If `100 mL "of" 1 M H_(2) SO_(4)` solution is mixed with `100 mL` of 98% `(W//W)` of `H_(2)SO_(4)` solution `(d = 0.1 g mL^(-1))`, then

Options

A. Concentration of solution becomes half.
B. Volume of solution beomces `200 mL`.
C. Mass of `H_(2)SO_(4)` is the solution is `98 g`
D. mass of `H_(2)SO_(4)` in the solution is `19.6 g`.

Answer

B, D

Explanation

`100 mL "of" 1 M H_(2) SO_(4) + 100 mL (98%, d = 0.1) H_(2) SO_(4)`
` implies 100 xx 1 M + 100 mL xx ((98 xx 10 xx 0.1)/(98))`
`implies 100 xx 1 M + 100 xx 1 M`
`:. [H_(2) SO_(4)] = (100 xx 1 + 100 xx 1)/(200 mL) = 1 M = 98 g // 1000 mL`
Mass of `H_(2) SO_(4) = (98 g xx 200 mL)/(1000 mL) = 19.6 g`
concentration of each component becomes half ot the initial value.