Question

2.66 g chloride of a metal when treated with silver nitrate solution give 2.87 g 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 1

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 chloride $=\frac{35.5}{143.5}\times 2.87g=0.71g$
Similarly, mass of chlorine in 5.74 g of silver chloride $=\frac{35.5}{143.5}\times 5.74=0.71g$
The mass of chlorine in first metal chloride = 0.71 g
So, the mass of metal in first metal chloride $=(2.66-0.71)=1.95g$
The mass of chlorine in second metal chloride = 1.42 g
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., 1.95 g are in the ratio of 0.71 : 1.42 or 1 : 2. It is a simple ratio. Thus, the results are in agreement with the law of multiple proportions.