Question

The vapors of 1 g of an element occupy 2.5625 L exerting a pressure of 0.5 atm at 1000 K. What is the molar mass (in $\text{g mol}^{-1}$) of the element? (Assume vapors follow ideal gas equation. Given $R = 0.082 \text{ L atm mol}^{-1} \text{K}^{-1}$)

• A. 64
• B. 16
• C. 32
• D. 128

Hint:

Gas Law \& Molar Mass:

• Use $PV = nRT$ and $n = \fracmM$ to derive $M = \fracmRTPV$.
• Ensure units match those used with the gas constant $R$.
• For $R = 0.082$, pressure must be in atm, volume in liters, and temperature in Kelvin.

Answer 1

The Correct Option is A

We use the ideal gas law: \[ PV = nRT \] Rewriting in terms of molar mass $M$, using $n = \frac{m}{M}$: \[ PV = \frac{m}{M}RT \Rightarrow M = \frac{mRT}{PV} \] Given:

• $P = 0.5~\text{atm}$
• $V = 2.5625~\text{L}$
• $m = 1~\text{g}$
• $T = 1000~\text{K}$
• $R = 0.082~\text{L atm mol}^{-1} \text{K}^{-1}$

Substitute into the formula: \[ M = \frac{1 \cdot 0.082 \cdot 1000}{0.5 \cdot 2.5625} = \frac{82}{1.28125} \] Divide: \[ M = 64~\text{g/mol} \] Hence, the molar mass is $64~\text{g/mol}$, matching option (1).

Answer 2

Step 1: Write down the given data
Mass of element, m = 1 g
Volume of vapor, V = 2.5625 L
Pressure, P = 0.5 atm
Temperature, T = 1000 K
Gas constant, R = 0.082 L·atm·mol⁻¹·K⁻¹

Step 2: Use the ideal gas equation to find moles
Ideal gas equation: PV = nRT
Rearranged to find moles, n = PV / RT
Substitute the values:
n = (0.5 × 2.5625) / (0.082 × 1000) = 1.28125 / 82 = 0.015625 mol

Step 3: Calculate the molar mass
Molar mass, M = mass / moles = m / n
M = 1 g / 0.015625 mol = 64 g/mol

Step 4: Conclusion
The molar mass of the element is 64 g/mol, calculated by applying the ideal gas law to the vapor and relating mass to the number of moles.