Question

3.A gas has a density of 2.03 g L-1 at 48 °C and P = 0.998 atm. What will be its density at 160 °C? Calculate the molar mass of the gas, assuming it obeys the ideal gas law.

4.A gas mixture contains 11.6 mol of C2H2 and 20.8 mol of H2. (a) Compute the mole fraction of C2H2 in the mixture. (b) A catalyst is then added to the mixture, and the C2H2 starts to react with the H2 to give C2H6:

Answer 1

3)

a)

Lets derive the equation to be used

we have below equation to be used:

p*V=n*R*T

p*V=(mass/molar mass)*R*T

p*molar mass=(mass/V)*R*T

p*molar mass=density*R*T

so,

density1*T1 = density2*T2

WE HAVE:

T1 = 48.0 oC

= (48.0+273) K

= 321 K

T2 = 160 oC

= (160 + 273) K

= 433 K

use:

density1*T1 = density2*T2

2.03*321 = density2*433

density2 = 1.50 g/L

Answer: 1.50 g/L

b)

P = 0.998atm

T= 48.0 oC

= (48.0+273) K

= 321 K

density = 2.03 g/L

Use:

p*molar mass=density*R*T

Put Values:

0.998 atm * MM = 2.03g/L * 0.08206 atm.L/mol.K *321.0 K

MM = 53.5799 g/mol

Answer: 53.6 g/mol

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