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What is the molecular weight (in g/mol) of a pure gaseous compound if the gas has a density of 5.21 g/l at -20⁰C and 1,467 torr? write your final answer in two decimal places.

User Airfang
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Final answer:

The molecular weight of a gaseous compound with a known density, temperature, and pressure can be determined using the Ideal Gas Law. By rearranging the equation to M = (density × R × T)/P and substituting the given values, we calculate the molecular weight to be approximately 55.86 g/mol.

Step-by-step explanation:

To determine the molecular weight of a gaseous compound given the data on density and conditions of temperature and pressure, the Ideal Gas Law can be used in conjunction with the density formula.

The Ideal Gas Law is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin. The density (ρ) can be defined as the mass (m) divided by the volume (V), so ρ = m/V. From the density formula, we can express m as ρV. Replacing 'n' in the Ideal Gas Law with m/M (where M is the molar mass) and rearranging the equation for M gives us M = mRT/(PV).

Firstly, let's convert the given temperature and pressure to the standard units used in the Ideal Gas Law:

  • Temperature in Kelvin: T = -20°C + 273.15 = 253.15 K
  • Pressure in atm (1 atm = 760 torr): P = 1467 torr / 760 torr/atm ≈ 1.93 atm

The density (ρ) is already given in g/L, which is the proper unit. Now we can rearrange the formula to solve for the molar mass (M): M = (ρ × R × T)/P. Substituting R = 0.0821 L·atm·mol⁻¹·K⁻¹ (the ideal gas constant) and the values for ρ, T, and P into the equation, we get:

M = (5.21 g/L × 0.0821 L·atm·mol⁻¹·K⁻¹ × 253.15 K) / (1.93 atm) ≈ 55.86 g/mol, rounded to two decimal places.

User Mrwalker
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