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`V_(1) mL` of `NaOH` of normality `X` and `V_(2) mL` of `Ba(OH)_(2)` of mormality `Y` are mixed together. The mixture is completely neutralised by `100 mL` of `0.1 N HCl`. If `V_(1)//V_(2)=1/4` and `X/Y=4`, what fraction of the acid is neutralised by `Ba(OH)_(2)`?A. `0.5`B. `0.25`C. `0.33`D. `0.67` |
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Answer» Correct Answer - A `"Meq.of"NaOH+Meq.of Ba(OH)_(2)=Meq.of HCl` `XV_(1)+YV_(2)=100xx0.1=10` `4Yxx(V_(2))/(4)+YV_(2)=10` `2YV_(2)=10` `:. V_(2)Y=5` `:. V_(1)X=5` `:.` Fraction of acid by `Ba(OH)_(2)` `=(5)/(10)=0.5` |
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