Respuesta :
The composition of the mixture by volume 62.1% He and 37.9% Ar.
PV = nRT
n = mass / mw
substitute and rearrange...
PV = (mass / mw) RT
mw = (mass / V) RT/P
and since density = mass / V
mw = (0.660g /L) x (0.08206 Latm/moleK) x (298K) / (755mmHg x 1atm/760mmHg)
mw = 17.61 g/mole
mole fraction He x molar mass He + mole fraction Ar x molar mass Ar = 17.61
and if we let χHe = mole fraction He.. then (1-χHe) = mole fraction Ar.. ie.
χHe x 4.003 + (1-χHe) x 39.95 = 17.61
-35.95 χHe = - 22.34
χHe = 0.621
χAr = 1-0.621 = 0.379
now.. what about volume well... if both gases are ideal, then PV= nRT ---> V/n = RT/P.. so at the constant T and P.. V/n = a constant.. ie.. V1/n1 = V2/n2.. ie.. V1/V2 = n1/n2..meaning this...
"volume ratio = mole ratio"
so. VHe / VAr = m0les H2 / m0les Ar.
and since mole fraction He = moles He / moles total and mole fraction Ar = moles Ar / moles total...
therefore..
mole fraction He / mole fraction Ar = [moles He / (moles total)] / [ moles Ar / (moles total)] = nHe / nAr
VHe / VAr = χHe / χAr = 0.621 / 0.379...
or if you prefer...
62.1% He and 37.9% Ar
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