At 752.3 torr and 20.35oC, what is the density of argon in g/L? Report your answer with 4 sig figs
I got 1.875 using d=PM/RT, but it says wrong. I don't know why...
I appreciate your help.
Show your work and I'll find your error. Did you use K = 273.15 + 20.35?
Did you use 0.08205 for R?
Did you convert 752.3 to atm?
I think the answer is more like 1.6 something. Perhaps you just pushed the wrong buttons.
yea, i got the K value wrong.
Now I get 1.643.
Thanks.
To calculate the density of a gas using the ideal gas law, you are on the right track with the formula:
d = PM / RT
where:
d is the density of the gas in g/L
P is the pressure of the gas in atm (torr can be converted to atm by dividing by 760)
M is the molar mass of the gas in g/mol
R is the ideal gas constant, which is 0.0821 L·atm/(mol·K)
T is the temperature of the gas in Kelvin (K = °C + 273.15)
In this case, you have the pressure (752.3 torr) and temperature (20.35°C), so let's first convert them to the correct units:
Pressure: 752.3 torr ÷ 760 torr/atm = 0.9908 atm
Temperature: 20.35°C + 273.15 = 293.5 K
Now, you need to find the molar mass of argon (Ar). The molar mass of Ar is approximately 39.95 g/mol.
Using the values you have:
d = PM / RT
= (0.9908 atm) * (39.95 g/mol) / (0.0821 L·atm/(mol·K) * 293.5 K)
Calculating this expression should give you the correct answer for the density of argon in grams per liter (g/L). Remember to round your answer to 4 significant figures.