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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. `90`B. `80.3`C. `30.38`D. `46.87` |
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Answer» Correct Answer - D Molality of `H_(2)SO_(4)` is 9 i.e. 9 mole of `H_(2)SO_(4)` in 1kg solvent 1kg solvent contain = 9 mole `H_(2)SO_(4)` 1kg solvent contain `= 9 xx 98 "wt" H_(2)SO_(4)` 1000kg solvent contain `= 9 xx 98//1000 xx 910` 910kg solvent contain `=802.62g` wt. of solution `=910 kg` wt. of solution `802.62+910=1712.62g` x `%` by wt `=("wt of solvent")/("wt of solution")xx100` `=(802.62)/(1712.62) xx 100=46.87` wt of solution `= 802.62 + 910` `=1712.62g` `x%` by wt `x%` ` = ("wt of solute")/("wt of solution") xx 100` `= (802.62)/(1712.62) xx100 = 46.87` . |
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