If 1.00 mole of a gas has a mass of 18 grams, and you have 15.0 grams of the gas, what would its volume be at STP?

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Molar mass of the gas  = 18 g per mol

The mass of gas available = 15 g

Therefore, the number of moles present is given by,

Number of moles = mass / molar mass

Number of moles = 15 g/ 18 g per mol

                         = 5/6 mol

                         = 0.83333 mol.

STP - Standard Temperature & Pressure

This is 0 C and 1 bar according to IUPAC definition. At these conditions, the volume of 1 mol of ideal gas is 22.71 L/mol.

If we assume that the given gas is an ideal, volume of 1 mol at STP is 22.71 L/mol. But according to Avogadro Law volume, V is,

V = kn   Where k is constant and n is the number of moles.

If the volume of 15 g of gas is V, then,

22.71 = k x 1

V = k x (5/6)  (since it has only 5/6 moles)

V = 22.71 x (5/6) = 18.925 L


Therefore, volume of 15 g of the gas is 18.925 L.

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