1. At 733 mmHg and 114.8oF, calculate the density of HBr gas. Complete solution
The molar mass of HBr is 80.9 g mol−1
R = gas constant = 0.0821 L atm K−1 mol−1
Converting units:
P = (733 mmHg) × (1 atm / 760 mmHg) = 0.9645 atm
T = (114.8oF + 459.67) × (5/9) = 319.15 K
Solution:
The ideal gas equation can be used.
PV = nRT
Moles (n) = Mass (m) / Molar mass (M)
Therefore, the ideal gas equation can be rewritten as:
PV = (m/M)RT
PVM = mRT
Density (d) = Mass (m) / Volume (V)
Rearrange the equation to make d the subject:
d = m / V = PM / RT
d = PM / RT
Therefore,
Density of HBr gas = (0.9645 atm × 80.9 g mol−1) / (0.0821 L atm K−1 mol−1 × 319.15 K) = 2.978 g/L
Density of HBr gas = 2.978 g/L = 2.98 g/L
Answer: The density of HBr gas is 2.98 g/L
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