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The molality of `1 L` solution with `x% H_(2)SO_(4)` is equal to 9. The weight of the solvent present in the solution is `910 g`. The value of `x` is:A. 90B. 80.3C. 40.13D. 9 |
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Answer» Correct Answer - B Molality `= (1000 xx "Weight of solute")/(Mw xx "Weight of solven")` `9 = (1000 xx W)/(98 xx 910)` `W = 802.6 gL^(-1)` `= 80.26 g` per `100 mL` |
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