The boiling point of an aqueous solution with NaCl is 101.02 ∘c. Calculate the molality concentration of the solution

Question

pls help me answering this im quite confused

Answer 1

Colligative properties:

Boiling point elevation.

∆T = imK

∆T = change in boiling point = 1.02º since normal b.p. is 100ºC

i = van't Hoff factor = 1 for NaCl since it dissociates into 2 particles, Na+ and Cl-

m = molality

K = boiling constant for H2O = 0.512º/m

Solving for m...

m = ∆T /(i)(K) = 1.02 / (2)(0.512)

m = 0.996 m