1. Find the mass, in grams, of a 240 mL sample of dryair at a pressure of 750 mmHg and a temperature of 19°C. Refer tothe chart on the previous page for dry air density.

Answer/ PV=NRT, so you get((750/760)*(240/1000))/ (.0821*292)= N
Then N*28.97(molar mass of air) is your answer, I got0.286 grams (3 significant figures) for my answer.

2. An unknown gas sample is collected in a 255 mL containerat 25°C and 755 mmHg. How many moles of gas are present?
answer/ Volume of gas = 255 mL = 0.255 L, temperature =25°C +273 = 298 k , pressure = 755 mmHg = 755/760 atm= 0.9934atm
N=PV/RT = (0.9934)x (0.255 L) / (0.0821) x (298 k) =0.01035

3. If the gas sample of question 2 was weighed and foundto have a mass of 0.456 grams, what is the molar mass of thegas?
answer/ Molar mass of gas = given mass of gas/ moles of gas= 0.456g/0.01035= 44.058 g/mol

4. What was the density in grams per liter of the gas ofquestion 2? Give mass of gas (in g) / volume of container ( in L)=answer/ 0.456g/0.255= 1.788 g/L

DATA
a) Mass of the flask, stopper, and air : 115.808 g
b) Temperature in the room: 20.2 °C
c) Mass of the flask, stopper, and lab gas : 115.698 g
d) Volume of the flask measured water in the cylinder : 257ml
e) Room pressure in inHg : 29.88 in Hg
f) Density of dry air from the chart: 2.24

Report and Calculations
g) Mass of the air in the flask ________________
h) Mass of the flask and stopper without air ________________
i) Mass of the lab gas sample ________________
j) Density of the lab gas in grams per liter ________________
k) Room pressure in atmospheres ________________
l) Volume of the flask in liters ________________
m) Temperature of the lab gas in Kelvins ________________
n) Number of moles of lab gas from the ideal gas law ________________
o) Molar mass of the lab gas ________________