Question

17) A 8.10 L container holds a mixture of two gases at 21 °C. The partial pressures of gas A and gas B, respectively, are 0.369 atm and 0.755 atm. If 0.120 mol of a third gas is added with no change in volume or temperature, what will the total pressure become?

18) A gas mixture is made by combining 8.0 g each of Ar, Ne, and an unknown diatomic gas. At STP, the mixture occupies a volume of 18.97 L. What is the molar mass of the unknown gas?

24) At a certain temperature and pressure, 1 L of CO2 gas weighs 2.05 g. What is the mass of 1 L of CO gas at the same temperature and pressure?

Answer 1

Answer:

17.

To determine the total pressure

Consider,

PV = nRT

Temperature T = 21 + 273.15

= 294.15 K

nA = PA*VT / R*T

substitute the known values

= (0.369*8.10)/(0.082*294.15)

= 0.1239 moles

nB = PB * VT / R * T

substitute values

= 0.755*8.10 / 0.082*294.15

= 0.2535 moles

nT = nA + nB + 0.120

= 0.1239 + 0.2535 + 0.120

= 0.4974 moles

XA = nA / nT

substitute values

= 0.1239/0.4974

= 0.2491

Now consider,

PA = Ptotal * XA

substitute values

Ptotal = PA / XA

= 0.369/0.2491

= 1.4813 atm

18.

To determine the molar mass of unknown gas

Let us consider,

PV = nRT

substitute known values

1*18.97 = n*0.0821*273.15

n = 0.8459 mole

Number of moles of Argon = Mass / Molar mass

= 8 / 39.95

= 0.20 mol

Number of moles of Ne = mass / molar mass

= 8 / 20.18

= 0.3964 mol

Number of moles of unknown gas = 0.8459 - 0.20 - 0.3964

= 0.2495 mol

The molar mass of an unknown gas is = mass / Moles of unknown

= 8 / 0.2495

= 32.1 g / mol

Please post the remaining question as separate post. I hope it works for you. Thank you.