Question

A stock solution of HNO3HNO3 is prepared and found to contain 12.7 M of HNO3HNO3. 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.

Answer 1

Answer: The concentration of the diluted solution is 0.635 M.

Explanation:

The given data is as follows.

         $M_{1}$ = 12.7 M,      $V_{1}$ = 25.0 ml = 0.025 L  (as 1 ml = 0.001 L)

         $M_{2}$ = ? ,       $V_{2}$ = 0.5 L

Therefore, we will calculate the molarity of the solution as follows.

          $M_{1}V_{1} = M_{2}V_{2}$

          $12.7 M \times 0.025 L = M_{2} \times 0.5 L$

            $M_{2}$ = 0.635 M

Thus, we can conclude that the concentration of the diluted solution is 0.635 M.