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.

2.66 g chloride of a gmetal when treated with silver nitrate solution

1.	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.
Answer» SOLUTION :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.