The percent of O in Cr₂O₃ : 31.58%
<h3>Further explanation</h3>
Given
Cr = 52.00 amu, O = 16.00 amu
Required
The percent of O
Solution
MW Cr₂O₃ = 2 x Ar Cr + 3 x Ar O
MW Cr₂O₃ = 2.52+3.16
MW Cr₂O₃ =152 amu
Use the universal gas formula
PV=nRT
where
P=pressure ( 0.980 atm)
V=volume (L)
T=temperature ( 23 ° C = 23+273.15 = 296.15 ° K)
n=number of moles of ideal gas (0.485 mol)
R=universal gas constant = 0.08205 L atm / (mol·K)
Substitute values,
Volume, V (in litres)
=nRT/P
=0.485*0.08205*296.15/0.980
= 12.0256 L
= 12.0 L (to three significant figures)
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