QJEEM21Chf238: JEE Main 2021 Chemistry - Class 12 JEEM21Chf238 Solution: AB 2 is 10% dissociated in water to A 2+ and B − . The boiling…

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AB₂ is 10% dissociated in water to A²⁺ and B^$-$. The boiling point of a 10.0 molal aqueous solution of AB₂ 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 = 100

^{\circ}

C]

Accepted Answer

AB₂

⇋

A⁺ + 2B^$-$

∴

For AB₂, n = 3

i = 1 + (n

-

1)

α

= 1 + (3

-

1)

\times

0.1

= 1.2

Now,

Δ

T_b = K_b (im)

\Rightarrow

T_b

-

T

_{b}^{o}

= 1.2

\times

0.5

\times

10

\Rightarrow

T_b

-

100 = 6

\Rightarrow

T_b = 106

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