X g of nitrobenzene on nitration gave 4.2 g | vedclass

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Question

(A) The chemical reaction is: $C_6H_5NO_2 + HNO_3 \xrightarrow{H_2SO_4} C_6H_4(NO_2)_2 + H_2O$ Molar mass of nitrobenzene $(C_6H_5NO_2)$ $= (6 	imes

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$3$

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(A) The chemical reaction is: $C_6H_5NO_2 + HNO_3 \xrightarrow{H_2SO_4} C_6H_4(NO_2)_2 + H_2O$ Molar mass of nitrobenzene $(C_6H_5NO_2)$ $= (6 	imes 12) + (5 	imes 1) + 14 + (2 	imes 16) = 123 \ g \ mol^{-1}$ Molar mass of $m-$dinitrobenzene $(C_6H_4N_2O_4)$ $= (6 	imes 12) + (4 	imes 1) + (2 	imes 14) + (4 	imes 16) = 168 \ g \ mol^{-1}$ Moles of $m-$dinitrobenzene produced $= \frac{4.2 \ g}{168 \ g \ mol^{-1}} = 0.025 \ mol$ Since $1 \ mol$ of nitrobenzene produces $1 \ mol$ of $m-$dinitrobenzene,moles of nitrobenzene required $= 0.025 \ mol$ Mass of nitrobenzene $(X)$ $= 0.025 \ mol 	imes 123 \ g \ mol^{-1} = 3.075 \ g$ The nearest integer is $3$.

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

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