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If `3*01xx10^(20)` molecules are removed from 98 mg of `H_(2)SO_(4)`, then the number of moles of `H_(2)SO_(4)` left areA. `0*1xx10^(-3)`B. `0*5xx10^(-3)`C. `1*66xx10^(-3)`D. `9*95xx10^(-2)` |
Answer» Correct Answer - B Moles of `H_(2)SO_(4)` removed `= (3*01xx10^(20))/(6*02xx10^(23))=0*5xx10^(-3)` Total moles of `H_(2)SO_(4) = (98xx10^(-3))/(98) = 1xx10^(-3)` Moles of `H_(2)SO_(4)` left `=1xx10^(-3)-0*5xx10^(-3)` `=0*5xx10^(-3)` moles |
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