`25 mL HNO_(3)`. If the volumes are mixed with `75 mL` of `4.0 M HNO_(3)`. If the volumes are additive, the molarity of the final mixture would be:

Options

A. 3.25M
B. 4.0 M
C. 3.75 M
D. 3.50 M

Answer

C

Explanation

Resulting Molarity `= (M_(1)V_(1)+M_(2)V_(2))/(V_(1)+V_(2))`
`= ( 3 xx 25 + 4 xx 75)/( 25 + 75)`