Question

10 drops of 0.200M [I-]solution are used to make a solution with 48 drops, what is the final molarity of the [I-] solution.

Answer 1

Answer:

0.042 M

Explanation:

Considering

$Molarity_{working\ solution}\times Volume_{working\ solution}=Molarity_{stock\ solution}\times Volume_{stock\ solution}$

Given  that:

$Molarity_{working\ solution}=?$

$Volume_{working\ solution}=48\ drops$

$Volume_{stock\ solution}=10\ drops$

$Molarity_{stock\ solution}=0.200\ M$

So,  

$Molarity_{working\ solution}\times 48\ drops=0.200\ M\times 10\ drops$

$Molarity_{working\ solution}=\frac{0.200\times 10}{48}\ M=0.042\ M$