In: Chemistry
At 100°C, The molality of solute is 2.13 m. If Kb is 0.52 K/m, the boiling point of solution will be ________
(a) 102°C
(b) 103°C
(c) 101°C
(d) 100°C
Answer : c) 101 °C
Explanation :
Given :
Molality (m) = 2.13 m
Kb = 0.52 K/m
Elevation in boiling point of solution is depends upon the molality.
T∆ (boiling) = Kb × m
Where,
T∆ is temperature elevation.
Kb is boiling point constant (K/m)
Therefore,
T∆ = 0.52 × 2.13
= 1.105
Boiling point of solution is
B.Pt = 100 + 1.105
= 101.105 °C
= 101.1 °C
The boiling point of solution is 101 °C.
c) 101 °C