QJEEAdv20Ch59f2: JEE Advanced 2020 Chemistry - Class 12 JEEAdv20Ch59f2 Solution: Consider the reaction sequence from P to Q shown below. The…

Aniline (P) is treated with NaNO₂ and HCl in cold condition to form benzene diazonium chloride. The process of conversion of primary aromatic amines into its diazonium salt is called diazotisation.



$\beta$ -naphthol couples with phenyldiazonium electrophile to produce an intense orange-red dye (Q) as major product.



Given that, volume of aniline (P) = 9.3 mL (density of P = 1.00 g/mL)

So, mass of aniline = 9.3 g

Molecular mass of aniline (C₆H₇N) = 93 g/mol

Therefore, moles of aniline = 9.3 / 93 = 0.1 mol of P.

Molecular mass of 2 napthol aniline orange dye (Q) = 248 g/mol

$\Rightarrow$ 0.1 mol of aniline (P) will produce 0.1 mol of compound (Q).

But, according to the question the major product Q from P is 75%.

Therefore, mass of 'Q' produced

= (0.1 $\times$ 248 $\times$ 0.75)g = 18.60 g