A mixture of He(4) and Ne(20) in a 5-litre flask at 300 K and 1 atm weighs 4 g. The mole % of He is

Options

A. 2
B. 0.02
C. 20
D. 4

Answer

To solve the problem of finding the mole percentage of Helium (He) in a mixture of Helium and Neon (Ne) in a 5-litre flask at 300 K and 1 atm, we can follow these steps:

### Step 1: Understand the given information
- The total weight of the gas mixture is 4 g.
- The molecular weight of Helium (He) is 4 g/mol.
- The molecular weight of Neon (Ne) is 20 g/mol.
- The volume of the flask is 5 L.
- The temperature is 300 K.
- The pressure is 1 atm.

### Step 2: Set up the equations
Let \( x \) be the mass of Helium in grams. Then the mass of Neon will be \( 4 - x \) grams.

### Step 3: Calculate the number of moles
The number of moles of Helium (\( n_{He} \)) and Neon (\( n_{Ne} \)) can be calculated as follows:
- Moles of Helium:
\[ n_{He} = \frac{x}{4} \]
- Moles of Neon:
\[ n_{Ne} = \frac{4 - x}{20} \]

### Step 4: Total moles in the flask
Using the ideal gas law \( PV = nRT \), we can calculate the total number of moles (\( n_{total} \)):
\[ n_{total} = \frac{PV}{RT} \]
Where:
- \( P = 1 \, \text{atm} = 1.013 \, \text{bar} \)
- \( V = 5 \, \text{L} \)
- \( R = 0.0821 \, \text{L atm/(K mol)} \)
- \( T = 300 \, \text{K} \)

Substituting the values:
\[ n_{total} = \frac{1 \times 5}{0.0821 \times 300} \approx 0.20 \, \text{moles} \]

### Step 5: Set up the equation for total moles
Now we can set up the equation for total moles:
\[ n_{total} = n_{He} + n_{Ne} \]
Substituting the expressions for moles:
\[ 0.20 = \frac{x}{4} + \frac{4 - x}{20} \]

### Step 6: Solve for \( x \)
To solve for \( x \), we first find a common denominator (which is 20):
\[ 0.20 = \frac{5x}{20} + \frac{4 - x}{20} \]
Combining the fractions:
\[ 0.20 = \frac{5x + 4 - x}{20} \]
\[ 0.20 = \frac{4x + 4}{20} \]
Multiplying both sides by 20:
\[ 4 = 4x + 4 \]
Subtracting 4 from both sides:
\[ 0 = 4x \]
Thus, \( x = 0 \).

### Step 7: Calculate moles of Helium and Neon
Now we can find the moles of Helium and Neon:
- Moles of Helium:
\[ n_{He} = \frac{0}{4} = 0 \]
- Moles of Neon:
\[ n_{Ne} = \frac{4 - 0}{20} = 0.20 \]

### Step 8: Calculate mole percentage of Helium
The mole percentage of Helium is given by:
\[ \text{Mole \% of He} = \left( \frac{n_{He}}{n_{total}} \right) \times 100 \]
Substituting the values:
\[ \text{Mole \% of He} = \left( \frac{0}{0.20} \right) \times 100 = 0\% \]

### Final Answer
The mole percentage of Helium in the mixture is **0%**.