SO _ 2 Cl _ 2 on reaction with excess of water results into acidic mixture SO _ 2 Cl _ 2 + 2 H _ 2 O → H _ 2 SO _ 4 + 2HCl 16 moles of NaOH is required for the complete neutralisation of the resultant acidic mixture.

Correct answer: C. 4 SO _ 2 Cl _ 2 + 2 H _ 2 O → 2HCl + H _ 2 SO _ 4 Let x moles of SO _ 2 Cl _ 2 are required, ∴ n( H _ 2 SO _ 4 ) = x, n(HCl) = 2x ⇒ n( H ^ + ) = 4x Moles of H ^ + formed = 4x Moles of OH ^ - from 16 moles of NaOH = 16 4x = 16 (for complete neutralisation) x = 4

JEE Main 202225 JulyMorning Shift

Exam: JEE Main
Year: 2022
Date: 25 July
Shift: Morning Shift
Subject: Chemistry
Chapter: Some Basic Concepts of Chemistry

JEE Main 2022 · 25 July · Morning Shift · Chemistry · Some Basic Concepts of Chemistry

SO _ 2 Cl _ 2 on reaction with excess of water results into acidic...

{SO}_{2}{Cl}_{2}

on reaction with excess of water results into acidic mixture

{SO}_{2}{Cl}_{2} + 2{H}_{2}O \to {H}_{2}{SO}_{4} + 2HCl

16

moles of

NaOH

is required for the complete neutralisation of the resultant acidic mixture. The number of moles of

{SO}_{2}{Cl}_{2}

used is

Options

A $16$
B $8$
C $4$
D $2$

Correct answer

C. $4$

Step-by-step solution

${SO}_{2}{Cl}_{2} + 2{H}_{2}O \to 2HCl + {H}_{2}{SO}_{4}$
Let $x$ moles of ${SO}_{2}{Cl}_{2}$ are required,
$∴ n\left({H}_{2}{SO}_{4}\right) = x, n\left(HCl\right) = 2x$
$⇒ n\left({H}^{+}\right) = 4x$
Moles of ${H}^{+}$ formed $= 4x$
Moles of ${OH}^{-}$ from $16$ moles of $NaOH = 16$
$4x = 16$ (for complete neutralisation)
$x = 4$

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