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(A) Molecular weight of aniline $(P)$ $= C_6H_7N = 6 	imes 12 + 7 	imes 1 + 14 = 93 \ g \ mol^{-1}$.Density of $P = 1.00 \ g \ mL^{-1}$,so mass of $

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$18.60$

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(A) Molecular weight of aniline $(P)$ $= C_6H_7N = 6 	imes 12 + 7 	imes 1 + 14 = 93 \ g \ mol^{-1}$.Density of $P = 1.00 \ g \ mL^{-1}$,so mass of $9.3 \ mL$ of $P = 9.3 \ g$.Moles of $P = \frac{9.3 \ g}{93 \ g \ mol^{-1}} = 0.1 \ mol$.The reaction is a coupling reaction where $1 \ mol$ of aniline $(P)$ produces $1 \ mol$ of the azo dye $(Q)$.Given the overall yield is $75 \%$,the actual moles of $Q$ formed $= 0.1 	imes 0.75 = 0.075 \ mol$.The molecular formula of $Q$ is $C_{16}H_{12}N_2O$.Molar mass of $Q = 16 	imes 12 + 12 	imes 1 + 2 	imes 14 + 16 = 192 + 12 + 28 + 16 = 248 \ g \ mol^{-1}$.Amount of $Q$ in grams $= 0.075 \ mol 	imes 248 \ g \ mol^{-1} = 18.6 \ g$.

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