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Question

Question #1
A 1335 mL sample of oxygen (O2) was collected by displacement of water at 24 degrees celsius under a total barometric pressure of 425 torr. What mass of dry oxygen was collected? The vapor pressure of water at 24 degrees celsius is 22 torr.

Question #2
We mix 129 g of oxygen gas with 24 g of argon gas at standard temperature and pressure. What will be the final volume of the gas mixture?

Question #3
A gas occupying 89 mL at standard conditions is heated to 25 degrees celsius while the pressure is reduced to 0.83035 atm. What is the new volume occupied by the gas?

Question #4
How many molecules are contained in 1442 mL of O2 gas at -10 degrees celsius and 508 torr?

Question #5
Calculate the volume occupied by 64.0 grams of CH4 at 127 degrees celsius under a pressure of 1581 torr.

Answer 1

1. 

Total pressure=vapor pressure of water + gas pressure of sample.

 

425 torr=oxygen pressure + 22.377 torr

 

402.62 torr=gas pressure of oxygen.

 

Turn torr into atm:  402.62 torr*(1 atm/760 torr) or 0.52976 atm

 

Use PV=nRT equation to find mass of oxygen (solve for n)

 

P=0.52976 atm, V=1.335 L, T=273.15 + 24 K or 297.15 K, n=?

 

n=(PV)/(RT) or (0.52976atm*1.335L)/(0.08206 L-atm/mol-K*297.15K)

 

n=0.029 mole

 

2.

At STP, P=1 atm, T=273.15 K, V=?

 

Volume=Volume of Oxygen Gas + Volume of Argon Gas

 

V=(nRT)/P

 

moles of Oxygen=129 g/31.999 g per mole or 4.03 moles O2 gas.

 

moles of Argon=24 g/15.7596 g per mole or 1.52 moles Ar gas.

 

add moles together and plug into formula V=nRT/P

 

Volume=(4.03 + 1.52 moles)*(0.08206 L-atm/mole-K)*273.15 K/1 atm

 

Volume=124.4 mL