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At` 25^@ `, the density of ` 15 M H_(2)SO_(4) `is ` 1.8 g cm^(-3) ` .Thus mass percentage of ` H_(2)SO_(4) ` in aqueous solution isA. ` 2% `B. ` 81.6% `C. ` 18% `D. ` 1.8% ` |
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Answer» Correct Answer - B `18 M H_(2)SO_(4)`means `18 mol L^(-1) 1000 mL of H_(2)SO_(4)` solution has`H_(2)SO_(4) = 18 mol =15xx98 g` Also, `1000 mL = 1000 xx 1.8 g H_(2)SO_(4)` solution thus ,`(1000xx1.8)g H_(2)SO_(4)` solution has `H_(2)SO_(4) = 15xx98 g` `100 g H_(2)SO_(4)` solution has `H_(2)SO_(4) = (15xx98)/(1000xx1.8)=(1470)/(1800)=81.6%` Thus , mass percentage of` H_(2)SO_(4) = 81.6%` |
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