`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 fin

Question

`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 fin

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

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - A
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`

4 views Answered 2023-02-05 15:29