For the equilibrium PCl_{5_{(g)}} rightleftha | vedclass

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(D) The molar mass of $PCl_5$ is $M_{theoretical} = 31 + 5 	imes 35.5 = 31 + 177.5 = 208.5 \ g/mol$.The observed molar mass $M_{observed} = 2 	imes

Vedclass · Accepted Answer

$30$

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(D) The molar mass of $PCl_5$ is $M_{theoretical} = 31 + 5 	imes 35.5 = 31 + 177.5 = 208.5 \ g/mol$.The observed molar mass $M_{observed} = 2 	imes 	ext{Vapour Density} = 2 	imes 80 = 160 \ g/mol$.The degree of dissociation $\alpha$ is given by the formula $\alpha = \frac{M_{theoretical} - M_{observed}}{M_{observed} 	imes (n - 1)}$,where $n$ is the number of moles of products formed from $1$ mole of reactant.For the reaction $PCl_5 ightleftharpoons PCl_3 + Cl_2$,$n = 2$.Substituting the values: $\alpha = \frac{208.5 - 160}{160 	imes (2 - 1)} = \frac{48.5}{160} \approx 0.303$.Converting to percentage,$\alpha \approx 30.3\% \approx 30\%$.

For the equilibrium $PCl_{5_{(g)}} \rightleftharpoons PCl_{3_{(g)}} + Cl_{2_{(g)}}$,the observed vapour density of the mixture is $80$. Given atomic masses $P = 31$ and $Cl = 35.5$,the degree of dissociation of $PCl_{5_{(g)}}$ is approximately....$\%$

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