Calculate the volume of 80% (by mass) of `H_2SO_4` (density = 1.80 g/mL] required to prepare 1 litre of 0.2 molar `H_2SO_4` [Relative atomic masses : H = 1, O = 16, S = 32]

Explanation

Applying the following relation and substituting the values, we get
Molarity = `("Percentage by mass x10 x d")/("molar mass") = (80 xx 10 xx 1.8)/(98) = 1440/98 = 14.69 "mol" L^(-1)`
Using molarity equation
`M_1 V_1 = M_2V_2`
`14.69 xx V_1= 0.2 xx 1000 mL`
` V_1 = (200)/(14.69) = 13.69 mL`
Thus volume of `H_2SO_4` required = 13.6 mL