What is the amount of gas, in moles, that occupies 60.82 L at 31.0 °C and 367 mm Hg according to the ideal gas law?

• A. 2.45 mol
• B. 0.98 mol
• C. 3.60 mol
• D. 1.20 mol

Answer: (D)

Verified step by step guidance

1. Identify the known variables from the problem: volume \(V = 60.82\ \text{L}\), temperature \(T = 31.0\ ^\circ\text{C}\), and pressure \(P = 367\ \text{mm Hg}\).
2. Convert the temperature from Celsius to Kelvin using the formula \(T(K) = T(^\circ C) + 273.15\), so \(T = 31.0 + 273.15\) K.
3. Convert the pressure from mm Hg to atm because the ideal gas constant \(R\) is commonly given in atm. Use the conversion \$1\ \text{atm} = 760\ \text{mm Hg}\(, so \)P(\text{atm}) = \frac{367}{760}$ atm.
4. Use the ideal gas law equation \(PV = nRT\) to solve for the number of moles \(n\). Rearrange the equation to \(n = \frac{PV}{RT}\).
5. Substitute the values of \(P\), \(V\), \(R\) (use \$0.0821\ \text{L}\cdot\text{atm}/\text{mol}\cdot\text{K}\(), and \)T\( into the equation and solve for \)n$.