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.

$13.5M\times 0.0250L=M_2\times 0.500L$

$M_2=0.675M$

Therefore, the concentration of the diluted solution is 0.675 M