Question

A stock solution of HNO3 is prepared and found to contain 13.5 M of HNO3. If 25.0 mL of the stock solution is diluted to a final volume of 0.500 L, the concentration of the diluted solution is ________ M.

(A) 0.270
(B) 1.48
(C) 0.675
(D) 270
(E) 675

Answer 1

Answer : The correct option is, (C) 0.675 M

Explanation :

Using neutralization law,

`M_1V_1=M_2V_2`

where,

`M_1` = concentration of
`HNO_3` = 13.5 M

`M_2` = concentration of diluted solution = ?

`V_1` = volume of
`HNO_3` = 25.0 ml = 0.0250 L

conversion used : (1 L = 1000 mL)

`V_2` = volume of diluted solution = 0.500 L

Now put all the given values in the above law, we get the concentration of the diluted solution.

`13.5M* 0.0250L=M_2* 0.500L`

`M_2=0.675M`

Therefore, the concentration of the diluted solution is 0.675 M