Question

Calculate percentage change in ${{M}_{avg}}$ of the mixture, if $PC{{l}_{5}}$ undergo \[50%\] decomposition is a closed vessel.
$A.\,\,50%$
\[B.\,\,66.66%\]
$C.\,\,33.65%$
$D.\,\,0%$

Answer 1

Hint:Average molecular mass of a mixture is the sum of products of each component’s moles present and corresponding molecular mass, whole divided by the total number of moles. So, to find ${{M}_{avg}}$ we only need the number of moles of each constituent present. Also, molecular mass of $PC{{l}_{5}}=208.5$

Complete step by step answer:
Consider that at the beginning, $2n$ moles of $PC{{l}_{5}}$ is present. \[50%\] decomposition means that after a certain time, only $n$ moles of $PC{{l}_{5}}$ is left. We can write the equation to get better clarity on this:
$PC{{l}_{5}}\rightleftharpoons PC{{l}_{3}}+C{{l}_{2}}$

Time t=0	$2n$	$0$	$0$
Time t=t	$n$	$n$	$n$

Therefore, after a certain time, as $n$ moles have dissociated, $n$ moles of $PC{{l}_{3}}$ and $n$ moles of $C{{l}_{2}}$ will be formed, in accordance with the stoichiometry of the equation.
Hence, average molecular mass at this point:
${{({{M}_{avg}})}_{t}}=\dfrac{\Sigma (number\text{ }of\text{ }moles\text{ }of\text{ }each\text{ }component\times molecular\text{ }mass\text{ }of\text{ }each\text{ }component)}{total\text{ }number\text{ }of\text{ }moles}$
Where ${{({{M}_{avg}})}_{t}}$ is the average molecular mass after time $t$
Molecular mass of $PC{{l}_{3}}=137.5$and that of $C{{l}_{2}}=71$
Substituting these values, we get:
${{({{M}_{avg}})}_{t}}=\dfrac{(208.5\times n)+(137.5\times n)+(71\times n)}{3n}$
Taking $n$ outside from the numerator, we can cancel it off with the $n$ in the denominator. Solving further, we get:
${{({{M}_{avg}})}_{t}}=\dfrac{208.5+137.5+71}{3}$
Therefore, we get: ${{({{M}_{avg}})}_{t}}=139$
Therefore, percentage change in ${{M}_{avg}}=\dfrac{{{({{M}_{avg}})}_{0}}-{{({{M}_{avg}})}_{t}}}{{{({{M}_{avg}})}_{0}}}\times 100$
Where ${{({{M}_{avg}})}_{0}}$ is the initial average molecular mass.
Initially, as only $PC{{l}_{5}}$ was present, ${{({{M}_{avg}})}_{0}}$ is equal to the molecular mass of $PC{{l}_{5}}=208.5$
Substituting the values, we get:
$%\text{ change}=\dfrac{208.5-139}{208.5}\times 100=\dfrac{69.5}{208.5}\times 100$
On solving, we get:
$%\text{change }=33.65%$
Hence, the correct option is option \[\left( C \right)\].

Additional information: Problems based on dissociation, equilibrium etc. are easily solved by the knowledge of the reaction involved. So, make sure to write the correct balanced equation first and then make a table as shown in the above problem to ease the process of problem solving.

Note:
Note that for computing the average molecular mass of the compound, we have also taken the average atomic masses of the individual elements present, such as chlorine. We can also do this question by taking the initial number of moles as $n$, in which case ${n}/{2}\;$ moles of each component would be formed after time $t$. But in this case too, the final answer we receive will be the same, as percentage change is a constant for the specified conditions. While computing average molecular mass, there is an alternate formula which will yield the same result.
${{({{M}_{avg}})}_{t}}=\Sigma \text{(mole fraction of each component}\times \text{molecular mass of each component)}$
Where the mole fraction of each component is computed as the number of moles of that component at equilibrium divided by the total number of moles. Through this method, we are actually just redefining and reasserting the use of mole fraction.