2.66 g chloride of a gmetal when treated with silver nitrate solution give 2.87g of silver chloride. 3.37 g of another chloride of the same metal give 5.74 g of silver chloride when treated with silver nitrate solution. Show that the results are in agreement with a law of chemical combination.

Explanation

Molecular mass of silver chloride
`=(108+35.5)=143.5g`
Mass of chlorine in 143.5 g of silver chloride =35.5 g
So, the mass of chlorine in 2.87g of silver chlorine.
`=(35.5)/(143.5)xx2.87g=0.71g`.
Similarly, mass of chlorine in 5.74g of silver chloride.
`=(35.5)/(143.5)=1.42g`
The mass of chlorine in first metal chloride =0.71g
So, the mass of metal in first metal chloride.
`=(2.66-0.71)=1.95g`
The mass of chlorine in second metal chloride=1.42g
So, the mass of metal in second metal chloride
`=(3.37-1.42)=1.95g`
In both the chlorides, the mass of the metal is same, but the masses of chlorine combining with the same mass of metal i.e., 1.95g are in the ratio of 0.71:1.42 or 1:2. it is a simpler atio. Thus, the results are in agreement with the law of multiple proportions.