Can someone help me with this chemistry problem please??

Question

A 5.00 liter sample of gas measured at 27.0 degree C and 1.25 atm of pressure has a mass of 10.13 g. What is the molecular weight of the gas in u??

Please show me how you get it also. Thank you!!
If you know, please answer, if you don't, please don't guess. I'd appreciated very much.

oh by the way here are the answer for the problem


a) 39.9
b) 0.898
c) 2.57
d) 2.50 x 10 exp -2

Answer 1

PV=nRT
P=1.25 atm
V=5l
T=27C=300K
R=0.0820574587 latm/K/mole
n=PV/RT=(1.25atm*5l)/(300K*0.0820574587 l atm/K/mole)
n=.254moles

.254moles weighs 10.13g
10.13g/.254moles=39.9g/mole

Answer 2

Find the moles of the gas solving the state equation of gases:

p*V = n*R*T

for n (number of moles):

n = p*V/R*T

where R = 0.082 L*Atm/mol*K and T = 27 + 273 = 300 K the absolute temperature:

n = 1.25*5/0.082*300, n = 0.254 mol (approx).

Now m = n*Mw, where m the mass and Mw the molecular weight of the gas. Solving for Mw we get:

Mw = m/n, Mw = 10.13/0.254, Mw = 39.87 g/mol or 39.9 approx.

So the right answer is (a)

Answer 3

Well, according to the equation of the ideal gas:
P*V = nRT
27,0º C + 273,15 = 300,2 K
n= # of moles
R= gas constant

(5,00L*1,25atm)/(0,08206L-atm/mol-K)(300,2 K) = n
n = 0,254 moles
If you have 10,13 g in 0,254 moles, then
10,13g/0,254 moles = 39,9g/mol

Answer 4

sure i think it's easy but may be kind of confusing read your questions carefully then you get it