Question

300.0 mL of a 0.335 M solution of NaI is diluted to 700.0 mL. What is the new concentration of the solution?

Answer 1

Answer: The new concentration of the solution is 0.143 M.

Explanation:

Given: $V_{1}$ = 300.0 mL,    $M_{1}$ = 0.335 M

$V_{2}$ = 700.0 mL,         $M_{2}$ = ?

Formula used is as follows.

$M_{1}V_{1} = M_{2}V_{2}$

Substitute values into the above formula as follows.

$M_{1}V_{1} = M_{2}V_{2}\\0.335 M \times 300.0 mL = M_{2} \times 700.0 mL\\M_{2} = 0.143 M$

Thus, we can conclude that the new concentration of the solution is 0.143 M.