At `27^(@)C`, hydrogen is leaked through a tiny hole into a vessel for `20 min`. Another unknown gas at the same temperature and pressure as that of hydrogen is leaked through the same hole for `20 min`. After the effusion of the gases, the mixture exerts a pressure of `6 atm`. The hydrogen content of the mixture is `0.7 mol`. If the volume of the container is `3 L`, what is the molecular weight of the unknown gas?

Explanation

To solve the problem step by step, we will use the ideal gas law and Graham's law of effusion.

### Step 1: Use the Ideal Gas Law
The ideal gas law is given by the equation:
\[ PV = nRT \]
Where:
- \( P \) = pressure (in atm)
- \( V \) = volume (in liters)
- \( n \) = number of moles
- \( R \) = ideal gas constant (0.0821 L·atm/(K·mol))
- \( T \) = temperature (in Kelvin)

First, convert the temperature from Celsius to Kelvin:
\[ T = 27 + 273 = 300 \, K \]

Now, substitute the known values into the ideal gas equation:
\[ 6 \, \text{atm} \times 3 \, \text{L} = n \times 0.0821 \, \text{L·atm/(K·mol)} \times 300 \, K \]

### Step 2: Calculate Total Moles in the Mixture
Rearranging the equation to solve for \( n \):
\[ n = \frac{PV}{RT} = \frac{6 \times 3}{0.0821 \times 300} \]

Calculating the right-hand side:
\[ n = \frac{18}{24.63} \approx 0.7308 \, \text{mol} \]

### Step 3: Determine Moles of the Unknown Gas
We know the number of moles of hydrogen gas is given as \( 0.7 \, \text{mol} \). To find the moles of the unknown gas, we subtract the moles of hydrogen from the total moles:
\[ n_{\text{unknown}} = n_{\text{total}} - n_{\text{H}_2} = 0.7308 - 0.7 = 0.0308 \, \text{mol} \]

### Step 4: Apply Graham's Law of Effusion
Graham's law states that the rates of effusion of two gases are inversely proportional to the square roots of their molar masses:
\[ \frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}} \]

Where:
- \( r_1 \) and \( r_2 \) are the rates of effusion of hydrogen and the unknown gas, respectively.
- \( M_1 \) and \( M_2 \) are the molar masses of hydrogen and the unknown gas, respectively.

Since the time of effusion is the same for both gases, we can relate the moles to the rates:
\[ \frac{n_{\text{H}_2}}{n_{\text{unknown}}} = \sqrt{\frac{M_{\text{unknown}}}{M_{\text{H}_2}}} \]

Substituting the known values:
\[ \frac{0.7}{0.0308} = \sqrt{\frac{M_{\text{unknown}}}{2}} \]

### Step 5: Solve for the Molecular Weight of the Unknown Gas
Squaring both sides:
\[ \left(\frac{0.7}{0.0308}\right)^2 = \frac{M_{\text{unknown}}}{2} \]

Calculating the left side:
\[ \left(\frac{0.7}{0.0308}\right)^2 \approx 49.8 \]

Now, rearranging to find \( M_{\text{unknown}} \):
\[ M_{\text{unknown}} = 2 \times 49.8 \approx 99.6 \, \text{g/mol} \]

### Final Answer
The molecular weight of the unknown gas is approximately **99.6 g/mol**.