What is the density of NH_3 gas at 435 K and 1.00 atm? (Use R = 0.0821 L·atm·mol^{-1}·K^{-1} and molar mass of NH_3 = 17.0 g/mol.)

• A. 0.477 g/L
• B. 0.320 g/L
• C. 0.640 g/L
• D. 1.22 g/L

Answer: (A)

Verified step by step guidance

1. Recall the ideal gas law equation: \(P V = n R T\), where \(P\) is pressure, \(V\) is volume, \(n\) is moles, \(R\) is the gas constant, and \(T\) is temperature in Kelvin.
2. Express the number of moles \(n\) in terms of mass \(m\) and molar mass \(M\): \(n = \frac{m}{M}\).
3. Rewrite the ideal gas law to solve for volume per mass (or density): starting from \(P V = n R T\), substitute \(n = \frac{m}{M}\) to get \(P V = \frac{m}{M} R T\).
4. Rearrange the equation to isolate density \(\rho = \frac{m}{V}\): \(\rho = \frac{P M}{R T}\).
5. Plug in the given values: \(P = 1.00\) atm, \(M = 17.0\) g/mol, \(R = 0.0821\) L\cdot atm\cdot mol^{-1}\cdot K^{-1}\(, and \)T = 435\( K into the formula \)\rho = \frac{P M}{R T}$ to find the density in g/L.