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The molarity of 98% by wt. `H_(2)SO_(4)` (d = 1.8 g/ml) isA. 6MB. 18MC. 10MD. 4M |
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Answer» Correct Answer - B `98% H_(2)SO_(4)` by weight `rArr 98 g` of `H_(2)SO_(4)` in 100 g of solution volume of solution `=(100g)/(1.8 g mL^(-1))=55.55 mL` `=0.056 L` Moles of `H_(2)SO_(4)=98//98=1 mol` Molarity `=(1 mol)/(0.056L) = 17.86 M` |
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