What volume of 95% `H_(2) SO_(4)` by weight `(d = 1.85 g mL^(-1))` and what mass of water must be taken to prepare `100 mL` of 15% solution of `H_(2) SO_(4)` `(d = 1.10 g mL^(-1))`

Answer

`9.4 cm^(3), 96.6 g`

Explanation

95 % `H_(2)SO_(4)` means that 95 g of `H_(2)SO_(4)` is present in 100 g of solution .
Vol. of solution `= (100)/(1.85) = 54.05`
Molarity `= (95 xx 1000)/(98 xx 54.05) = 17.93 M`
Similarly , 15% `H_(2)SO_(4)` means 15 g of `H_(2)SO_(4)` is present in 100 g of solution.
Vol. of solution `= (100)/(1.10) = 90.91`
Molarity ` = (15 xx 1000)/(98 xx 90.91) = 1.68 M`
Applying molarity equaiton.
`M_(1)V_(1) = M_(2)V_(2)`
`(95% H_(2)SO_(4)) (15% H_(2)SO_(4))`
`17.93 xx V_(1) = 1.68 xx 100`
`therefore " " V_(1) = (1.68 xx 100)/(17.03) = 9.37 cm^(3) = 9.4 cm^(3)`
`therefore ` Volume of 95% `H_(2)SO_(4)` required ` = 9.4 cm^(3)`
Mass of `100 cm^(3)` of `15% H_(2)SO_(4)` to be prepared .
` = 100 xx 1.10 = 110g `
Mass of `9.4 cm^(3) ` of 95 `H_(2)SO_(4)`
`therefore` Mass of water to be taken ` = 110-17.4 = 92.6 g`.