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QJEEM22Ch8ae3: JEE Main 2022 Chemistry - Class 11 JEEM22Ch8ae3 Solution: 1 L aqueous solution of H 2 SO 4 contains 0.02 m mol H 2 SO 4 . …

1 L aqueous solution of H2SO4 contains 0.02 m mol H2SO4. 50% of this solution is diluted with deionized water to give 1 L solution (A). In solution (A), 0.01 m mol of H2SO4 are added. Total m mols of H2SO4 in the final solution is ___________ × 103 m mols.

Step-by-Step Verified Solution

nH2SO4\mathrm{n}_{\mathrm{H}_2 \mathrm{SO}_4} in SolnA=50%\mathrm{Sol}^{\mathrm{n}} \mathrm{A}=50 \% of original solution

=0.01 m mol=0.01 \mathrm{~m} \mathrm{~mol} .

nH2SO4\mathrm{n}_{\mathrm{H}_2 \mathrm{SO}_4} in Final solution =0.01+0.01=0.01+0.01

=0.02 m mol=0.02 \,\mathrm{m\,mol}

=0.00002×103m mol=0.00002 \times 10^3 \mathrm{m\,mol}

The answer 0

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