Answer to Question #97354 in General Chemistry for stanley Eze

Question #97354
two oxides of a meta contained respectively 92.59% and 96.15% of the metal. show those facts to agree with the law of multiple proportions.
1
Expert's answer
2019-10-28T07:49:35-0400

Q97354


Solution:


The law of multiple proportions states that: If two elements form more than one compound between them, then the ratios of the masses of the second element which combine with a fixed mass of the first element will be ratios of small whole numbers. ---(1)


So, let us denote the unknown metal as "M" and let it's two oxides have formulas "MO_x" and "MO_y" .


"\\therefore MO_x" and "MO_y" contain 92.59% and 96.15% of the metal "M" respectively. (given) ---(2)


Let us assume the atomic weight of the metal "M" to be "'p'gms."

Thus, molecular weight of "MO_x=(p+16x)gms."

And molecular weight of "MO_y=(p+16y)gms."


"M+ xO\\to MO_x"

"\\implies 1 mol" or "'p'gms." of the metal reacts with "16x gms." of Oxygen. ---(3)


"M+ yO\\to MO_y"

"\\implies 1 mol" or "'p'gms." of the metal reacts with "16y gms." of Oxygen. ---(4)


From (1), (2) and (3);

For the given facts to agree with the Law of Multiple Proportions, we need to show that for a fixed amount of the metal "('p' gms.)" , the ratio of Oxygen masses in the two oxides is a whole number.

"i.e." "(16x)\/(16y)\\iff (x\/y)" is a whole number.


(2)"\\implies" "p\/(p+16x)=92.59" %

"\\implies 1\/(1+(16x\/p))=0.9259"

"\\implies 16x\/p=(0.9259)^{-1}-1" ---(5)

And,"\\implies p\/(p+16y)=96.15" %

"\\implies 1\/(1+(16y\/p))=0.9615"

"\\implies 16y\/p=(0.9615)^{-1}-1" ---(6)


Dividing (5) by (6), we get;

"x\/y=(0.0741*0.9615)\/(0.9259*0.0385)"

"=2" (Answer)

The ratio x/y is 2, which is a whole number.

Hence Proved.


Thus, the given facts agree with the Law of Multiple Proportions.






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