Question

Calculate percentage change in $M_{avg}$ of the mixture, if $PC{l}_5$ undergo $50\%$ decomposition is a closed vessel.

• A. $50\%$
• B. $66.66\%$
• C. $33.65\%$
• D. $Zero$

Answer 1

The correct option is C $33.65\%$

This is because dissociation of one molecule of $PC{l}_5$ gives one molecule of $PC{l}_3$ and one molecule of $C{l}_2$.

Also, $M_{PC{l}_5}=M_{PC{l}_3}+M_{C{l}_2}$

Let, initially 2n moles of $PC{l}_5$ be present.

Then, $M_{avg}=\frac{2n\times 208.5}{2n}=208.5g/mol$

After $50$% dissociation, the mixture contains n moles of $PC{l}_5$, n moles of $PC{l}_3$ and n moles of $C{l}_2$.

Then $M_{avg}=\frac{n\times 208.5+n\times 137.5+n\times 71}{3n}=138.33g/mol$

The percentage change in $M_{avg}$ of the mixture $=\frac{208.5-138.33}{208.5}\times 100=33.65$%