Question

A lab group made a copper amine stock solution with a concentration of 0.02 M. A group member transfers 9.27 mL of this solution to a 25 mL volumetric flask and dilutes to the line with water. What is the concentration of the resulting solution?

Answer 1

Answer : The concentration of the resulting solution is, 0.0074 M

Explanation :

Formula used :

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the initial molarity and volume of stock solution.

$M_2\text{ and }V_2$ are the final molarity and volume of diluted solution.

We are given:

$M_1=0.02M\\V_1=9.27mL\\M_2=?\\V_2=25mL$

Putting values in above equation, we get:

$0.02M\times 9.27mL=M_2\times 25mL\\\\M_2=0.0074M$

Hence, the concentration of the resulting solution is, 0.0074 M