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Determine the volume occupied by 3.2grams of oxygen at 76cm of Hg and 27^oC. (R=8.314x107 ergs/g mol-k) (a) 2.461 cm3 (b) 2461 cm3 (c) 246.1 cm3 (d) 24.61 cm3

 Determine the volume occupied by 3.2grams of oxygen at 76cm of Hg and 27^oC. (R=8.314x107 ergs/g mol-k)
(a) 2.461 cm3
(b) 2461 cm3
(c) 246.1 cm3
(d) 24.61 cm3

Grade:12th pass

1 Answers

rushwant sekar
13 Points
4 years ago
p=76*13.6*980 dyne/cm2
t=27c=300k
r=8.314*10^7
v=rt/p
when u substitute the values and simplify.. you get 
=24610 cm^3 mol^-1
since 1 mol of oxygen corresponds to 32g.therefore 32g of oxygen occupies
=24610 cm^3
1 gram of oxygen corresponds to =24610/32 cm^3
therefore 3.2g of oxygen corresponds to 
=24610*3.2/32
=2461 cm^3
thus the volume occupied by 3.2g of oxygen at 76cm of mercury at 27 celsius is = 2461 cm^3

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