Answer:
3.2 L
Explanation:
Given data:
Mass of oxygen = 3.760 g
Pressure of gas = 88.4 Kpa (88.4×1000 = 88400 Nm⁻²)
Temperature = 19°C (19+273.15 = 292.15 K)
R = 8.314 Nm K⁻¹ mol⁻¹
Volume occupied = ?
Solution:
Number of moles of oxygen:
Number of moles = mass/ molar mass
Number of moles = 3.760 g/ 32 g/mol
Number of moles = 0.12 mol
The given problem will be solve by using general gas equation,
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant
T = temperature in kelvin
V = nRT/P
V = 0.12 mol × 8.314 Nm K⁻¹ mol⁻¹ × 292.15 K /88400 Nm⁻²
V = 291.472 Nm /88400 Nm⁻²
V = 0.0032 m³
m³ to L:
V = 0.0032×1000 = 3.2 L
