The average density of the universe as a whole is estimated as `3 xx 10^(-29)g//m` L . If we assume that the entire mass is only H atoms, what is the average volume of space that contains one H atom?

Options

A. 15.998 L
B. `1.8 xx 10^(-5) L`
C. 55.561 L
D. `3.6 xx 10^(-5) L`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Given Information
We are given the average density of the universe as \(3 \times 10^{-29} \, \text{g/mL}\) and we need to find the average volume of space that contains one hydrogen atom.

### Step 2: Identify the Mass of One Hydrogen Atom
The mass of one hydrogen atom is approximately \(1.673 \times 10^{-24} \, \text{g}\).

### Step 3: Use the Density Formula
The formula for density is given by:
\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \]
From this, we can rearrange the formula to find the volume:
\[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} \]

### Step 4: Substitute the Values
Now, we can substitute the values we have into the formula:
\[ \text{Volume} = \frac{1.673 \times 10^{-24} \, \text{g}}{3 \times 10^{-29} \, \text{g/mL}} \]

### Step 5: Perform the Calculation
Calculating the volume:
\[ \text{Volume} = \frac{1.673}{3} \times 10^{5} \, \text{mL} \]
Calculating \( \frac{1.673}{3} \):
\[ \frac{1.673}{3} \approx 0.55767 \]
Thus,
\[ \text{Volume} \approx 0.55767 \times 10^{5} \, \text{mL} \approx 55.767 \, \text{mL} \]

### Step 6: Convert to Liters
Since \(1 \, \text{L} = 1000 \, \text{mL}\), we convert the volume from mL to L:
\[ \text{Volume in Liters} = \frac{55.767 \, \text{mL}}{1000} \approx 0.055767 \, \text{L} \]

### Step 7: Final Answer
Thus, the average volume of space that contains one hydrogen atom is approximately:
\[ \text{Volume} \approx 0.0558 \, \text{L} \text{ or } 55.8 \, \text{mL} \]