`100 ml` of `1 M H_(2) SO_(4)` solution
`(d_("solution") = 1.5 gm//ml)` is mixed with 400 ml of water `(d_("water") = 1 gm //ml)` then molarity of final solution `(d_("final solution") = 1.25 gm//ml)` is -

Options

A. 0.227 M
B. 2.5 M
C. 0.4 M
D. 2.27 M

Answer

A

Explanation

Mass of `H_(2)SO_(4)` solution `= 100 xx 1.5 = 150 gm`
Mass of water `= 400`
Mass (total) `= 550 gm`
Moles of `H_(2)SO_(4) = 0.1` mole
Volume finally `= (550)/(1.25) = 440 mL`
`M = (0.1)/(440) xx 1000 = 0.227 M`