Calculate the density, g/cm3, of ethane C2H6 at 108.0oC and 809 torr.?

Question

enter an answer in exponential notation use e, example: 1.47e-8

Answer 1

Ethane is a gas and in these conditions can be treated as an ideal gas. We need to assume an amount. Any amount will do because the density of a substance is independent of the quantity of the substance. Let's assume 1 mole of C2H6. This has a mass of 30.1 g.

m=30.1 g
n= 1.00 mol
T = 381.1 K
P = 809 Torr = 1.06 atm

We need the volume to calculate the density:

V=nRT/P= (1.00 mol) (0.08206 L-atm/mol-K)(381.1K)/1.06 atm

V = 29.5 L = 29500 ml = 29500 cm^3

D= m/V = 30.1 g / 29500 cm^3 = 1.02 x 10^-3 g/cm^3

Answer 2

108C = 381K The molar volume is 22.4L for a perfect gas at STP. Atomic weights: C=12 H=1 C2H6=30

30gC2H6/22.4LC2H6 x 273K/381K x 809torr/760torr = 1.0 g per L to two sig figs.