In basic medium CrO _ 4 ^ 2 - oxidises S _ 2 O _ 3 ^ 2 - to form SO _ 4 ^ 2 - and itself changes into Cr (OH) _ 4 ^ - .The volume of 0 . 154 MCrO _ 4 ^ 2 - required to react with 40mL of 0 . 25 MS _ 2 O _ 3 ^ 2 - is

Correct answer: 0 CrO _ 4 ^ 2 - + S _ 2 O _ 3 ^ 2 - → SO _ 4 ^ 2 - + Cr (OH) _ 4 ^ + 3 gm equi. of CrO _ 4 ^ 2 - = S _ 2 O _ 3 ^ 2 - 0 . 154 × 3 × v = 0 . 25 × 40 × 8 v = 173 . 16 = 173ml Hence answer is (173)

JEE Main 202125 FebruaryMorning Shift

Exam: JEE Main
Year: 2021
Date: 25 February
Shift: Morning Shift
Subject: Chemistry
Chapter: Redox Reactions

JEE Main 2021 · 25 February · Morning Shift · Chemistry · Redox Reactions

In basic medium CrO _ 4 ^ 2 - oxidises S _ 2 O _ 3 ^ 2 - to form SO...

In basic medium

{CrO}_{4}^{2 -}

oxidises

{S}_{2}{O}_{3}^{2 -}

to form

{{SO}_{4}}^{2 -}

and itself changes into

Cr{\left(OH\right)}_{4}^{-}

.The volume of

0 . 154{MCrO}_{4}^{2 -}

required to react with

40mL

0 . 25{MS}_{2}{{O}_{3}}^{2 -}

is _____

mL .

(Rounded-off to the nearest integer)

Correct answer

Step-by-step solution

{CrO}_{4}^{2 -} + {S}_{2}{O}_{3}^{2 -} \to {SO}_{4}^{2 -} + Cr{\left(OH\right)}_{4}^{+ 3} $gm equi. of$ {CrO}_{4}^{2 -} = {S}_{2}{O}_{3}^{2 -} $$ 0 . 154 \times 3 \times v = 0 . 25 \times 40 \times 8 $$ v = 173 . 16 = 173ml $Hence answer is \left(173\right)$

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