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Certain non-metal X forms two oxides I and II . The mass percentage of oxygen in I(X4 O6) is 43.7% which is the same as that of X in the 2nd oxide.Find the formula of 2nd oxide. (Based on law of multiple proportions...... pls xplain each step carefully)
Dear Ajay,
let 'm' be atomic mass of X. let formula of second oxide be XuOv
mass %age of oxygen in I=96/(96+4m)=24/(m+24)=0.437,m+24=24/0.437,m=31(approx.)
31u/(31u +16v)=0.437,31u/0.437=31u+16v
31/0.437=31+16(v/u)
v/u=2.5,therefore II is X2O5
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