`44 g` of an oxide of nitrogen gives `224 L` of nitrogen and 60 g of another oxide of nitrogen gives `22*4 L` of nitrogen at S.T.P. The data illustr

Question

`44 g` of an oxide of nitrogen gives `224 L` of nitrogen and 60 g of another oxide of nitrogen gives `22*4 L` of nitrogen at S.T.P. The data illustr

`4*4 g` of an oxide of nitrogen gives `2*24 L` of nitrogen and 60 g of another oxide of nitrogen gives `22*4 L` of nitrogen at S.T.P. The data illustrated
A. Law of conservation of mass
B. Law of constant proportion
C. Law of multiple proportions
D. Law of reciprocal proportions.

Monjur Alahi

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2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Correct Answer - C
In first oxidie,
Mass of `2*24` L of nitrogen at STP `=2*8g`
`:.` Mass of oxygen `=4*4-2*8=1*6g`
In second oxide
Mass of `22*4`L of nitrogen at STP = 28 g
`:.` Mass of oxygen = 60-28=32g
`:.` In second oxide `22*8g` of nitrogen combines with `3*2g` of oxygen.
Keeping the mass of nitrogen same in both oxides, the different masses of oxygen which combines with `2*8g` of nitrogen are `1*6 g` : `3*2` g or 1:2
This is a simple whole number ratio. This illustrates the law of multiple proportions.