The normality of H _ 2 SO _ 4 in the solution obtained on mixing 100mL of 0 . 1M H _ 2 SO _ 4 with 50mL of 0 . 1MNaOH is____ × 10 ^ - 1 N .

Correct answer: 0 No. of equivalents of H _ 2 SO _ 4 = M × V × n - factor = 100 × 0 . 1 × 2 = 20 No. of equivalents of NaOH = 50 × 0 . 1 = 5 No. of equivalents of H _ 2 SO _ 4 left = 20 - 5 = 15 ⇒ 150 × x = 15 x = (1)/(10) = 0 . 1N = 1 × 10 ^ - 1 N

JEE Main 202227 JulyEvening Shift

Exam: JEE Main
Year: 2022
Date: 27 July
Shift: Evening Shift
Subject: Chemistry
Chapter: Redox Reactions

JEE Main 2022 · 27 July · Evening Shift · Chemistry · Redox Reactions

The normality of H _ 2 SO _ 4 in the solution obtained on mixing...

The normality of

{H}_{2}{SO}_{4}

in the solution obtained on mixing

100mL

0 . 1M{H}_{2}{SO}_{4}

with

50mL

0 . 1MNaOH

is____

\times {10}^{- 1}N

Correct answer

Step-by-step solution

No. of equivalents of ${H}_{2}{SO}_{4} = M \times V \times n - factor = 100 \times 0 . 1 \times 2 = 20$
No. of equivalents of $NaOH = 50 \times 0 . 1 = 5$
No. of equivalents of ${H}_{2}{SO}_{4}$ left $= 20 - 5 = 15$
$⇒ 150 \times x = 15$
$x = \dfrac{1}{10} = 0 . 1N = 1 \times {10}^{- 1}N$

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