JEE MAIN - Chemistry (2022 - 25th June Morning Shift - No. 18)
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 ___________ $$\times$$ 103 m mols.
Answer
0
Explanation
$\mathrm{n}_{\mathrm{H}_2 \mathrm{SO}_4}$ in $\mathrm{Sol}^{\mathrm{n}} \mathrm{A}=50 \%$ of original solution
$=0.01 \mathrm{~m} \mathrm{~mol}$.
$\mathrm{n}_{\mathrm{H}_2 \mathrm{SO}_4}$ in Final solution $=0.01+0.01$
$=0.02 \,\mathrm{m\,mol}$
$=0.00002 \times 10^3 \mathrm{m\,mol}$
The answer 0
$=0.01 \mathrm{~m} \mathrm{~mol}$.
$\mathrm{n}_{\mathrm{H}_2 \mathrm{SO}_4}$ in Final solution $=0.01+0.01$
$=0.02 \,\mathrm{m\,mol}$
$=0.00002 \times 10^3 \mathrm{m\,mol}$
The answer 0
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