JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Solutions

A 1 molal

K_{4} Fe (CN)_{6}

solution has a degree of dissociation of 0.4. Its boiling point is equal to that of another solution which contains 18.1 weight percent of a non electrolytic solute A. The molar mass of

A

is ______ u. (Round off to the nearest integer).
[Density of water

= 1.0 g cm^{- 3}

]

Answer: 85

Solution

Potassium ferrocyanide dissociates into five ions, four

K^{+}

and one

[Fe (CN)_{6}]^{4 -}

, so with a degree of dissociation of

0.4

the van't Hoff factor is

i = 1 + α (n - 1) = 1 + 0.4 (5 - 1) = 2.6

and its elevation of boiling point is

Δ T_{b} = i K_{b} m = 2.6 K_{b}

for the one molal solution. In

100 g

of the second solution there are

18.1 g

of the non-electrolyte and

81.9 g

of water, so its molality is

\frac{18.1}{0.0819 M}

and its elevation is

K_{b} \frac{18.1}{0.0819 M}

. Equal boiling points mean equal elevations:

2.6 = \frac{18.1}{0.0819 M}

M = \frac{18.1}{2.6 \times 0.0819} = 85.0

The molar mass of the non-electrolytic solute is

85 u

JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Integer answer

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JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Solutions

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