Question

0.110 L of c₂H₄ at 360 torr and 28°c how many grams of gas

Answer 1

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