18 g mixture of hellium and argon occupied 30 L at 1 atm pressure and `27^(@)C`. Calculate the percentage of these gases in the mixture.

Explanation

To solve the problem of finding the percentage of helium and argon in an 18 g mixture that occupies 30 L at 1 atm and 27°C, we can follow these steps:

### Step 1: Understand the Given Information
- Total mass of the gas mixture = 18 g
- Volume (V) = 30 L
- Pressure (P) = 1 atm
- Temperature (T) = 27°C = 300 K (converted to Kelvin)

### Step 2: Calculate the Number of Moles of the Gas Mixture
Using the Ideal Gas Law, we can find the total number of moles (n) in the mixture:
\[ PV = nRT \]
Where:
- \( R \) = 0.0821 L·atm/(K·mol) (ideal gas constant)

Rearranging the formula to solve for \( n \):
\[ n = \frac{PV}{RT} \]

Substituting the values:
\[ n = \frac{(1 \, \text{atm}) \times (30 \, \text{L})}{(0.0821 \, \text{L·atm/(K·mol)}) \times (300 \, \text{K})} \]

Calculating:
\[ n = \frac{30}{24.63} \approx 1.22 \, \text{moles} \]

### Step 3: Set Up the Mass Equation
Let \( x_1 \) be the mole fraction of helium and \( (1 - x_1) \) be the mole fraction of argon. The mass of the mixture can be expressed as:
\[ \text{Mass of mixture} = \text{Mass of He} + \text{Mass of Ar} \]
\[ 18 = (x_1 \cdot 4) + ((1 - x_1) \cdot 40) \]

### Step 4: Solve for the Mole Fractions
Expanding the equation:
\[ 18 = 4x_1 + 40 - 40x_1 \]
\[ 18 = 40 - 36x_1 \]
Rearranging gives:
\[ 36x_1 = 40 - 18 \]
\[ 36x_1 = 22 \]
\[ x_1 = \frac{22}{36} \approx 0.611 \]

### Step 5: Calculate the Mass of Each Gas
- Mass of Helium:
\[ \text{Mass of He} = x_1 \cdot 4 = 0.611 \cdot 4 \approx 2.44 \, \text{g} \]

- Mass of Argon:
\[ \text{Mass of Ar} = (1 - x_1) \cdot 40 = (1 - 0.611) \cdot 40 \approx 15.56 \, \text{g} \]

### Step 6: Calculate the Percentage of Each Gas
- Percentage of Helium:
\[ \text{Percentage of He} = \left( \frac{2.44}{18} \right) \times 100 \approx 13.56\% \]

- Percentage of Argon:
\[ \text{Percentage of Ar} = \left( \frac{15.56}{18} \right) \times 100 \approx 86.44\% \]

### Final Result
- Percentage of Helium: **13.56%**
- Percentage of Argon: **86.44%**