A 0.750 g sample of an unknown gas occupies 0.300 L at 47°C and 1.10 atm. What is the molar mass of the gas? (R = 0.0821 L·atm·mol⁻¹·K⁻¹)

Correct answer

A. 58.4 g/mol

Principle or equation

The ideal gas equation is PV = nRT, where n = m/M (moles = mass / molar mass). Rearranging gives M = mRT / (PV). Temperature must be in kelvin: T(K) = T(°C) + 273.15.

Why this answer is correct

Convert temperature: T = 47 + 273.15 = 320.15 K. Use M = mRT / (PV) = (0.750 g × 0.0821 L·atm·mol⁻¹·K⁻¹ × 320.15 K) / (1.10 atm × 0.300 L). Numerator = 0.750 × 0.0821 × 320.15 ≈ 19.72 g·L·atm·mol⁻¹. Denominator = 0.330 L·atm. M ≈ 19.72 / 0.330 ≈ 59.8 g/mol, closest to 58.4 g/mol (rounding variations).

Example

For a 1.00 g gas at 300 K and 1.00 atm in 0.500 L: M = (1.00 × 0.0821 × 300) / (1.00 × 0.500) = 24.63 / 0.500 = 49.3 g/mol.