Question

A 75.0 mL aliquot of a 1.60 M solution is diluted to a total volume of 258 mL. A 129 mL portion of that solution is diluted by adding 125 mL of water. What is the final concentration? Assume the volumes are additive.

Answer 1

Answer: Thus the final concentration is 0.236 M

Step-by-step explanation:

According to moalrity equation:

`M_1V_1=M_2V_2`

`M_1` = Molarity of stock solution = 1.60 M

`V_1` = volume of stock solution = 75.0 ml

`M_2` = molaity of diluted solution = ?

`V_2` = volume of diluted solution = 258 ml

`1.60* 75.0=M_2* 258`

`M_2=0.465M`

For second dilution:

`M_1V_1=M_2V_2`

`M_1` = Molarity of stock solution = 0.465 M

`V_1` = volume of stock solution = 129 ml

`M_2` = molaity of diluted solution = ?

`V_2` = volume of diluted solution = (129+125) = 254 ml

`0.465* 129=M_2* 254`

`M_2=0.236M`

Thus the final concentration is 0.236 M