The density of `O_(2)` is 16 at STP. At what temperature (in `.^@C`) its density will be 14 ? Consider that the pressure remais constant.

Answer

To solve the problem, we will use the relationship between density and temperature at constant pressure. The formula we will use is:

\[ \frac{d_1}{T_1} = \frac{d_2}{T_2} \]

Where:
- \(d_1\) is the initial density
- \(T_1\) is the initial temperature in Kelvin
- \(d_2\) is the final density
- \(T_2\) is the final temperature in Kelvin

### Step 1: Identify the given values
- Initial density, \(d_1 = 16 \, \text{g/L}\)
- Final density, \(d_2 = 14 \, \text{g/L}\)
- Initial temperature at STP, \(T_1 = 0^\circ C = 273 \, \text{K}\)

### Step 2: Set up the equation
Using the formula:

\[ \frac{d_1}{T_1} = \frac{d_2}{T_2} \]

Substituting the known values:

\[ \frac{16}{273} = \frac{14}{T_2} \]

### Step 3: Cross-multiply to solve for \(T_2\)
Cross-multiplying gives us:

\[ 16 \cdot T_2 = 14 \cdot 273 \]

### Step 4: Calculate \(T_2\)
Now, we can solve for \(T_2\):

\[ T_2 = \frac{14 \cdot 273}{16} \]

Calculating the right side:

\[ T_2 = \frac{3822}{16} = 238.875 \, \text{K} \]

### Step 5: Convert \(T_2\) to Celsius
To convert the temperature from Kelvin to Celsius:

\[ T_2^\circ C = T_2 - 273 = 238.875 - 273 = -34.125^\circ C \]

### Final Answer
Thus, the temperature at which the density of oxygen will be 14 g/L is approximately:

\[ \boxed{-34.13^\circ C} \]