Question

The concentration of the alkane ${\mathit{C}}_{\mathbf{20}}{\mathit{H}}_{\mathbf{42}}$(FM 282.56) in a particular sample of rainwater is $\mathbf{0}\mathbf{.}\mathbf{2}\mathbf{}\mathit{p}\mathit{p}\mathit{b}$. Assume that the density of rainwater is close to$\mathbf{1}\mathbf{.}\mathbf{00}\mathit{g}\mathbf{/}\mathit{m}\mathit{L}$and find the molar concentration of ${\mathit{C}}_{\mathbf{20}}{\mathit{H}}_{\mathbf{42}}$.

Short answer

The molar concentration of$C_{20}{H}_{42}$ is$7.078\times {10}^{-10}$

Step-by-step solution

Conversion of the alkane in g/L

Convert the concentration of the alkane from ppb to mg/L

$0.2\mathrm{ppb}=0.2\times {10}^{-3}\mathrm{mg}/\mathrm{L}=2\times {10}^{-4}\mathrm{mg}/\mathrm{L}$

$\mathrm{Img}={10}^{-3}\mathrm{g}$the mass of$C_{20}{H}_{42}$is$2\times {10}^{-7}\mathrm{g}/\mathrm{L}$

The value of the molar concentration

The molar concentration of $C_{20}{H}_{42}$will be as follows:

$$\begin{gathered} \mathrm{Molar}\mathrm{concentration}=\frac{\mathrm{number}\mathrm{of}}{\mathrm{volume}\mathrm{in}\mathrm{L}}n \\ =\frac{\mathrm{mass}/\mathrm{formula}\mathrm{mass}}{\mathrm{volume}\mathrm{in}\mathrm{L}}n \\ =2\times {10}^{-7}g/282.56L \\ =7.078\times {10}^{-10} \end{gathered}$$

Hence, the molar concentration of $C_{20}{H}_{42}$is$7.078\times {10}^{-10}$