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`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 MB. 2.5 MC. 0.4 MD. 2.27 M |
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Answer» 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` |
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