3.6 gram of oxygen of adsorbed on 1.2 g of metal powder. What volume of oxygen adsorbed per gram of the adsorbent at 1 atm and 273 K ?

Options

A. `0.19 Lg^(-1)`
B. `1L g^(-1)`
C. `2.1 Lg^(-1)`
D. none of these

Answer

To solve the problem of determining the volume of oxygen adsorbed per gram of the adsorbent at 1 atm and 273 K, we can follow these steps:

### Step 1: Calculate the mass of oxygen adsorbed per gram of metal powder.
Given:
- Mass of oxygen adsorbed = 3.6 g
- Mass of metal powder = 1.2 g

We can find the mass of oxygen adsorbed per gram of metal powder using the formula:
\[ \text{Mass of oxygen per gram of metal} = \frac{\text{Mass of oxygen}}{\text{Mass of metal}} = \frac{3.6 \, \text{g}}{1.2 \, \text{g}} = 3 \, \text{g} \]

### Step 2: Use the ideal gas equation to find the volume of oxygen.
The ideal gas equation is given by:
\[ PV = nRT \]
Where:
- \( P \) = pressure (1 atm)
- \( V \) = volume (what we want to find)
- \( n \) = number of moles of gas
- \( R \) = ideal gas constant (0.0821 L·atm/(K·mol))
- \( T \) = temperature (273 K)

First, we need to calculate the number of moles of oxygen adsorbed. The molar mass of oxygen (O₂) is approximately 32 g/mol.

Using the formula:
\[ n = \frac{\text{mass}}{\text{molar mass}} = \frac{3.6 \, \text{g}}{32 \, \text{g/mol}} = 0.1125 \, \text{mol} \]

### Step 3: Substitute the values into the ideal gas equation to find the volume.
Rearranging the ideal gas equation to solve for volume \( V \):
\[ V = \frac{nRT}{P} \]

Substituting the known values:
\[ V = \frac{0.1125 \, \text{mol} \times 0.0821 \, \text{L·atm/(K·mol)} \times 273 \, \text{K}}{1 \, \text{atm}} \]

Calculating:
\[ V = \frac{0.1125 \times 0.0821 \times 273}{1} \]
\[ V \approx 2.5 \, \text{L} \]

### Step 4: Calculate the volume of oxygen adsorbed per gram of adsorbent.
To find the volume of oxygen adsorbed per gram of metal powder:
\[ \text{Volume per gram of adsorbent} = \frac{V}{\text{mass of metal}} = \frac{2.5 \, \text{L}}{1.2 \, \text{g}} \approx 2.08 \, \text{L/g} \]

### Final Answer:
The volume of oxygen adsorbed per gram of the adsorbent is approximately **2.08 L/g**.

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