Question

A gas mixture is made by combining 8.8 g each of Ar, Ne, and an unknown diatomic gas. At STP, the mixture occupies a volume of 19.89 L.

Answer 1

Ar + Ne + X2

note that at STP, by definition, 1 mol of gases occupy 22.4 L

then

mol = 19.89/22.4 = 0.887946 mol of gases

MW of Ar = 39.9

mol of Ar = mass/MW = 8.8/39.9 = 0.220551 mol of Ar

Mw of Ne = 20.1

mol of Ne = mass/MW = 8.8/20.1= 0.437810mol of Ar

total mol = 0.887946

total mol = mol of Ar + mol of Ne + mol of X2

0.437810+0.220551 + mol of X2 = 0.887946

mol of X2 = 0.887946 - (0.437810+0.220551 ) = 0.229585mol of X2

since

MW = mass/mol = 8.8/0.229585 =38.3300 g/mol

since tit is diatomic

X = 38.3300/2 = 19.9, this ust be F; or fluorine