Question

/2 b) If 1.000 mol of gas occupies 22.711L at STP, and the mass of the vapour in the flask is 530 mg, what is the molar mass of the vapour?

Answer 1

$\because$PV = nRT (Ideal gas equation)

$\therefore$ ( P₁V₁ / n₁T₁ ) = ( P₂V₂ / n₂T₂ )

STP conditions : - P₁ = 760 mm Hg , T₁ = 273 K , V₁ = 22.711 L , n₁ = 1 mol

Laboratory conditions : - P₂ = 650 mm Hg , T₂ = 97° C = (97 + 273) = 370 K , V₂ = 0.1454 L , n₁ = (m/M) mol = (0.530/M) mol (where m is given mass and M is molar mass, both in grams)

Substituting the above values, we get : -

760 x 22.711 650 x 0.1454 0.530 1 x 273 x 370 М

760 x 22.711 650 x 0.1454 x M 0.530 x 370 1 x 273

760 x 22.711х0.530 х 370 М 1x 273 х 650x 0.1454

M = 131.19 grams