mixture hydrogen

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Tutorial questions (5th May 2014)
Chemistry NS104
1. A mixture of hydrogen H2 (1.01 g) and chlorine Cl2 (17.73 g) in a container
at 300 K has a total gas pressure of 98.8 kPa. What is the partial pressure of
hydrogen and chlorine in the mixture?
ANSWER: Moles of H2 =n= mass/molar mass = 1.02/2.02 = 0.500 mol
Moles of Cl2 =n= mass/molar mass = 17.73/70.90 = 0.250 mol
Total moles = 0.500 + 0.250 = 0.750 mol.
Total pressure = 98.8 kPa.
Partial pressure of each gas is proportional to its mole fraction in the mixture
Therefore partial pressure of H2 and Cl2 is calculated with the following
formula :
P= Xn
X
Ptotal
Where mole fraction is calculated as :
XH = nH / ntotal = (0.500/0.750) =0.66
Xcl2 = ncl2 / ntotal = (0.250/0.750) = 0.33
PH= XH X Ptotal = 0.66 x 98.8 = 65.2kPa
PCl2= XCl2 X Ptotal = 0.33 x 98.8 = 32.9kPa
2. What is the density (in g/L) of a sample of SO2 at 932 mmHg and 65°C ?
(R= 0.0821 L . atm/ K . mol , 1atm= 760 mmHg)
First convert units :
T= 65C (converter to K) = 338K
P= 932 mmHg (converted to atm) = 1.22atm
Than, calculate the Molar mass of SO2, which is ≈ 64.1 g/ mol
The formula for density (accroding ideal gas law) is :
/
d= P X M / R X T = 1.23 atm X 64.1 g/ mol 0.0821 L atm/ K mol .
338 K
d= 2.84 g/L
𝑴=
𝒎𝑹𝑻
𝑷𝑽
=
𝒎𝑹𝑻
𝑽𝑷
=
𝒎
𝑽
𝒙
p=
𝑹𝑻
𝑷
𝑴𝑷
𝑹𝑻
Because
Prepared by Jasmin Sutkovic
5th May 2014
=𝒑𝒙
𝒎
𝑽
=𝒑
𝑹𝑻
𝑷
, so we get
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