Question

What is the boiling point of an aqueuous solution of

Answer 1

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)

$\pi =icRT$

$c=\frac{\pi }{RT}=\frac{10.50 atm}{0.0821 atm L/mol K\times 298.15 K}$

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 = $\Delta T_b$

$\Delta T_b=iK_b\times m$

$\Delta T_b=T_b-T$

T = Boiling point of the pure solvent

$T_b$ = boiling point of the solution

$K_b$ = Molal elevation constant

We have :

$K_b=0.52^oC/m$ (given)

m = 0.429 mol/kg

T = 100° C (water)

$\Delta T_b=1\times 0.52^oC/m\times 0.429 mol/kg$

$\Delta T_b=0.223^oC$

$\Delta T_b=T_b-T$

$T_b=\Delta T_b+T=0.223^oC+100^oC=100.223^oC$

100.223°C is the boiling point of an aqueous solution.