Question

You prepare a 2nd solution (solution B) by pipetting 10.00 ml of your 1st solution (solution A) into a 50.00 ml volumetric flask and diluting with deionized water to the 50.00 ml mark.

If the concentration of solution B is 0.1045 M NaCl, what was the NaCl concentration of solution A?

Answer 1

Concentration solution A was 0.5225 M

Step-by-step explanation:

10.00 mL of solution A was diluted to 50.00 mL and yields 50.00 mL of solution B

According to laws of dilution-
`C_(A)V_(A)=C_(B)V_(B)`

where,
`C_(A)` and
`C_(B)` are concentration of solution A and B respectively

`V_(A)` and
`V_(B)` are volumes of solution A and B respectively

Here
`C_(B)` = 0.1045 M,
`V_(B)` = 50.00 mL and
`V_(A)` = 10.00 mL

Hence,
`C_(A)=(C_(B)V_(B))/(V_(A))=((0.1045M* 50.00mL))/(10.00mL)=0.5225M`

So, concentration solution A was 0.5225 M