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JEE Advanced levelSingle correct optionDifficulty 3 of 3qc217

A solution containing $0.535 g$ of $NH X_{4} Cl$ is boiled with excess NaOH, and the ammonia evolved is absorbed completely in excess Nessler's reagent. The brown product is $HgO \cdot Hg (NH X_{2}) I$ (molar mass $559 g/mol$ ; $NH X_{4} Cl$ is $53.5 g/mol$ , one product unit per N). The mass of brown solid is

A $11.2 g$
B $2.80 g$
C $0.559 g$
D $5.59 g$

Answer: D

Moles of

NH X_{4} Cl

=

0.535/53.5 = 0.0100

, giving 0.0100 mol

NH X_{3}

and so 0.0100 mol of the iodide of Millon's base. Mass =

0.0100 \times 559 = 5.59 g

.

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