The density of a gaseous mixture of He and `N_2` is found to be `(10)/(22.4) g//1` at STP. The percentage composition of He and `N_2` in this mixture respectively will be

Options

A. 75%, 25%
B. 25%, 75%
C. 30%, 70%
D. 40%, 60%

Answer

To solve the problem of finding the percentage composition of helium (He) and nitrogen (N₂) in a gaseous mixture with a given density, we can follow these steps:

### Step 1: Understand the Given Information
The density of the gaseous mixture is given as \( \frac{10}{22.4} \, \text{g/L} \) at STP. At STP, 22.4 L corresponds to 1 mole of gas.

### Step 2: Calculate the Total Mass and Moles
Since the density is \( \frac{10}{22.4} \, \text{g/L} \), we can infer that:
- The total mass of the mixture is 10 g.
- The total volume is 22.4 L, which corresponds to 1 mole of the gas mixture.

### Step 3: Define Variables for Moles
Let:
- \( x \) = number of moles of helium (He)
- \( 1 - x \) = number of moles of nitrogen (N₂)

### Step 4: Relate Mass to Moles
The mass of helium can be expressed as:
\[ \text{Mass of He} = x \times 4 \, \text{g} \]
The mass of nitrogen can be expressed as:
\[ \text{Mass of N₂} = (1 - x) \times 28 \, \text{g} \]

### Step 5: Set Up the Mass Equation
The total mass of the mixture is the sum of the masses of helium and nitrogen:
\[ 4x + 28(1 - x) = 10 \]

### Step 6: Simplify the Equation
Expanding the equation:
\[ 4x + 28 - 28x = 10 \]
Combine like terms:
\[ -24x + 28 = 10 \]
Rearranging gives:
\[ -24x = 10 - 28 \]
\[ -24x = -18 \]
Dividing by -24:
\[ x = \frac{18}{24} = \frac{3}{4} \]

### Step 7: Calculate Moles of Nitrogen
Now, substituting back to find the moles of nitrogen:
\[ 1 - x = 1 - \frac{3}{4} = \frac{1}{4} \]

### Step 8: Calculate the Masses
Now we can calculate the mass of each gas:
- Mass of He:
\[ \text{Mass of He} = x \times 4 = \frac{3}{4} \times 4 = 3 \, \text{g} \]
- Mass of N₂:
\[ \text{Mass of N₂} = (1 - x) \times 28 = \frac{1}{4} \times 28 = 7 \, \text{g} \]

### Step 9: Calculate the Percentage Composition
Now, we can find the percentage composition of each gas:
- Percentage of He:
\[ \text{Percentage of He} = \left( \frac{3}{10} \right) \times 100 = 30\% \]
- Percentage of N₂:
\[ \text{Percentage of N₂} = \left( \frac{7}{10} \right) \times 100 = 70\% \]

### Final Answer
The percentage composition of helium and nitrogen in the mixture is:
- Helium (He): 30%
- Nitrogen (N₂): 70%

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