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

Question

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

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`

Monjur Alahi

5 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 - 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`

5 views Answered 2023-02-05 15:29