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Question

11.6 g of an organic compound having formula CnH2n+2 is burnt in excess of O2(g) initially taken in a 22.41 litre steel vessel. Before reaction, the gaseous mixture was at 273 K with pressure reading 2 atm. After complete combustion and loss of considerable amount of heat, the mixture of product and excess of O2 had a temperature of 546 K and 4.6 atm pressure. The formula of organic compound is:

A
C2H6
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B
C3H8
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C
C5H12
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D
C4H10
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Solution

The correct option is D C4H10
CnH2n+2+(3n+12)O2nCO2+(n+1)H2O
Let initial pressure of CnH2n+2 is P, then increase in pressure :
=P[(2n+1)1(3n+12)]
=(n12)P
For a constant volume of the steel vessel, TP
So,
At 546 K pressure is 4.6 atm
at 5462=273 K pressure is 4.62=2.3 atm;
Increase in pressure= (2.32) atm=0.3 atmR=ideal gas constant=0.0821 L.atm mol1 K1V=volume of steel vessel=22.41 L
P=nRTV=11.6M×(0.0821×27322.41)
P=11.6M
where M = Molecular weight of organic compound
M=n×12+(2n+2)×1M=14n+2
Since,
(n12)P=0.3
Putting P=11.614n+2
(n12)×11.6(14n+2)=0.3
n114n+2=0.611.6
n=4
Compound is C4H10

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