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

Options

A. 9.42
B. 10.42
C. 11.42
D. 12.42

Answer

B

Explanation

Given `H_2SO_4` is 93% by volume
wt. of `H_2SO_4` = 93g
Volume of solution = 100ml
`:.` weight of solution = 100`xx` 1.84 gm = 184 gm
wt. of water = 184 - 93 = 91 gm
`"Molarity"="Moles"/("wt. of water in kg")=(93xx1000)/(98xx91)=10.42`