Question

1.00 mL of a 3.55 × 10–4 M solution of oleic acid is diluted with 9.00 mL of petroleum ether, forming solution A. Then 2.00 mL of solution A is diluted with 8.00 mL of petroleum ether, forming solution B. What is the concentration of solution B?

Answer 1

Concentration of solution B is
`7.10* 10^(-6)M`

Step-by-step explanation:

Solution A: Total volume of solution A = (9.00+1.00) mL = 10.00 mL

According to law of dilution,
`C_(1)V_(1)=C_(2)V_(2)`

where
`C_(1)` and
`C_(2)` are initial and final concentration respectively.
`V_(1)` and
`V_(2)` are initial and final volume respectively.

Here
`C_(1)` =
`3.55* 10^(-4)M`,
`V_(1)` = 1.00 mL,
`V_(2)` = 10.00 mL

So,
`C_(2)=(C_(1)V_(1))/(V_(2))` =
`(3.55* 10^(-4)M* 1.00mL)/(10.00mL)=3.55* 10^(-5)M`

So, concentration of solution A =
`3.55* 10^(-5)M`

Solution B: Total volume of solution B = (2.00+8.00) mL = 10.00 mL

Similarly as above,
`C_(2)=(C_(1)V_(1))/(V_(2))` =
`(3.55* 10^(-5)M* 2.00mL)/(10.00mL)=7.10* 10^(-6)M`

So, concentration of solution B =
`7.10* 10^(-6)M`