0.110 L of c2H4 at 360 torr and 28°c how many grams of gas
Using ideal gas equation,
PV = nRT
Where, P = pressure = 360 torr = 360/760 = 0.474 atm
V = Volume = 0.110 L
n = moles
R = Gas constant = 0.0821 atm-L/mol-K
T = Temperature = 28 °c = 28 +273 = 301 K
Putting all the values in ideal gas equation,
0.474 × 0.110 = n× 0.0821 × 310
n = 0.00205 mol
Mass = mole(n) × molar mass
Molar mass of C2H4 = 28 g
Mass = 0.00205 × 28
= 0.0573 g
Mass = [0.0573] g
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