Question #43445

What is the density of helium gas at 25 degree celsius and 760 torr?
R= 0.082 Lxatm/molxk
AW= 4 g/mole
a) 0.16 g/L
b) 6.1 g/L
c) 124 g/L
d) 0.082 g/L

Expert's answer

According to the ideal gas law,

PV = nRT

n = m/M - number of moles

d = m/V - density

Hence,

P = (^{m}/_{MV})RT

PM = dRT

d = PM/RT

760 torr = 1 atm

d = (1 atm x 4 g/mol)/(0.082 Lxatm/molxK x 298 K) = 0.164 g/L

So, the correct option is (a): 0.16 g/L

PV = nRT

n = m/M - number of moles

d = m/V - density

Hence,

P = (

PM = dRT

d = PM/RT

760 torr = 1 atm

d = (1 atm x 4 g/mol)/(0.082 Lxatm/molxK x 298 K) = 0.164 g/L

So, the correct option is (a): 0.16 g/L

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