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`2.66 g` of chloride of a metal when treated with silver nitrate solution, gave `2.87 g` of silver chloride `3.37 g` of another chloride of the same metal gave `5.74 g` of silver chloride when treated with silver nitrate solution. Show that the results illustrate the law of multiple proportions. |
Answer» `AgNO_(3)+MClrarrMNO_(3)+underset(("white ppt"))(AgCl)` Molecular mass of `AgCl=(108+35.5)=143.5g` In the first sample `143.5 g` of `AgCl` contain `Cl = 35.5 g` `2.87 g` of `AgCl` contain `Cl=((35.5g))/((143.5g))xx(2.87g)=0.71g` Now, chlorine `(Cl)` present in `AgCl` has actually been obtained from metal chloride `:.` Mass of metal (M) in the sample `= (2.66-0.71) = 1.95 g` In the second sample `143.5 g` of `AgCl` contain `Cl = 35.5 g` `5.74 g` of `AgCl` contain `Cl=((35.5g))/((143.5g))xx(5.74g)=0.42g` Mass of metal (M) in the sample `= (3.37 - 1.42) = 1.95 g` From the above data, it is clear that in both the samples of metal chlorides, mass of metal `(M) = 1.95 g` `:.` Ratios of the masses of chlorine `(Cl)` which combine with a fixed mass of metal `= 0.71 : 1.42 or 1:2`. As the ratio is a simple whole number ratio therefore, the law of multiple proportions is illustrated. |
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