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.

Step-by-step 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* 54.0=M_2* 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* 109=M_3* (109+161)`

`0.40* 109=M_3* 270`

`M_3=0.16M`

Thus the final concentration is 0.16 M