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A `15mL` sample of `0.20 M" " MgCl_(2)` is added to `45ML` of `0.40 M AlCl_(3)` What is the molarity of `Cl` ions in the final solutionA. `1.0M`B. `0.60M`C. `0.35M`D. `0.30M` |
Answer» Correct Answer - A `"Molarity of" Cl^(-)` `=(M_(1)V_(1)+M_(2)V_(2))/("Total vol")` `=(15 xx .2 xx 2 + 45 xx .45 xx 3)/(15 + 45) = (60)/(60) = 1M` . |
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