A gas having molecular mass `84.5 g mol^(-1)` enclosed in a flask at `27^(@)C` has a pressure of `1.5 "bar"`. Calculate the density of the gas under the same conditions.

Explanation

To calculate the density of the gas, we can use the formula:

\[ \rho = \frac{PM}{RT} \]

where:
- \(\rho\) = density of the gas
- \(P\) = pressure of the gas
- \(M\) = molar mass of the gas
- \(R\) = ideal gas constant
- \(T\) = temperature in Kelvin

### Step-by-step Solution:

1. **Convert the Temperature to Kelvin**:
The temperature is given as \(27^\circ C\). To convert this to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273 \]
\[ T = 27 + 273 = 300 \, K \]

2. **Identify the Given Values**:
- Molar mass \(M = 84.5 \, g \, mol^{-1}\)
- Pressure \(P = 1.5 \, bar\)
- Ideal gas constant \(R = 0.083 \, L \, bar \, mol^{-1} \, K^{-1}\)
- Temperature \(T = 300 \, K\)

3. **Substitute the Values into the Density Formula**:
Now we can substitute the known values into the density formula:
\[ \rho = \frac{PM}{RT} \]
\[ \rho = \frac{(1.5 \, bar)(84.5 \, g \, mol^{-1})}{(0.083 \, L \, bar \, mol^{-1} \, K^{-1})(300 \, K)} \]

4. **Calculate the Density**:
First, calculate the numerator:
\[ 1.5 \, bar \times 84.5 \, g \, mol^{-1} = 126.75 \, g \, bar \, mol^{-1} \]
Next, calculate the denominator:
\[ 0.083 \, L \, bar \, mol^{-1} \, K^{-1} \times 300 \, K = 24.9 \, L \, bar \, mol^{-1} \]
Now, substitute these values back into the equation:
\[ \rho = \frac{126.75 \, g \, bar \, mol^{-1}}{24.9 \, L \, bar \, mol^{-1}} = 5.09 \, g \, L^{-1} \]

5. **Final Answer**:
The density of the gas is:
\[ \rho = 5.09 \, g \, L^{-1} \]