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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. |
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Answer» \(n_{H_2SO_4}\) in Soln A = 50% of original solution = 0.01 m mol \(n_{H_2SO_4}\) in Final solution = 0.01 + 0.01 = 0.02 mmol = 0.00002 x 103 mmol The answer is 0 |
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