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The mass of 70% `H_(2)SO_(4)` required for neutralization of one mole of NaOH is:A. 49 gB. 98 gC. 70 gD. `34*3 g` |
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Answer» Correct Answer - C `2NaOH + H_(2)SO_(4) to Na_(2)SO_(4)+ H_(2)O` Pure `H_(2)SO_(4)` required for 2 mole of NaOH `=(49xx100)/(70) = 70g` |
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