Question

How many ppm of C₂₉H₆₀are in 23 µM C₂₉H₆₀?

Short answer

The number of ppm of $C_{29}{H}_{60}$in a $23\mu M$ solution is $9.4\mathrm{ppm}$ .

Step-by-step solution

Defining parts per million and molarity

The term "parts per million" refers to the concentration of a certain substance in a combination of chemicals.It is expressed in parts per million (ppm).

ppm$=\frac{\mathrm{Mass}\mathrm{of}\mathrm{solute}}{\mathrm{Mass}\mathrm{of}\mathrm{solution}}\times {10}^6$

$1ppm$is the same as $1\mu g/\mathrm{mL}$or $1\mathrm{mg}/\mathrm{L}$in solution concentrations.

Molarity: Molarity is one of the numerous factors used to express a solution's concentration.

Molarity$=\frac{\mathrm{no}.\mathrm{of}\mathrm{moles}\mathrm{of}\mathrm{solute}}{\mathrm{volume}\mathrm{of}\mathrm{solution}\mathrm{in}\mathrm{L}}$

Determining the number of ppm in 23 μM C29H60  

The concentration of $C_{29}{H}_{60}$in the solution is $23\mu M$, that is, $23\times {10}^{-6}M$.

Recall that,

Molarity $=\frac{\mathrm{no}\mathrm{of}\mathrm{moles}\mathrm{of}\mathrm{solute}}{\mathrm{volume}\mathrm{of}\mathrm{solution}\mathrm{in}\mathrm{L}}$

No. of moles of $C_{29}{H}_{60}=\frac{weightof{C}_{29}{H}_{60}}{molecularweightof{C}_{29}{H}_{60}}$.

Thus, the molarity of $C_{29}{H}_{60}$ is

Molarity of ${\mathrm{C}}_{29}{\mathrm{H}}_{60}=\frac{{\displaystyle \frac{\mathrm{mass}\mathrm{of}{\mathrm{C}}_{29}{\mathrm{H}}_{60}}{\mathrm{molar}\mathrm{mass}\mathrm{of}{\mathrm{c}}_{29}{\mathrm{H}}_{60}}}}{\mathrm{volume}\mathrm{of}\mathrm{solution}\mathrm{in}\mathrm{L}}$

The molecular weight of $C_{29}{H}_{60}$is as follows:

$MW\hspace{0.33em}of\hspace{0.33em}{C}_{29}{H}_{60}=(29\times Atomic\hspace{0.33em}weight\hspace{0.33em}of\hspace{0.33em}C)+(60\times Atomic\hspace{0.33em}weight\hspace{0.33em}ofH)$

$\begin{array}{l}=(29\mathrm{x}12.01)+(60\mathrm{x}1.008)\\ =348.29+60.48\\ =408.77\mathrm{g}/\mathrm{mol}\end{array}$.

Suppose the volume of a solution is 1L and weight $C_{29}{H}_{60}$needed to make a $23\mu M$solution is x .

Molarity of $C_{29}{H}_{60}$:

$$\begin{gathered} C_{29}{H}_{60}=\frac{{\displaystyle \frac{x}{408.77}}}{1L} \\ 23\times {10}^{-6}M=\frac{{\displaystyle \frac{x}{406.77}}}{1L} \\ x=23\times {10}^{-6}mol/L\times 408.77g/mol \\ =9401.71\times {10}^{-6}g/L \end{gathered}$$.

Thus, the molarity of ${\mathrm{C}}_{29}{\mathrm{H}}_{60}$is $9401.71\times {10}^{-6}g/L$.

As $1ppm=1mg/L$,

$$\begin{gathered} 9401.7\times {10}^{-6}\mathrm{g}/\mathrm{L}=9401.7\times {10}^{-6}\times {10}^3\mathrm{mg}/\mathrm{L} \\ =9401.7\times {10}^{-3}\mathrm{mg}/\mathrm{L} \\ =9.4\mathrm{mg}/\mathrm{L} \\ =9.4\mathrm{ppm} \end{gathered}$$.

Hence, the number of ppm of ${\mathrm{C}}_{29}{\mathrm{H}}_{60}$in a $23\mu M$solution is 9.4ppm.