rushwant sekar
Last Activity: 7 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
