Question

117 g of NaCl is added to 222 g of water in a saucepan. At what does temperature does water boil at 101.325 kPa? Ebullioscopy constant for water = 0.52 K kg mol^-1 and b.p. = 100°C

(a) 98.3°C

(b) 102.8°C

(c) 104.7°C

(d) 101.5°C

This question was posed to me in a national level competition.

My query is from Colligative Properties and Determination of Molar Mass topic in chapter Solutions of Chemistry – Class 12

Answer 1

The correct option is (c) 104.7°C

To elaborate: Given,

Weight of solvent, w1 = 222 g

Weight of solute, w2 = 117 g

Kb = 0.53 K kg mol^-1

Now, addition of a non-volatile solute causes elevation in boiling point, ΔTb

ΔTb = (kb x 1000 x w2)/(M2 x w1)

On substituting, ΔTb = (0.52 x 1000 x 117)/(58.5 x 222) = 4.7°C

New boiling temperature = 100 + 4.7 = 104.7°C.