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 become 200 ml
C. mass of `H_2SO_4` in the solution is 98 gm
D. mass of `H_2SO_4` in the solution is 19.6 gm

Answer

A,B,D

Explanation

(A)Molarity of second solution is `=(10xxdxxx)/M=1M`
(B)Volume=100+100=200 ml
( D)Mass of `H_2SO_4=(200xx1)/1000xx98=19.6` gm