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` in g per 100 mL is

Options

A. `90`
B. `80.3`
C. `30.38`
D. `46.87`

Answer

D

Explanation

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` .