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25 ml of a solution of `Na_(2)CO_(3)` having a specific gravity of 1.25 required 32.9 ml of a solution of HCl containing 109.5 grams of the acid per litre for complete neutralization. Calculate the volume of 0.84N-`H_(2)SO_(4)` that will be completely neutralized by 125 grams of the `Na_(2)CO_(3)` solutionA. 460 mlB. 540 mlC. 480 mlD. 470 ml |
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Answer» Correct Answer - D `N_(1)V_(1)=N_(2)V_(2)` `Nxx25=(109.25xx32.9)/(36.5)impliesN=(109.5xx32.9)/(36.5xx25)` `N_(3)V_(3)=N_(4)V_(4)" "(V_(3)=(m)/(d),V_(3)=(125)/(1.25))` `(109.5xx32.9)/(36.5xx25)xx100=0.84xxV impliesV=470ml` |
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