The molal boiling point constant for water is 0.52°c/m. at what temperature will a mixture of 45.0 g of nacl and 0.500 kg of water boil?

Temperature of boiling point for solution is 101.60°C

Explanation:

This colligative property is boiling point elevation.

ΔT = Kb . m . i

ΔT = T° boiling point for solution - T° boiling point for pure solvent

Kb = The molal boiling point constant

m = molality

i = Van't Hoff factor ( for NaCl i = 2)

NaCl → Na⁺  +  Cl⁻

Let's calculate molality (mol of solute in 1kg of solvent)

Mol of salt = Salt mass / Salt Molar mass

Mol of salt = 45 g / 58.45 g/m → 0.769 moles

0.769 moles/0.5 kg = 1.54 m

T° boiling point for solution - 100°C = 0.52°C/m . 1.54m . 2

T° boiling point for solution = (0.52°C/m . 1.54m . 2) + 100°C

T° boiling point for solution =  101.60°C