4.008g of pure KClO3 was quantitatively decomposed to give 2.438g of KCl and oxygen. KCl was dissolved was in H2O and treated with AgNO3 sol. The result was a precipitate of AgCl weighing 4.687g. Under further treatment AgCl was found to contain 3.531g of Ag. What are the atomic weights of Ag, Cl & K relative to O=16?

Asked by Vidushi412 | 26th Sep, 2018, 02:01: PM

Expert Answer:

Given:
 
Weight of KClO3 =4.008 g
 
Weight of KCl + Oxygen =2.438 g
 
Weight of AgCl =3.531 g 
 
Atomic weight of Ag = ?
 
No. space of space moles space of space KCl space equals fraction numerator Weight space of space KCl over denominator Molecular space weight space of space KCl end fraction

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals fraction numerator 2.438 over denominator 74.55 end fraction

No. space of space moles space of space KCl space equals space 0.0327 space moles

No. space of space moles space of space AgCl space equals space 0.0327 space moles

Mass space of space Ag space in space AgCl space equals space 3.531 space straight g

therefore space Mass space of space Cl space in space AgCl space equals 4.687 minus space 3.531

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 1.156 space straight g

Mass space of space straight K space in space KCl space equals space 2.438 space minus 1.156

space space space space space space space space space space space space space space space space space space space space space space space space space space equals 1.282 space straight g

No. space of space moles space of space AgCl space equals space No. space of space moles space of space Ag space equals space No. ofmoles space of space Cl

No. space of space moles space of space Ag space equals space fraction numerator Mas space of space Ag space over denominator Atomic space weight space of space Ag end fraction

space space space space space space space space space space space space space space space space space space space space 0.033 space equals space fraction numerator 3.531 over denominator Atomic space weight space of space Ag end fraction space space space space space

Atomic space weight space of space Ag space equals space 107 space

Also comma

Atomic space weight space of space Cl space equals fraction numerator 1.156 over denominator 0.033 end fraction

Atomic space weight space of space Cl space equals space 35.03

Atomic space weight space of space straight K space equals space fraction numerator 1.282 over denominator 0.033 end fraction

Atomic space weight space of space straight K space equals 38.84
Atomic weight of Ag = 107
 
Atomic weight of Cl = 35.03
 
Atomic weight of K =38.84

Answered by Varsha | 26th Sep, 2018, 06:45: PM