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Standard XII

Chemistry

Electronegativity and Hybridisation

9.9 g of an a...

Question

$9.9 g$ of an amide with molecular formula $C_{4} H_{5} N_{x} O_{y}$ on heating with alkali liberated $1.7 g$ of ammonia. If the percentage of oxygen is $32.33$ %, then the ratio of $N$ and $O$ atoms in the compound is

2 : 1

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1 : 2

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2 : 5

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2 : 3

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Solution

The correct option is A $1 : 2$
Since the molecular formula of the amide is $C_{4} H_{5} N_{x} O_{y}$ , therefore, on heating with alkali, 1 mole of the amide will produce x moles of $N H_{3}$
=17x g of $N H_{3}$ .
Now, 1.7g of $N H_{3}$ is obtained from amide=9.9g
Therefore 17xg of $N H_{3}$ will be obtained from amide
$= \frac{9.9}{1.7} \times 17 x g = 99 x g$
Therefore molecular weight of amide=99x
% of N= $\frac{14 x}{99 x} \times 100 = \frac{1400}{99}$
Their Number of N atoms (x) in one molecule of amide
$= \frac{1400}{99 \times 14} = 1.01$
% of O = 32.33 (given)
Therefore number of O atoms (y) in one molecule of amide
$= \frac{32.33}{16} = 2.02$
Ratio of N and O atoms, i.e.,
$x : y = 1.01 : 2.02 = 1 : 2$

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9.9 g of an amide with molecular formula C4H5NxOy on heating with alkali liberated 1.7g of ammonia. If the percentage of oxygen is 32.33%, then the ratio of N and O atoms in the compound is

$9.9 g$ of an amide with molecular formula $C_{4} H_{5} N_{x} O_{y}$ on heating with alkali liberated $1.7 g$ of ammonia. If the percentage of oxygen is $32.33$ %, then the ratio of $N$ and $O$ atoms in the compound is