i really need help in this, i have slept at 3 in the morning last night doing this, and im dying right now
Mass of Co2 - 0.40g
Volume - 66.30 cm3
Pressure - 100kpa
Temp 25c (+273k)
Ideal Gas Law : PV=nrt
how do i find the molar mass?
I know how to do this and i tried, but my answer is 157.23 which is impossible, its sooooo big! i must have done something wrong. plz help
help ib chemistry! plz help ideal gas law!?
2015-09-19 2:01 am
回答 (2)
2015-09-19 4:21 am
Ideal gas equation......
PV = nRT ........ ......ideal gas equation
PV = mRT / M ....... n=m/M, where m is mass, M is molar mass
M = mRT / (PV) ..... solve for M
M = 0.40g x 8.314LkPa/molK x 298K / 100. kPa / 0.066L
M = 149 g/mol
If the mass is 0.118g instead of 0.40g, then you will get much closer to the correct molar mass of CO2.
M = 0.118g x 8.314LkPa/molK x 298K / 100. kPa / 0.066L
M = 44.1 g/mol
PV = nRT ........ ......ideal gas equation
PV = mRT / M ....... n=m/M, where m is mass, M is molar mass
M = mRT / (PV) ..... solve for M
M = 0.40g x 8.314LkPa/molK x 298K / 100. kPa / 0.066L
M = 149 g/mol
If the mass is 0.118g instead of 0.40g, then you will get much closer to the correct molar mass of CO2.
M = 0.118g x 8.314LkPa/molK x 298K / 100. kPa / 0.066L
M = 44.1 g/mol
2015-09-19 2:26 am
n / V = P / RT = (100 kPa) / ((8.3144621 L kPa/K mol) x (25 + 273) K) = 0.04036 mol/L
(0.40 g) / (0.06630 L) = 6.03318 g/L
(6.03318 g/L) / (0.04036 mol/L) = 149 g/mol
So Ricki is correct, and something is very wrong with your numbers.
(0.40 g) / (0.06630 L) = 6.03318 g/L
(6.03318 g/L) / (0.04036 mol/L) = 149 g/mol
So Ricki is correct, and something is very wrong with your numbers.
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