JEE MAIN - Chemistry (2021 - 17th March Evening Shift - No. 19)
A 1 molal K4Fe(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
Explanation
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Effective molality = 0.6 + 1.6 + 0.4 = 2.6 m
As elevation in boiling point is a colligative property which depends on the amount of solute. So, to have same boiling point, the molality of two solutions should be same.
Molality of non-electrolyte solution = molality of $${K_4}[Fe{(CN)_6}]$$ = 2.6 m
Now, 18.1 weight per cent solution means 18.1 g solute is present in 100 g solution and hence, (100 $$-$$ 18.1) = 81.9 g water.
$$Molality = {{(Mass\,of\,solute/Molar\,mass\,of\,solute)} \over {Mass\,of\,solvent\,in\,kg}} \times 1000$$
Now, $$2.6 = {{18.1/M} \over {81.9/1000}}$$
where, M is the molar mass of non-electrolyte solute
Molar mass of solute, M = 85
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