Question

A gas mixture is made by combining 6.5 g each of Ar, Ne, and an unknown diatomic gas. At STP, the mixture occupies a volume of 12.91 L What is the molar mass of the unknown gas? Identify the unknown gas. Number g/mol OOOOO

Answer 1

At STP. (P) Pressure = I am (T) Temperature = 273K We know, PV = nRT ģ every term has their usual significance. I atm X 12.91L = nx 0. 082 L.atm. kromol x 273k (given volume) 3. n = 0.5767 mole. Here, n= total no. of moles of mixture gireo- n= no. of moles of Art No. of moles of Net no. of moles unknown diatomic gas. Now, we know, no. of moles - given mass Molar mass 6.5g i. 6.5 g of Ar= 40g/mole Molar mass of Ar= 40 almot] 0.1625 mole Ar. i 6. 5 g of Ne = 6. 5g 20 g/mol = 0.325 mole Ne.

Hence, 0.5767 - 0.1625 + 0.325 + no. of moles of diatomie unknown. gas 7 as value of n calculated ir no. of moles of diatomic gas = 0.0892 more Hence, 0.0892 mole of the gas = 6.5g. i i mole 99 99 99 = 6.5 0.0892 = 72.867 g Hence, molar mass of the unknown gas is 72.867 g/mol Here, 72.867 ~ 7 (molar mass of 2) cl. (as calculated mass is near Hences unknown gas is the to the molar mass of el).