Question

1. A mixture containing 2.81 g each of CH4 (g), C2H4 (g) and C4H10 (g) is contained in a 1.50 L flask at a temperature of 35 oC. What is the partial pressure (in atm) of C4H10 (g)?

2. A sample of gas contains 62.0 moles at 327 K in 4.00 L of volume. What is the pressure (in atm) of the gas?

3. What is the molar mass of 572 mg of vapor in 0.500 L at 824 torr and 298 K?

Answer 1

1)first calculate mole of each

mole of CH₄ =mass in gm/molar mass = 2.81/16.04 = 0.175187mole

mole of C₂H₄ =mass in gm/molar mass = 2.81/ 28.05 =0.100178mole

mole of C₄H₁₀ =mass in gm/molar mass = 2.81/58.12 =0.048348mole

total mole = 0.175187+0.100178+0.048348 = 0.323713 mole

Use ideal gas equation & calculate total pressure

PV = nRT

P = nRT/V

= 0.3237130.08205Latm mol^-1K^-1308.15K/1.5L

total pressure = 5.456atm

partial pressure of C₄H₁₀ = mole of C₄H₁₀total pressure/total mole of gas mixture

= 0.0483485.456/0.323713

partial pressure of C₄H₁₀ = 0.81494atm

2)

Use ideal gas equation & calculate total pressure

PV = nRT

P = nRT/V

= 620.08205Latm mol^-1K^-1327K/4L

P = 415.87atm

3)824 torr = 1.08421 atm

PV = nRT

n = PV/RT

= 1.08421atm0.5L/0.08205Latm mol^-1K^-1298K

n = 0.02217 mole

no. of mole = mass/molar mass

molar mass = mass/no. of mole   

= 572/0.02217

= 258000mg = 25.8 gm

molar mass of vapour = 25.8 gm/mole