In a very good vacuum system in the laboratory, the vacuum attained was `10^-13 atm`. If the temperature of the system was `300 K`, the number of molecules present in a volume of `1 cm^3` is.

Options

A. `2.4 xx 10^6`
B. `24`
C. `2.4 xx 10^9`
D. zero

Explanation

To find the number of molecules present in a volume of 1 cm³ at a pressure of \(10^{-13}\) atm and a temperature of 300 K, we can use the Ideal Gas Law, which is given by the equation:

\[ PV = nRT \]

Where:
- \(P\) = pressure (in Pascals)
- \(V\) = volume (in cubic meters)
- \(n\) = number of moles
- \(R\) = universal gas constant (\(8.314 \, \text{J/(mol K)}\))
- \(T\) = temperature (in Kelvin)

### Step 1: Convert the pressure from atm to Pascals
Given:
\[ P = 10^{-13} \, \text{atm} \]

To convert atm to Pascals, we use the conversion factor:
\[ 1 \, \text{atm} = 1.01 \times 10^5 \, \text{Pa} \]

Thus,
\[ P = 10^{-13} \, \text{atm} \times 1.01 \times 10^5 \, \text{Pa/atm} = 1.01 \times 10^{-8} \, \text{Pa} \]

### Step 2: Convert the volume from cm³ to m³
Given:
\[ V = 1 \, \text{cm}^3 \]

To convert cm³ to m³, we use the conversion:
\[ 1 \, \text{cm}^3 = 10^{-6} \, \text{m}^3 \]

Thus,
\[ V = 1 \, \text{cm}^3 = 10^{-6} \, \text{m}^3 \]

### Step 3: Use the Ideal Gas Law to find the number of moles \(n\)
Rearranging the Ideal Gas Law to solve for \(n\):
\[ n = \frac{PV}{RT} \]

Substituting the values we have:
- \(P = 1.01 \times 10^{-8} \, \text{Pa}\)
- \(V = 10^{-6} \, \text{m}^3\)
- \(R = 8.314 \, \text{J/(mol K)}\)
- \(T = 300 \, \text{K}\)

\[ n = \frac{(1.01 \times 10^{-8}) \times (10^{-6})}{(8.314) \times (300)} \]

Calculating the denominator:
\[ 8.314 \times 300 = 2494.2 \, \text{J/mol} \]

Now substituting back:
\[ n = \frac{1.01 \times 10^{-14}}{2494.2} \approx 4.05 \times 10^{-18} \, \text{mol} \]

### Step 4: Convert moles to molecules
To find the number of molecules, we use Avogadro's number (\(N_A = 6.022 \times 10^{23} \, \text{molecules/mol}\)):
\[ \text{Number of molecules} = n \times N_A \]
\[ \text{Number of molecules} = (4.05 \times 10^{-18}) \times (6.022 \times 10^{23}) \approx 2.44 \times 10^{6} \]

### Final Answer
The number of molecules present in a volume of 1 cm³ at a pressure of \(10^{-13}\) atm and a temperature of 300 K is approximately:
\[ \boxed{2.44 \times 10^{6}} \]