A gas cylinder was found unattended in a public place. The investigating team took the collected samples from it. The density of the gas was found to be 2.380 `gL^(−1)` at `27^o`C and 1 atm pressure. Hence the molar mass of the gas is:

Options

A. 46
B. 50
C. 59
D. 40

Answer

To find the molar mass of the gas using the given density, temperature, and pressure, we can use the ideal gas law and the relationship between density and molar mass.

### Step-by-Step Solution:

1. **Identify the Ideal Gas Law**:
The ideal gas law is given by the equation:
\[ PV = nRT \]
where:
- \( P \) = pressure (in atm)
- \( V \) = volume (in liters)
- \( n \) = number of moles
- \( R \) = gas constant (0.0821 L·atm/(K·mol))
- \( T \) = temperature (in Kelvin)

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

3. **Relate Molar Mass to Density**:
The number of moles \( n \) can be expressed as:
\[ n = \frac{\text{mass}}{\text{molar mass}} = \frac{m}{M} \]
where \( m \) is the mass of the gas and \( M \) is the molar mass.

Rearranging the ideal gas law gives:
\[ PV = \frac{m}{M}RT \]
Rearranging further, we can express molar mass \( M \) in terms of density \( d \):
\[ M = \frac{dRT}{P} \]

4. **Substitute the Known Values**:
Given:
- Density \( d = 2.380 \, g/L \)
- Pressure \( P = 1 \, atm \)
- Gas constant \( R = 0.0821 \, L·atm/(K·mol) \)
- Temperature \( T = 300 \, K \)

Now substituting these values into the equation for molar mass:
\[ M = \frac{(2.380 \, g/L) \times (0.0821 \, L·atm/(K·mol)) \times (300 \, K)}{1 \, atm} \]

5. **Calculate Molar Mass**:
Performing the multiplication:
\[ M = \frac{(2.380) \times (0.0821) \times (300)}{1} \]
\[ M = \frac{58.596}{1} \approx 58.6 \, g/mol \]

6. **Final Answer**:
The molar mass of the gas is approximately:
\[ \boxed{58.6 \, g/mol} \]
This can also be rounded to \( 59 \, g/mol \).