JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Solutions

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A B_{2}

10%

dissociated in water to

A^{2 +}

and

B^{-}

. The boiling point of a 10.0 molal aqueous solution of

AB_{2}

is ____

^{\circ} C

. (Round off to the nearest Integer).
[Given : Molal elevation constant of water

K_{b} = 0.5 K kg mol^{- 1}

, boiling point of pure water

= 10 0^{\circ} C

]

Answer: 106

Solution

Each formula unit that breaks up yields three ions, so with a degree of dissociation

α

the van't Hoff factor is

i = 1 - α + 3 α = 1 + 2 α = 1 + 2 (0.10) = 1.20

The elevation of boiling point is then

Δ T_{b} = i K_{b} m = 1.20 \times 0.5 \times 10.0 = 6.0 K

Added to the boiling point of pure water, the solution boils at 106

^{\circ}

JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Integer answer

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