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(C) The reaction is $Cl_{2(g)} \rightleftharpoons 2Cl_{(g)}$.Let the number of moles of $Cl_2$ molecules and $Cl$ atoms be $n$ each.The total number o

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$5$

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(C) The reaction is $Cl_{2(g)} ightleftharpoons 2Cl_{(g)}$.Let the number of moles of $Cl_2$ molecules and $Cl$ atoms be $n$ each.The total number of moles in the mixture is $n + n = 2n$.The mole fraction of $Cl_2$ is $X_{Cl_2} = \frac{n}{2n} = 0.5$.The mole fraction of $Cl$ is $X_{Cl} = \frac{n}{2n} = 0.5$.Given the total pressure $P = 1 \ atm$,the partial pressures are:$P_{Cl_2} = X_{Cl_2} 	imes P = 0.5 	imes 1 = 0.5 \ atm$.$P_{Cl} = X_{Cl} 	imes P = 0.5 	imes 1 = 0.5 \ atm$.The equilibrium constant $K_P$ is given by:$K_P = \frac{(P_{Cl})^2}{P_{Cl_2}} = \frac{(0.5)^2}{0.5} = 0.5$.We are given $K_P = x 	imes 10^{-1}$,so $0.5 = x 	imes 10^{-1}$.Therefore,$x = 5$.

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