4 grams of an ideal gas occupies 5.6035 litres of volume at 546 K and 2 atm, pressure. What is its molecular weight ?

Options

A. 4
B. 16
C. 32
D. 64

Explanation

To find the molecular weight of the ideal gas, we can use the ideal gas equation and the relationship between moles, mass, and molecular weight. Here's a step-by-step solution:

### Step 1: Write down the ideal gas equation
The ideal gas equation is given by:
\[ PV = nRT \]
Where:
- \( P \) = Pressure (in atm)
- \( V \) = Volume (in liters)
- \( n \) = Number of moles
- \( R \) = Universal gas constant (0.0821 L·atm/(K·mol))
- \( T \) = Temperature (in Kelvin)

### Step 2: Rearrange the equation to find the number of moles
We can express the number of moles \( n \) as:
\[ n = \frac{PV}{RT} \]

### Step 3: Substitute the known values into the equation
Given:
- \( P = 2 \) atm
- \( V = 5.6035 \) L
- \( R = 0.0821 \) L·atm/(K·mol)
- \( T = 546 \) K

Substituting these values into the equation:
\[ n = \frac{(2 \, \text{atm})(5.6035 \, \text{L})}{(0.0821 \, \text{L·atm/(K·mol)})(546 \, \text{K})} \]

### Step 4: Calculate the number of moles
Calculating the right side:
1. Calculate the numerator: \( 2 \times 5.6035 = 11.207 \)
2. Calculate the denominator: \( 0.0821 \times 546 = 44.8666 \)

Now, calculate \( n \):
\[ n = \frac{11.207}{44.8666} \approx 0.249 \, \text{moles} \]

### Step 5: Use the relationship between mass, moles, and molecular weight
The relationship is given by:
\[ n = \frac{m}{M} \]
Where:
- \( m \) = mass (in grams)
- \( M \) = molecular weight (in g/mol)

Rearranging gives us:
\[ M = \frac{m}{n} \]

### Step 6: Substitute the known mass and calculated moles to find molecular weight
Given:
- \( m = 4 \) grams

Substituting the values:
\[ M = \frac{4 \, \text{g}}{0.249 \, \text{moles}} \approx 16.06 \, \text{g/mol} \]

### Step 7: Round to the nearest whole number
The molecular weight is approximately 16 g/mol.

### Final Answer:
The molecular weight of the ideal gas is **16 g/mol**.

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