tex]Molality=\frac{\text{Moles of solute}}{\text{mass of solvent in kg}}[/tex]
Moles of sugar = 1.25 mol
Mass of water = 0.750 kg
Molality of the solution ;
[tex]m=\frac{1.25 mol}{0.750 kg}=1.67 m[/tex]
1.67 m is the molality of a solution .
Freezing point of the resulting solution
[tex]\Delta T_b=T_b-T'[/tex]
[tex]\Delta T_b=K_b\times m[/tex]
where,
[tex]\Delta T_b[/tex] =elevation in boiling point =
T' = Boiling point of pure solvent
[tex]T_b[/tex] = boiling point of solution
[tex]K_b[/tex] = boiling point constant
m = molality
we have :
[tex]K_b[/tex] of water = 0.512 °C/m ,
m = 1.67 m
[tex]\Delta T_b=0.512^oC\times 1.67 m[/tex]
[tex]\Delta T_b=0.85^oC[/tex]
Boiling point of pure water = T' = 100°C
Boiling point of solution = [tex]T_b[/tex]
[tex]T_b=T'+\Delta T_b=100^oC+0.85^oC=100.85^oC[/tex]
The boiling point of a solution is 100.85°C.
Freezing point of the resulting solution
[tex]\Delta T_f=T-T_f[/tex]
[tex]\Delta T_f=K_f\times m[/tex]
where,
[tex]\Delta T_f[/tex] =depression in freezing point =
T = Freezing point of pure solvent
[tex]T_f[/tex] = Freezing point of solution
[tex]K_f[/tex] = freezing point constant
m = molality
we have :
[tex]K_f[/tex] of water = 1.86°C/m ,
m = 1.67 m
[tex]\Delta T_f=1.86^oC\times 1.67 m[/tex]
[tex]\Delta T_f=3.1^oC[/tex]
Freezing point of pure water = T = 0°C
Freezing point of solution = [tex]T_f[/tex]
[tex]T_f=T-\Delta T_f=0^oC-3.1^oC=-3.1^oC[/tex]
The freezing point of a solution is -3.1°C.
