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0.22 g of organic compound C(x)H(y)O whi...

0.22 g of organic compound `C_(x)H_(y)O` which occupied 112 mL at NTP and on combustion gave 0.44 g `CO_(2)`. If the percentage of oxygen is 36.45%, then the ratio of x to y in the compound is:

A

`1:1`

B

`1:2`

C

`1:3`

D

`1:4`

Text Solution

Verified by Experts

The correct Answer is:
B
Molar mass of compound `=(wRT)/(PV)= (0.22 xx0.0821 xx273 xx1000)/(1xx 112) = 44`
0.44g `CO_2` will contain carbon `=12/44xx0.44g`
This is the mass of carbon present in 0.22 gof the compound.
% of C in the compound `= 12/44 xx0.44xx100/0.22 =54.54%`
% of H = 100 - 54.54 – 36.45 = 9.01%
Then, the ratio of number of atoms of .C. and .H. = `(54.54)/12 : (9.01)/1=1:2`
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