Question

A chemist dissolves 0.400 mol benzene in 800 grams of Substance X, then finds the boiling point of Substance X has increased by 1.4°C. What is the
molal boiling point elevation constant (Kb) for Substance X?
A) 3.2°C/m
B) 0.70°C/m
C) 2.8°C/m
D) 1.40cm

Answer 1

Answer:

Explanation:

The  molal boiling point elevation constant (Kb) of a substance is defined as change in boiling point / its molality.

For substance X, the change in boiling point = 1.4°C

Its molality = 0.4 mol / 800 g = 0.4 mol / 0.8 kg = 0.5 m

So Kb = 1.4 / 0.5 = 2.8°C/m

The answer is C).

Answer 2

Answer:

Explanation:

molality=mol/mass=0.4/0.8=0.5mol/kg

molal boiling point elevation constant (Kb) for Substance X

= delta bp/molality

= 1.4/0.5

= 2.8°C/m