At `30^(@)C` and 720 mm of Hg, the density of a gas is 1.5 `g// l t`. Calculate molecular mass of the gas. Also find the number of molecules in 1 cc of the gas at the same temperature.

Options

A. `2.29 xx 10^19`
B. `3.32 xx 10^18`
C. `5.66 xx 10^18`
D. `8.49 xx 10^19`

Answer

To solve the problem, we will follow these steps:

### Step 1: Convert Temperature to Kelvin
Given temperature = 30°C
To convert to Kelvin, use the formula:
\[ T(K) = T(°C) + 273 \]
\[ T(K) = 30 + 273 = 303 \, K \]

### Step 2: Convert Pressure to Atmospheres
Given pressure = 720 mm of Hg
To convert mm of Hg to atmospheres, use the conversion:
\[ 1 \, atm = 760 \, mm \, Hg \]
\[ P(atm) = \frac{720 \, mm \, Hg}{760 \, mm \, Hg} = 0.9474 \, atm \]

### Step 3: Use the Ideal Gas Law
The ideal gas equation is given by:
\[ PV = nRT \]
Where:
- \( P \) = pressure in atm
- \( V \) = volume in liters
- \( n \) = number of moles
- \( R \) = universal gas constant = 0.0821 L·atm/(K·mol)
- \( T \) = temperature in Kelvin

### Step 4: Relate Density to Molar Mass
Density (\( d \)) is given as 1.5 g/L.
We can express the number of moles \( n \) as:
\[ n = \frac{m}{M} \]
Where:
- \( m \) = mass of the gas in grams
- \( M \) = molar mass of the gas in g/mol

From the density, we know:
\[ d = \frac{m}{V} \Rightarrow m = d \cdot V \]
Substituting this into the ideal gas equation gives:
\[ P \cdot V = \frac{d \cdot V}{M} \cdot R \cdot T \]
Cancelling \( V \) from both sides:
\[ P = \frac{d \cdot R \cdot T}{M} \]
Rearranging gives:
\[ M = \frac{d \cdot R \cdot T}{P} \]

### Step 5: Substitute Values to Find Molar Mass
Substituting the known values:
- \( d = 1.5 \, g/L \)
- \( R = 0.0821 \, L·atm/(K·mol) \)
- \( T = 303 \, K \)
- \( P = 0.9474 \, atm \)

\[ M = \frac{1.5 \cdot 0.0821 \cdot 303}{0.9474} \]
Calculating this gives:
\[ M = \frac{37.38615}{0.9474} \approx 39.38 \, g/mol \]

### Step 6: Calculate the Number of Molecules in 1 cc
1 cc = 1 mL = \( 10^{-3} \, L \)
Using the ideal gas law again:
\[ PV = nRT \]
Substituting \( V = 10^{-3} \, L \):
\[ P \cdot 10^{-3} = n \cdot R \cdot T \]
Rearranging gives:
\[ n = \frac{P \cdot V}{R \cdot T} \]
Substituting the values:
\[ n = \frac{0.9474 \cdot 10^{-3}}{0.0821 \cdot 303} \]
Calculating gives:
\[ n \approx \frac{0.0009474}{24.7263} \approx 0.0000383 \, moles \]

### Step 7: Calculate the Number of Molecules
Using Avogadro's number (\( N_A = 6.022 \times 10^{23} \, molecules/mol \)):
\[ \text{Number of molecules} = n \cdot N_A \]
\[ \text{Number of molecules} = 0.0000383 \cdot 6.022 \times 10^{23} \]
Calculating gives:
\[ \text{Number of molecules} \approx 2.31 \times 10^{19} \, molecules \]

### Final Results
- Molecular mass of the gas = 39.38 g/mol
- Number of molecules in 1 cc of the gas = \( 2.31 \times 10^{19} \)