Question

A certain liquid X has a normal boiling point of 133.60°C and a boiling point elevation constant Kb= 2.46°C kg mol^-1.Calculate the bolling point of a solution made of 52.2g of benzamide (C7H7NO) dissolved in 350. g of X.

Answer 1

136.63 °C

Step-by-step explanation:

ΔTb=Tb solution - Tb pure

Where; Tb pure = 133.60°C

molar mass of solute = 121.14 g/mol

number of moles of solute; 52.2g/121.14 g/mol = 0.431 moles

molality = 0.431 moles/350 * 10^-3 = 1.23 molal

Then;

ΔTb = Kb * m * i

Kb = 2.46°C kg mol^-1

m = 1.23 molal

i = 1

ΔTb = 2.46 * 1.23 * 1

ΔTb = 3.03 °C

Hence;

Tb solution = ΔTb + Tb pure

Tb solution = 3.03 °C + 133.60°C

Tb solution = 136.63 °C