A gas of 1.00 g mass occupies 1.00 L at 298 K and 1 atm. Estimate the molar mass from PV = nRT (use n = PV/RT).

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Multiple Choice

A gas of 1.00 g mass occupies 1.00 L at 298 K and 1 atm. Estimate the molar mass from PV = nRT (use n = PV/RT).

Explanation:
Understanding how the ideal gas law connects mass, volume, pressure, and temperature helps you find the molar mass from these measurements. Start from PV = nRT, which lets you solve for the number of moles: n = PV/(RT). Plug in the values: P = 1 atm, V = 1.00 L, T = 298 K, R = 0.082057 L atm/(mol K). So n ≈ (1 × 1.00) / (0.082057 × 298) ≈ 1 / 24.45 ≈ 0.0409 mol. The molar mass M is the mass per mole, so M = m/n = 1.00 g / 0.0409 mol ≈ 24.5 g/mol. This shows that at room temperature and 1 atm, a 1.00 g sample occupying 1.00 L corresponds to roughly 0.041 moles, giving a molar mass around 24.5 g/mol. The other numerical options are farther from this calculated value.

Understanding how the ideal gas law connects mass, volume, pressure, and temperature helps you find the molar mass from these measurements. Start from PV = nRT, which lets you solve for the number of moles: n = PV/(RT).

Plug in the values: P = 1 atm, V = 1.00 L, T = 298 K, R = 0.082057 L atm/(mol K). So n ≈ (1 × 1.00) / (0.082057 × 298) ≈ 1 / 24.45 ≈ 0.0409 mol.

The molar mass M is the mass per mole, so M = m/n = 1.00 g / 0.0409 mol ≈ 24.5 g/mol.

This shows that at room temperature and 1 atm, a 1.00 g sample occupying 1.00 L corresponds to roughly 0.041 moles, giving a molar mass around 24.5 g/mol. The other numerical options are farther from this calculated value.

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