A certain public water supply contains 0.10ppb (part per billion) of chloroform `(CHCl_(3))`. How many molecules of `CHCl_(3)` would be obtained in 0.478mL drop of this water? (Assuming d=1g//Ml)

Options

A. `4xx10^(-13)xxN_(A)`
B. `10^(-3)xxN_(A)`
C. `4xx10^(-10)xxN_(A)`
D. none of these

Answer

To solve the problem of how many molecules of chloroform (CHCl₃) are present in a 0.478 mL drop of water containing 0.10 ppb of chloroform, we can follow these steps:

### Step 1: Calculate the mass of the water in grams
Given that the density of water is 1 g/mL, we can find the mass of 0.478 mL of water.

\[ \text{Mass of water} = \text{Density} \times \text{Volume} = 1 \, \text{g/mL} \times 0.478 \, \text{mL} = 0.478 \, \text{g} \]

### Step 2: Determine the mass of CHCl₃ in the water
From the problem, we know that 0.10 ppb means that there are 0.10 grams of CHCl₃ in 10^9 grams of water. Therefore, we can set up a proportion to find the mass of CHCl₃ in 0.478 grams of water.

\[ \text{Mass of CHCl₃} = \left( \frac{0.1 \, \text{g}}{10^9 \, \text{g}} \right) \times 0.478 \, \text{g} = 0.0478 \times 10^{-9} \, \text{g} \]

### Step 3: Calculate the number of moles of CHCl₃
To find the number of moles of CHCl₃, we use the formula:

\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]

The molar mass of CHCl₃ is approximately 119.5 g/mol. Thus,

\[ \text{Number of moles of CHCl₃} = \frac{0.0478 \times 10^{-9} \, \text{g}}{119.5 \, \text{g/mol}} \approx 0.0004 \times 10^{-9} \, \text{mol} = 4 \times 10^{-13} \, \text{mol} \]

### Step 4: Calculate the number of molecules of CHCl₃
Using Avogadro's number (approximately \(6.022 \times 10^{23}\) molecules/mol), we can find the number of molecules:

\[ \text{Number of molecules} = \text{Number of moles} \times \text{Avogadro's number} \]

\[ \text{Number of molecules of CHCl₃} = 4 \times 10^{-13} \, \text{mol} \times 6.022 \times 10^{23} \, \text{molecules/mol} \approx 2.4 \times 10^{11} \, \text{molecules} \]

### Final Answer
Thus, the number of molecules of CHCl₃ in a 0.478 mL drop of water is approximately \(2.4 \times 10^{11}\) molecules.

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