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A sample of `H_(2)SO_(4)` `("density" 1.787 g mL^(-1))` is labelleed as `80%` by weight. What is molarity of acid? What volume of acid has to be used to make `1"litre of" 0.2MH_(2)SO_(4)`? |
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Answer» `H_(2)SO_(4)` is `80%` by weight `:. Wt. of H_(2)SO_(4)= 80 g` W.t. of solution `= 100 g` `:.` Volume of solution `=(100)/(1.787)mL` `=(100)/(1.787xx1000)"litre"` `:. M_(H_(2)SO_(4))={(80)/(98xx(100)/(1.787xx1000))}=14.59` Let `V mL` of this `H_(2)SO_(4)` are used to prepare `1 "litre"` of `0.2MH_(2)SO_(4)`, then `mM` of conc. `H_(2)SO_(4)=mM` of dil. `H_(2)SO_(4)` (`mM` does not change on dilution) `Vxx14.59= 1000xx0.2` `V= 13.71 mL` |
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