X g of nitrobenzene on nitration gave 4.2 g of m-dinitrobenzene. X = ______ g. (nearest integer) [Given: molar mass (in g ) C : 12, H : 1, O : 16, N : 14] [2025]

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Solution

Q.
X g of nitrobenzene on nitration gave 4.2 g of m-dinitrobenzene.

X = ______ g. (nearest integer)

[Given: molar mass (in g ${mol}^{- 1}$ ) C : 12, H : 1, O : 16, N : 14] [2025]

Ans.
(3)

$Moles of m -dinitrobenzene = \frac{Mass of m -dinitrobenzene}{Molar mass of m -dinitrobenzene}$

$= \frac{4.2}{168} mol = 0.025 mol$

$As per reaction stoichiometry, moles of nitrobenzene = moles of m -dinitrobenzene = 0.025 mol$

$Mass of nitrobenzene = moles of nitrobenzene \times molar mass of nitrobenzene$

$= 0.025 \times 123 g = 3.075 g$

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