A piece of `Al` wieghing `2.7 g` is titrated with `75.0 mL` of `H_(2) SO_(4)` (specific gravity `1.8 mL^(-1)` and 24.7% `H_(2) SO_(4)` by weight). After the metal is completely dissolved, the solution is diluted to `400 mL`. Calculate the molarity of free `H_(2) SO_(4)` solution.

Answer

A, C

Explanation

mEq of `Al =` mEq of `H_(2) SO_(4)` reacted
mEq of `Al = (2.7)/(9) xx 1000 = 3000`
`(Ew of Al = (Mw)/(3) = (27)/(3) = 9 g)`
`(Al rarr Al^(3+) + 3e^(ɵ))`
`N of H_(2) SO_(4) = (% "by weight" xx 10 xx d)/(Ew_(2)) = (24.7 xx 10 xx 1.18)/(98//2)`
`= 5.95`
mEq of `H_(2) SO_(4) = N xx` Volume in `mL = 5.95 xx 75 = 446.25`
mEq of `Al` added = 300
mEq of `H_(2)SO_(4)` left after reaction `= 446.25 - 300 = 146.25`
Solution is diluted to `400 mL`
`:. N_(H_(2)SO_(4))` left `= (mEq)/(mL) = (146.25)/(400) = 0.0367`
`M_(H_(2)SO_(4))` left `= (N)/(n) = (0.0367)/(2) = 0.183`