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25mL of 3M HCI were added to 75 mL of 0....

25mL of `3M HCI` were added to 75 mL of `0.05M HCI`. The molarity of HCI in the resulting solution is approximately

A

`0.55M`

B

`0.35M`

C

`0.787M`

D

`3.05M`

Text Solution

Verified by Experts

The correct Answer is:
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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