A mixture of Ne and Ar kept in a closed vessel at 250 K has a total K.E.=3 kJ. The total mass of Ne and Ar is 30 g. Find mass % of Ne in gaseous mixture at 250 K.

Options

A. 61.63
B. 38.37
C. 0.5
D. 28.3

Explanation

To solve the problem step by step, we will use the given data and apply the relevant equations to find the mass percentage of Neon (Ne) in the mixture of Neon and Argon (Ar).

### Step 1: Write down the given data
- Total K.E. (K) = 3 kJ = 3000 J
- Total mass of Ne and Ar = 30 g
- Temperature (T) = 250 K

### Step 2: Use the kinetic energy formula
The total kinetic energy of a gas can be expressed as:
\[ K = \frac{3}{2} nRT \]
where:
- \( n \) = number of moles
- \( R \) = universal gas constant = 8.314 J/(mol·K)
- \( T \) = temperature in Kelvin

### Step 3: Rearranging the formula to find \( n \)
We can rearrange the formula to solve for \( n \):
\[ n = \frac{2K}{3RT} \]

### Step 4: Substitute the known values
Substituting the known values into the equation:
\[ n = \frac{2 \times 3000}{3 \times 8.314 \times 250} \]

### Step 5: Calculate \( n \)
Calculating the above expression:
\[ n = \frac{6000}{3 \times 8.314 \times 250} \]
\[ n = \frac{6000}{6235.5} \approx 0.962 \text{ moles} \]

### Step 6: Set up the mass equations
Let:
- Mass of Ne = \( x \) grams
- Mass of Ar = \( 30 - x \) grams

Using the molar masses:
- Molar mass of Ne = 20 g/mol
- Molar mass of Ar = 40 g/mol

The number of moles of Ne and Ar can be expressed as:
\[ n_{Ne} = \frac{x}{20} \]
\[ n_{Ar} = \frac{30 - x}{40} \]

### Step 7: Write the total moles equation
The total number of moles can be expressed as:
\[ n_{Ne} + n_{Ar} = n \]
Substituting the expressions for moles:
\[ \frac{x}{20} + \frac{30 - x}{40} = 0.962 \]

### Step 8: Clear the fractions
To eliminate the fractions, multiply through by 40:
\[ 2x + (30 - x) = 40 \times 0.962 \]
\[ 2x + 30 - x = 38.48 \]
\[ x + 30 = 38.48 \]
\[ x = 38.48 - 30 \]
\[ x = 8.48 \text{ grams} \]

### Step 9: Calculate the mass percentage of Ne
The mass percentage of Ne can be calculated as:
\[ \text{Mass \% of Ne} = \left( \frac{\text{mass of Ne}}{\text{total mass}} \right) \times 100 \]
Substituting the values:
\[ \text{Mass \% of Ne} = \left( \frac{8.48}{30} \right) \times 100 \]
\[ \text{Mass \% of Ne} = 28.27 \approx 28.3\% \]

### Final Answer
The mass percentage of Neon (Ne) in the gaseous mixture at 250 K is approximately **28.3%**.

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