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Question

The mole fraction of $$C_2H_4$$ in the mixture is :


A
0.25
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B
0.75
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C
0.45
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D
0.55
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Solution

The correct option is A $$0.75$$
The balanced reactions are as follows:
$${C_3H_8}+{5O_2}\rightarrow {3CO_2}+{4H_2O(l)}$$
$${C_3H_4}+{3O_2}\rightarrow {2CO_2}+{4H_2O(l)}$$
Initially,
$$PV=nRT$$
$$4.93\times V=(a+b)RT$$ .....(i)
After combustion, pressure is due to the total moles of $$CO_2$$.
$$11.8\times V=(3a+2b)RT$$ .....(ii)
Divide equation (ii) by equation (i), we get
$$\dfrac {11.08}{4.93}=2.25=\dfrac {3a+2b}{a+b}$$
$$2.25a+2.25b=3a+2b$$
$$0.25b=0.75a$$
$$b=3a$$
$$\chi_{C_2H_8}$$ or $$\chi_a=\dfrac {a}{a+b}=\dfrac {a}{a+3a}=\dfrac {1}{4}=0.25$$
$$\chi_{C_2H_4}$$ or $$\chi_b=1-0.25=0.75$$
Hence, the mole fraction of $$C_2H_4$$ in the mixture is $$0.75$$.

Chemistry

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