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 remains same
B. Volume of solution becomes 200ml
C. mass of `H_(2)SO_(4)` in the solution is 98gm
D. mass of `H_(2)SO_(4)` in the solution is 19.6gm

Answer

A, B, D

Explanation

(A) Molarity of second solution is `=(10 xx d xx x)/(M) =1M`
(B) Volume `=100 +100 = 200ml`
(D) Mass of `H_(2)SO_(4) = (200 xx 1)/(1000) xx 98 = 19.6gm`