Question

The concentration of the alkane C20H42 (FW 282.55) in a particular sample of rainwater is 0.40 ppb. Assume that the density of rainwater is close to 1.00 g/mL and find the molar concentration of C20H42.

Answer 1

Molar concentration =
`1.4 * 10^(-9) M`

Step-by-step explanation:

Concentration of alkane (
`C_(20)H_(42)` = 0.40 ppb

1 ppb =
`10^-3\ mg/L`

Concentration of alkane (
`C_(20)H_(42)` in mg/L

=
`4.00 * 10^(-4)\ mg/L`

Concentration of alkane (
`C_(20)H_(42)`) is
`4.00 * 10^(-4)\ mg/L` that means
`4.00 * 10^(-4)` is present in 1 L

Density of water = 1.00 g/mL

Molar concentration =
`(Mass\ in\ g)/(Molar mass * Volume\ in\ L)`

Formula weight = 282.55

Mass of
`C_(20)H_(42)` =
`4.00 * 10^(-4)\ mg/L` =
`4.00 * 10^(-7)\ g/L`

Here, formula mass is equal to molar mass.

Molar concentration =
`(4.00 * 10^(-7))/(282.55 * 1) = 1.4 * 10^(-9) M`