Ptotal = pN2 + pAr
You know Ptotal and pAr. Solve for pN2.
Then PV = nRT, substitute and solve for n N2 then use mols N2 = n = grams/molar mass and solve for grams.
A 255mL gas sample contains argon and nitrogen at a temperature of 65 degrees celsius. The total pressure of the sample is 725mmHg and the partial pressure of argon is 231 mmHg. what mass of nitrogen is present in the sample?
2 answers
Ptot= pN2 + pAR; (725mmHg-231mmHg=494mmHg)
494mmHg*(1atm/760mmHg)= 0.65atm
65celsius+273 = 338K
V=0.255L
R= 0.0821 Latm/mol K
Ideal Gas Law: PV=nRT
n N2 = (PV/RT) plug in above values to equation
n= (0.65atm * 0.255L) / (0.0821 Latm/molK * 338K)
moles N2 (n) = 0.00597 mols
mass N2= (0.00597 mols)/ (28.02 g/mol)= 0.167 g N2 present in sample
494mmHg*(1atm/760mmHg)= 0.65atm
65celsius+273 = 338K
V=0.255L
R= 0.0821 Latm/mol K
Ideal Gas Law: PV=nRT
n N2 = (PV/RT) plug in above values to equation
n= (0.65atm * 0.255L) / (0.0821 Latm/molK * 338K)
moles N2 (n) = 0.00597 mols
mass N2= (0.00597 mols)/ (28.02 g/mol)= 0.167 g N2 present in sample
1 answer