Respuesta :
Answer:
100.223°C is the boiling point of an aqueous solution.
Explanation:
Osmotic pressure of the solution = π = 10.50 atm
Temperature of the solution =T= 25 °C = 298 .15 K
Concentration of the solution = c
van'y Hoff factor = i = 1 (non electrolyte)
[tex]\pi =icRT[/tex]
[tex]c=\frac{\pi }{RT}=\frac{10.50 atm}{0.0821 atm L/mol K\times 298.15 K}[/tex]
c = 0.429 mol/L = 0.429 mol/kg = m
(density of solution is the same as pure water)
m = molality of the solution
Elevation in boiling point = [tex]\Delta T_b[/tex]
[tex]\Delta T_b=iK_b\times m[/tex]
[tex]\Delta T_b=T_b-T[/tex]
T = Boiling point of the pure solvent
[tex]T_b[/tex] = boiling point of the solution
[tex]K_b[/tex] = Molal elevation constant
We have :
[tex]K_b=0.52^oC/m[/tex] (given)
m = 0.429 mol/kg
T = 100° C (water)
[tex]\Delta T_b=1\times 0.52^oC/m\times 0.429 mol/kg[/tex]
[tex]\Delta T_b=0.223^oC[/tex]
[tex]\Delta T_b=T_b-T[/tex]
[tex]T_b=\Delta T_b+T=0.223^oC+100^oC=100.223^oC[/tex]
100.223°C is the boiling point of an aqueous solution.
