# Chemistry gas problem 1?

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At STP, 0.280 L of a gas weighs 0.400 g. Calculate the molar mass of the gas.

• molar mass = g / mole right?. you know grams, if you could find moles you could calculate molar mass…

assume the gas is ideal…

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….

V = nRT/P = 1 mole x .0821 Latm/moleK x 273 K/ 1 atm = 22.41 L

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

finally

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)

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