A 4.2 mol sample of Kr has a volume of 953 mL. How many moles of Kr are in a 4.9

Question

A 4.2 mol sample of Kr has a volume of 953 mL. How many moles of Kr are in a 4.90 L sample at the same temperature and pressure?

_______mol

A flexible container at an initial volume of 8.15 L contains 8.51 mol of gas. More gas is then added to the container until it reaches a final volume of 11.7 L. Assuming the pressure and temperature of the gas remain constant, calculate the number of moles of gas added to the container.

____________mol

Calculate the volume, in liters, occupied by 1.27 mol of oxygen gas at STP.

__________L

If 1.63 moles of an ideal gas has a pressure of 9.31 atm, and a volume of 38.05 L, what is the temperature of the sample?

__________K

A sample of carbon dioxide is contained in a 500.0 mL flask at 0.924 atm and 19.9 °C. How many molecules of gas are in the sample?

__________molecules

A mixture of He, Ne, and Ar has a pressure of 11.3 atm at 28.0 °C. If the partial pressure of He is 1.39 atm and that of Ar is 3.24 atm, what is the partial pressure of Ne?

___________atm

A 8.15-L container holds a mixture of two gases at 27 °C. The partial pressures of gas A and gas B, respectively, are 0.446 atm and 0.620 atm. If 0.180 mol of a third gas is added with no change in volume or temperature, what will the total pressure become?

_____________atm

Explanation / Answer

1) A 4.2 mol sample of Kr has a volume of 953 mL. How many moles of Kr are in a 4.90 L sample at the same temperature and pressure?

Solution :- 4.2 mol = 953 ml = 0.953 L

Therefore 4.90 L = ?

4.90 L * 4.2 mol / 0.953 L = 21.6 mol gas

So it has 21.6 mol gas

2) A flexible container at an initial volume of 8.15 L contains 8.51 mol of gas. More gas is then added to the container until it reaches a final volume of 11.7 L. Assuming the pressure and temperature of the gas remain constant, calculate the number of moles of gas added to the container.

Solution :- 8.15 L = 8.51 mol

                  11.7 L = ? mol

11.7 L *8.51 mol /8.15 L = 12.22 mol

So the moles of the gas added = 12.22 mol – 8.51 mol = 3.71 mol

So the moles of gas added = 3.71 mol

3) Calculate the volume, in liters, occupied by 1.27 mol of oxygen gas at STP.

Solution :- at STP 1 mol gas = 22.4 L

Therefore 1.27 mol gas * 22.4 L / 1 mol = 28.4 L

So the volume of gas = 28.4 L

4) If 1.63 moles of an ideal gas has a pressure of 9.31 atm, and a volume of 38.05 L, what is the temperature of the sample?

Solution :-

n=1.63 mol , P = 9.31 atm , V = 38.05 L , Temperature T= ?

PV= nRT

PV/nR = T

9.31 atm * 38.05 L /1.63 mol * 0.08206 L atm per mol K = T

2648 K = T

So the temperature = 2648 K

5) A sample of carbon dioxide is contained in a 500.0 mL flask at 0.924 atm and 19.9 °C. How many molecules of gas are in the sample?

Solution :- V= 500 ml = 0.500 L

P = 0.924 atm

T= 19.9 C +273 = 292.9 K

Lets calculate the moles of the gas

PV= nRT

n= PV/RT

n= 0.924 atm *0.500 L / 0.08206 L atm per mol K * 292.9 K

n = 0.01922 mol

now lets convert moles of CO2 to molecules

0.01922 mol * 6.02*10^23 molecules / 1 mol = 1.16*10^22 molecules of CO2

So it contains 1.16*10^22 molecules.

6) A mixture of He, Ne, and Ar has a pressure of 11.3 atm at 28.0 °C. If the partial pressure of He is 1.39 atm and that of Ar is 3.24 atm, what is the partial pressure of Ne?

Solution :- total pressure = sum of partial pressure of the gases in the mixture

Total pressure = P He + P Ar P Ne

P Ne = total pressure – ( P He + P Ar)

         = 11.3 atm – ( 1.39 atm + 3.24 atm)

         = 6.67 atm

So the partial pressure of the Ne = 6.67 atm

7) A 8.15-L container holds a mixture of two gases at 27 °C. The partial pressures of gas A and gas B, respectively, are 0.446 atm and 0.620 atm. If 0.180 mol of a third gas is added with no change in volume or temperature, what will the total pressure become?

Solution :-

Lets first calculate the moles of the gas present in the container

PV= nRT

n= PV/RT

n= (0.446 atm + 0.620 atm ) * 8.15 L / 0.08206 L atm per mol K * 300 K

n= 0.3529 mol gases

now total moles after adding the 0.180 mol of third gas are 0.3529 mol + 0.180 mol = 0.5329 mol

now lets calculate the final pressure

0.5329 mol * 1.066 atm / 0.3529 mol = 1.61 atm

So the final pressure = 1.61 atm

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