Question

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

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

Solutions

Expert Solution

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

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