wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
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.

Open in App
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 chloride =35.5143.5×2.87g=0.71g
Similarly, mass of chlorine in 5.74 g of silver chloride =35.5143.5×5.74=0.71g
The mass of chlorine in first metal chloride = 0.71 g
So, the mass of metal in first metal chloride =(2.660.71)=1.95g
The mass of chlorine in second metal chloride = 1.42 g
So, the mass of metal in second metal chloride =(3.371.42)=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.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Laws of Chemical Combination
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon