Question
The concentration of the alkane C20H42 (FM =282.56) in a particular rainwater sample is 0.2ppb.Assume that the density of rainwater is close to 1.00g/Ml and find the molar concentration of C20H42.
Answers
GPT-4o mini
To convert the concentration of C\(_{20}\)H\(_{42}\) from parts per billion (ppb) to molar concentration (M), we first need to understand what 0.2 ppb means in terms of mass in a given volume of rainwater.
1. **Understanding ppb:**
- 1 ppb means 1 microgram (µg) of solute in 1 liter (L) of solution.
- Thus, 0.2 ppb means 0.2 µg of C\(_{20}\)H\(_{42}\) in 1 L of rainwater.
2. **Convert µg to grams:**
\[
0.2 \, \text{µg} = 0.2 \times 10^{-6} \, \text{g} = 2.0 \times 10^{-7} \, \text{g}
\]
3. **Calculate the number of moles of C\(_{20}\)H\(_{42}\):**
To find the number of moles (\(n\)), we use the formula:
\[
n = \frac{\text{mass}}{\text{molar mass}}
\]
where the molar mass of C\(_{20}\)H\(_{42}\) is given as 282.56 g/mol.
Plugging in the values:
\[
n = \frac{2.0 \times 10^{-7} \, \text{g}}{282.56 \, \text{g/mol}} \approx 7.07 \times 10^{-10} \, \text{mol}
\]
4. **Convert moles to molarity (M):**
Molarity (M) is defined as the number of moles of solute per liter of solution. Since we have \(7.07 \times 10^{-10}\) mol of C\(_{20}\)H\(_{42}\) in 1 L of rainwater, the molarity is:
\[
\text{Molarity} = 7.07 \times 10^{-10} \, \text{mol/L} = 7.07 \times 10^{-10} \, \text{M}
\]
Thus, the molar concentration of C\(_{20}\)H\(_{42}\) in the rainwater sample is approximately **7.07 × 10⁻¹⁰ M**.
1. **Understanding ppb:**
- 1 ppb means 1 microgram (µg) of solute in 1 liter (L) of solution.
- Thus, 0.2 ppb means 0.2 µg of C\(_{20}\)H\(_{42}\) in 1 L of rainwater.
2. **Convert µg to grams:**
\[
0.2 \, \text{µg} = 0.2 \times 10^{-6} \, \text{g} = 2.0 \times 10^{-7} \, \text{g}
\]
3. **Calculate the number of moles of C\(_{20}\)H\(_{42}\):**
To find the number of moles (\(n\)), we use the formula:
\[
n = \frac{\text{mass}}{\text{molar mass}}
\]
where the molar mass of C\(_{20}\)H\(_{42}\) is given as 282.56 g/mol.
Plugging in the values:
\[
n = \frac{2.0 \times 10^{-7} \, \text{g}}{282.56 \, \text{g/mol}} \approx 7.07 \times 10^{-10} \, \text{mol}
\]
4. **Convert moles to molarity (M):**
Molarity (M) is defined as the number of moles of solute per liter of solution. Since we have \(7.07 \times 10^{-10}\) mol of C\(_{20}\)H\(_{42}\) in 1 L of rainwater, the molarity is:
\[
\text{Molarity} = 7.07 \times 10^{-10} \, \text{mol/L} = 7.07 \times 10^{-10} \, \text{M}
\]
Thus, the molar concentration of C\(_{20}\)H\(_{42}\) in the rainwater sample is approximately **7.07 × 10⁻¹⁰ M**.