Question

A 54.0 mL aliquot of a 1.60 M solution is diluted to a total volume of 218 mL. A 109 mL portion of that solution is diluted by adding 161 mL of water. What is the final concentration? Assume the volumes are additive g

Answer 1

Answer: The final concentration is 0.16 M.

Explanation:

According to the dilution law:

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of stock solution = 1.60 M

$V_1$ = volume of stock  solution = 54.0 ml

$M_2$ = molarity of diluted solution = ?

$V_2$ = volume of diluted solution = 218 ml

Putting these values:

$1.60\times 54.0=M_2\times 218$

$M_2=0.40M$

Now 109 ml of this diluted solution is further diluted by adding 161 mL of water.

Again applying dilution law:

$0.40\times 109=M_3\times (109+161)$

$0.40\times 109=M_3\times 270$

$M_3=0.16M$

Thus the final concentration is 0.16 M