1 L aqueous solution of H 2 SO 4 contains 0.02 m mol H 2 SO 4 .50 % of this solution is diluted with deionized water to give 1 L solution (A). In solution (A), 0.01 m mol of H 2 SO 4 are added. Total m mols of H 2 SO 4 in the final solution is × 103 m mols.

Question

1 L aqueous solution of H 2 SO 4 contains 0.02 m mol H 2 SO 4 .50 % of this solution is diluted with deionized water to give 1 L solution (A). In solution (A), 0.01 m mol of H 2 SO 4 are added. Total m mols of H 2 SO 4 in the final solution is × 103 m mols. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

n H 2 SO 4 in Sol n A =50 % of original solution =0.01 m mol. n H 2 SO 4 in Final solution =0.01+0.01 =0.02 mmol =0.00002 × 103 mmol The answer 0

Q. 1L aqueous solution of H2​SO4​ contains 0.02 mmolH2​SO4​.50% of this solution is diluted with deionized water to give 1L solution (A). In solution (A), 0.01m mol of H2​SO4​ are added. Total m mols of H2​SO4​ in the final solution is _________×103mmols.

nH2​SO4​​ in SolnA=50% of original solution =0.01mmol. nH2​SO4​​ in Final solution =0.01+0.01 =0.02mmol =0.00002×103mmol The answer 0

$n_{H_{2} S O_{4}}$ in $S o l^{n} A = 50%$ of original solution $= 0.01 m m o l$ .
$n_{H_{2} S O_{4}}$ in Final solution $= 0.01 + 0.01$
$= 0.02 mm o l$
$= 0.00002 \times 1 0^{3} mm o l$
The answer $0$