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Find the molality of `H_(2)SO_(4)` solution whose specific gravity is `1.98 mL^(-1)` and `93%` by volume `H_(2)SO_(4)` |
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Answer» Correct Answer - 9 `H_(2)SO_(4)` is `93%` by volume `:.` weight of `H_(2)SO_(4) = 93g` and volume of solution `= 100mL` `:.` wt.of solution `= 1.98 xx 100 = 198 g` `:.` wt. of water = wt. of solution -wt.of `H_(2)SO_(4)` `193 - 93 = 105g` Now, molarity `= ("Moles of" H_(2)SO_(4))/("Wt.of water"("in" kg))` |
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