The molality of 1 L solution with x% H(2...

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 100mL is:

A

90

B

80.3

C

40.13

D

9

Verified by Experts

The correct Answer is:

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`

Similar Questions

Explore conceptually related problems

The weight of solute present in 200 ml of 0.1 M H_(2)SO_(4)

One litre of 16 M H_2SO_4 has been diluted to 100L. The normality of the resulting solution is :

Calculate molality of 1 litre solution of 93% H_(2) SO_(4) by volume. The density of solution is 1.84 g mL^(-1) .

Calculate the normality of NaOH when 2 g is present in 800 mL solution.

The number of H^(+) ions present in 1 ml of a solution whose pH is 13 is :

12 g of H_(2)SO_(4) are dissolved in water to make 1200 ml of solution. The normality of the solution is