May 2024
Intermediate to advanced
872 pages
32h 3m
English
book
JEE Advanced Chemistry | Physical Chemistry - I | Includes JEE 2021-2024 Questions | Revised 2nd Edition | - Pearson
by Ananya Ganguly
Content preview from JEE Advanced Chemistry | Physical Chemistry - I | Includes JEE 2021-2024 Questions | Revised 2nd Edition | - Pearson
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Solved ProblemS
Problem 1
Annalysis showed that (a) 0.2 g of oxide was formed by heating 0.12 g of a metal in air. (b) The carbonate and nitrate of the metal contained 28.5 per cent and 16.2 per cent of metal, respectively. Applying the law of constant proportions, calculate the amount of metallic oxide when 1 g each of carbonate and nitrate is heated in air.
Solution
Wt. of the oxide = 0.2 g and the wt. of metal = 0.12 g
∴Wt. of O2 = (0.2 − 0.12) = 0.08 g
So, the ratio of metal to O2 in the oxide
= 
In 100 g of carbonate, the amount of metal is 28.5 g
In 1 g of carbonate, the amount of metal is
=
Now, according to the law of definite proportions ...
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