Calculate the molality of 90% `H_2SO_4` (weight/volume). The density of solution is `1.80 g mL^(-1)`.

Options

A. 91.8
B. 9.18
C. 46.6
D. 23.4

Answer

A

Explanation

Suppose, we have 1 litre of the given solution. Mass of 1L of the given solution = density x volume = `1.80 xx 1000= 1800` g Amount of `H_2SO_4` present in it
`=(1800 xx 90)/100 = 1620 g`
`therefore` Mass of water present =1800 - 1620 = 180 g
`therefore w =(m xx M. xx W)/1000`
`therefore m =(w xx 1000)/(M. xx W) = (1620 xx 1000)/(98 xx 180) = 91.8 m`