9.3 g of pure aniline upon diazotisation fol | vedclass

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Question

(A) The reaction sequence is as follows:Aniline $(C_6H_5NH_2)$ $\xrightarrow{NaNO_2 + HCl, T < 5^{\circ}C}$ Benzenediazonium chloride $(C_6H_5N_2^+Cl^

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(A) The reaction sequence is as follows:Aniline $(C_6H_5NH_2)$ $\xrightarrow{NaNO_2 + HCl, T Benzenediazonium chloride + Phenol $(C_6H_5OH)$ $ightarrow$ $p$-Hydroxyazobenzene (Orange dye,$C_{12}H_{10}N_2O$)From the stoichiometry,$1 	ext{ mole}$ of aniline produces $1 	ext{ mole}$ of orange dye.Molar mass of aniline $(C_6H_5NH_2)$ = $93 \ g \ mol^{-1}$.Moles of aniline = $\frac{9.3 \ g}{93 \ g \ mol^{-1}} = 0.1 \ mol$.Molar mass of orange dye ($p$-Hydroxyazobenzene,$C_{12}H_{10}N_2O$) = $(12 	imes 12) + (10 	imes 1) + (2 	imes 14) + 16 = 144 + 10 + 28 + 16 = 198 \ g \ mol^{-1}$.Since $1 	ext{ mole}$ of aniline gives $1 	ext{ mole}$ of dye,$0.1 	ext{ mole}$ of aniline will produce $0.1 	ext{ mole}$ of dye.Mass of orange dye = $0.1 \ mol 	imes 198 \ g \ mol^{-1} = 19.8 \ g$.Rounding to the nearest integer,we get $20 \ g$.

$9.3 \ g$ of pure aniline upon diazotisation followed by coupling with phenol gives an orange dye. The mass of orange dye produced (assume $100\%$ yield/conversion) is . . . . . $g$. (nearest integer)

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