# Chemistry gas problem 1 [Solved] 1 mole of gas at STP occupies 22.4L.
0.280 L / 22.4 L = 0.0125 moles of gas.
0.400g / 0.0125 moles = 32g/mol. Molar mass of the gas.

At STP 1 mol of an ideal gas occupies 22.4 L
0.280 L / 22.4 L = 0.0125 mol
0.400 g / 0.0125 mol = 32.0 g/mol (molar mass = 32.0)

At STP 1 mol of an ideal gas occupies 22.4 L
0.280 L / 22.4 L = 0.0125 mol
0.400 g / 0.0125 mol = 32.0 g/mol (molar mass = 32.0)

PV = nRT
n = moles = mass / mw
PV = (mass / mw) RT
mw = mass x RT / PV
= 0.400 g x (.0821 Latm/moleK) x (273K) / (1 atm x 0.280 L)
= 32.g /mole

*********** alternately ***********
you could remember that 1 mole of an ideal gas has a volume of 22.41 L at STP from here.

so 0.280 L x 1 mole / 22.41 L = 0.0125 mole

so 0.280 L x 1 mole / 22.41 L = 0.0125 mole

1 mole of gas at STP occupies 22.4L.
0.280 L / 22.4 L = 0.0125 moles of gas.
0.400g / 0.0125 moles = 32g/mol. Molar mass of the gas.

assume the gas is ideal

molar mass = 0.400g / 0.0125 mole = 32.0 g/mole

1 mole of gas at STP occupies 22.4L.
0.280 L / 22.4 L = 0.0125 moles of gas.
0.400g / 0.0125 moles = 32g/mol. Molar mass of the gas.

At STP 1 mol of an ideal gas occupies 22.4 L
0.280 L / 22.4 L = 0.0125 mol
0.400 g / 0.0125 mol = 32.0 g/mol (molar mass = 32.0)

This post is last updated on hrtanswers.com at Date : 1st of September – 2022