Test Index

Amines Part 3

Section: Chemistry

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Question : 19 of 48

Marks: +1, -0

Assuming that the reaction proceeds to completion, then 137 mg of 4-nitrotoluene will produce

\_\_\_\_

B

(Given molar mass in

\mathrm{g\,mol}^{-1}\ \mathrm{H}: 1,\ \mathrm{C}: 12,\ \mathrm{N}: 14,\ \mathrm{O}: 16,\ \mathrm{Br}: 80

[23 Jan 2026 Shift 1]

208
228
146
301

Solution:

Concept:

The reaction goes to completion, so the number of moles of reactant equals the number of moles of product.

Explanation:

Molar mass of 4‑nitrotoluene is

137\ \mathrm{g\,mol^{-1}}

.

Given mass is

137\ \mathrm{mg}

, which is

0.137\ \mathrm{g}

.

Moles of 4‑nitrotoluene =

\frac{0.137\ \mathrm{g}}{137\ \mathrm{g\,mol^{-1}}} = 0.001\ \mathrm{mol}

.

Moles of product B formed =

0.001\ \mathrm{mol}

(1:1 ratio).

Molar mass of B is

228\ \mathrm{g\,mol^{-1}}

.

Mass of B =

0.001\ \mathrm{mol} \times 228\ \mathrm{g\,mol^{-1}} = 0.228\ \mathrm{g}

.

Convert to mg:

0.228\ \mathrm{g} = 228\ \mathrm{mg}

.

Answer:

228\ \mathrm{mg}

(Option B).

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