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JEE Main 202228 JulyEvening Shift

Exam: JEE Main
Year: 2022
Date: 28 July
Shift: Evening Shift
Subject: Chemistry
Chapter: Some Basic Concepts of Chemistry

JEE Main 2022 · 28 July · Evening Shift · Chemistry · Some Basic Concepts of Chemistry

2L of 0 . 2 MH _ 2 SO _ 4 is reacted with 2L of 0 . 1MNaOH...

2L

of

0 . 2{MH}_{2}{SO}_{4}

is reacted with

2L

of

0 . 1MNaOH

solution, the molarity of the resulting product

{Na}_{2}{SO}_{4}

in the solution is____millimolar.

Correct answer

0

Step-by-step solution

${H}_{2}{SO}_{4} + 2NaOH \to {Na}_{2}{SO}_{4} + 2{H}_{2}O$
$\begin{cases}0 . 4mole & 0 . 2mole & - \\ 0 . 3mol & - & 0 . 1mol\end{cases}$
Molarity = \dfrac{Numberofmoles}{Volumeinlitres} $</p><p>Molarity of$ {Na}_{2}{SO}_{4} $is$ \dfrac{0 . 1}{4} = 0 . 025M $</p><p>$ = 25 \times {10}^{- 3}M $</p><p>Molarity of the resulting product$ {Na}_{2}{SO}_{4} $in the solution is$ 25 millimoles

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