1.

25mL of `3M HCI` were added to 75 mL of `0.05M HCI`. The molarity of HCI in the resulting solution is approximatelyA. `0.55M`B. `0.35M`C. `0.787M`D. `3.05M`

Answer» Correct Answer - C
Molarity of resulting solution
`M_("Total") = (M_(A)V_(1))/(V_("Total")) +(M_(2)V_(2))/(V_("Total"))`
`= (25 xx 3)/(100) +(75 xx 0.05)/(100)`
`= 0.75 +0.0375 = 0.7875M`


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