Part A) The density of a sample of N2O gas is 3.00 g/L at 298 K.What is the pres
ID: 950407 • Letter: P
Question
Part A) The density of a sample of N2O gas is 3.00 g/L at 298 K.What is the pressure of the gas (in mmHg)?
Part B) An experiment shows that a 110 mL gas sample has a mass of 0.173 g at a pressure of 751 mmHg and a temperature of 30 C.
What is the molar mass of the gas?
Part C) A gas has a density of 2.07 g/L at a temperature of 24 C and a pressure of 0.787 atm . Calculate the molar mass of the gas.
Express the molar mass to three significant figures and include the appropriate units.
Part D) Calculate the density of argon gas at a pressure of 749 mmHg and a temperature of 50 C.
Express the density to three significant figures and include the appropriate units.
Explanation / Answer
Part A) The density of a sample of N2O gas is 3.00 g/L at 298 K.What is the pressure of the gas (in mmHg)?
Answer :
density = MM P /RT
where density = 3 g/L
MM = molar mass of N2O = 44.01 g/mole
P = ? atm
R = 0.0821 L atm/K/mole
T = 298 K
putting the values...
3 = 44.01 X P / 0.0821 X 298
3 = 44.01 X P / 24.466
P = 3 X 24.466 / 44.01 = 73.398 / 44.01 = 1.668 atm
now 1 atm = 760 mm Hg
so 1.668 atm = 760 X 1.668 = 1267.68 mm Hg
2) An experiment shows that a 110 mL gas sample has a mass of 0.173 g at a pressure of 751 mmHg and a temperature of 30 C.
What is the molar mass of the gas?
Answer :
PV = nRT
PV = (mass / molar mass)RT
PV/RT = mass / molar mass
molar mass = mass / (PV/RT)
Solution:
molar mass = 0.173 g / ([751/760 atm][110/1000 mL]/[0.0821][303 K])
molar mass = 39.67 g/mol
Or we can solve another way
number of moles of gas = PV / Rt
= 751 x .110 / 62.363 x 30+273= 0.00433 moles
which weigh .173 g so mass of 1 mole = .173 / 0.00436
=39.67 g / mole
Part C) A gas has a density of 2.07 g/L at a temperature of 24 C and a pressure of 0.787 atm . Calculate the molar mass of the gas.
Answer : density = 2.07 g/L, temperature = 24 C =(24.0 + 273.15) = 297K.
pressure = 0.787 atm,V= 0.001 L (which is 1 mL converted to liters)
PV = nRT
n = PV / RT = (0.787 x 0.001) / (0.082057 x 297 )
n = 3.22 x 10¯5 mol
Molar mass = (2.07 g/L) / (3.22 x 10¯5 mol)
Molar mass = 64.28 g/mol
Part D) Calculate the density of argon gas at a pressure of 749 mmHg and a temperature of 50 C.
Express the density to three significant figures and include the appropriate units.
PV = nRT
where
P = pressure
V = volume
n = number of moles of gas
R = Gas constant (0.0821 L·atm/mol·K)
T = absolute temperature
If we solve the equation for volume, we get:
V = (nRT)/P
we know everything we need to find the volume now except the number of moles of gas. To find this, remember the relationship between number of moles and mass.
n = m/MM
where
n = number of moles of gas
m = mass of gas
MM = molecular mass of the gas
This is helpful since we needed to find the mass and we know the molecular mass of oxygen gas. If we substitute for n in the first equation, we get:
V = (mRT)/(MMP)
Divide both sides by m:
V/m = (RT)/(MMP)
But density is m/V, so flip the equation over to get:
m/V = (MMP)/(RT) = density of the gas.
Now we need to insert the values we know.
MM of argon gas is 39.948 grams/mole
P = 74mmHg =749/760=0.9855 atm
T = 50 °C, but we need absolute temperature.
TK = TC + 273
T = 50 + 273 = 323 K
m/V = (39.948 g/mol x 0.9855 atm)/(0.0821 L·atm/mol·K x 323 K)
m/V = 39.368/26.51g/L
m/V = 1.48 g/L
Answer: The density of the argon gas is 1.48 g/L.