A sample of a breathing mixture for divers contained 34.3% helium, He; 51.7% nitrogen, N2; and...
A sample of a breathing mixture for divers contained 34.3% helium, He; 51.7% nitrogen, N2; and 14.0% oxygen, O2 (by mass). What is the density of this mixture at 23°C and 772 mmHg?
d= g/L
Homework Answers
34.3% He mean 34.3g of He present in 100g of gases
no of moles of He (nHe) = W/G.A.Wt
= 34.3/4 = 8.575 moles
51.7% He mean 51.7g of N2 present in 100g of gases
no of moles of N2 (nN2) = W/G.M.Wt
= 51.7/28 = 1.85moles
14% O2 mean 14g of O2 present in 100g of gases
no of moles of O2( nO2) = W/G.M.Wt
= 14/32 = 0.4375 moles
Total no of moles of mixture (n) = nHe + nN2 + nO2
= 8.575 + 1.85 + 0.4375
= 10.8625moles
Total 100g of gases mixture
T = 23+273 = 296K
P = 722/760 = 0.95atm
n = 10.8625moles
PV = nRT
V = nRT/P
= 10.8625*0.0821*296/0.95
= 277.87L
mass of gases mixture = 100g
The density of gases mixture = mass/volume in L
= 100/277.87 = 0.36g/L >>>>answer
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