JEE MAIN - Chemistry (2023 - 25th January Morning Shift - No. 17)
The osmotic pressure of solutions of PVC in cyclohexanone at 300 K are plotted on the graph.
The molar mass of PVC is ____________ g mol$$^{-1}$$ (Nearest integer)
_25th_January_Morning_Shift_en_17_1.png)
(Given : R = 0.083 L atm K$$^{-1}$$ mol$$^{-1}$$)
Answer
41500
Explanation
$\begin{aligned} & \pi=\mathrm{M}^{\prime} \mathrm{RT}=\left(\frac{\mathrm{W} / \mathrm{M}}{\mathrm{V}}\right) \mathrm{RT} \\\\ & \Rightarrow \ \pi=\left(\frac{\mathrm{W}}{\mathrm{V}}\right)\left(\frac{1}{\mathrm{M}}\right) \mathrm{RT}=\mathrm{C}\left(\frac{\mathrm{RT}}{\mathrm{M}}\right) \\\\ & \Rightarrow \ \frac{\pi}{\mathrm{C}}=\frac{\mathrm{RT}}{\mathrm{M}} =\mathrm{Constant}\end{aligned}$
If we assume graph between $\frac{\pi}{\mathrm{C}}$ and $\mathrm{C}$, then right graph should be like this in the question,
_25th_January_Morning_Shift_en_17_2.png)
$$ \therefore $$ Question's graph is wrong.
Assuming $$\pi $$ vs C graph,
$\begin{aligned} & \pi=C R T \\\\ & \Rightarrow \pi=\frac{\text { mole }}{\text { volume }} \times R T \\\\ & \Rightarrow \pi=\frac{\text { mole }}{\text { volume }} \times \frac{\mathrm{M}}{\mathrm{M}} \times \mathrm{RT} \\\\ & \Rightarrow \pi=\frac{\text { mass }}{\text { volume }} \times \frac{\mathrm{RT}}{\mathrm{M}} \\\\ & \Rightarrow \pi(\mathrm{atm})=\frac{R T}{M} \times C\left(\mathrm{g} \mathrm{lit}^{-1}\right)\end{aligned}$
$\begin{aligned} & \text {So, Slope }=\frac{\mathrm{RT}}{\mathrm{M}}=\frac{0.083 \times 300}{\mathrm{M}}=6 \times 10^{-4} \\\\ & \therefore \mathrm{M}=\frac{0.083 \times 300}{6 \times 10^{-4}}=\frac{830 \times 300}{6} \\\\ & =41,500 \mathrm{g} / \mathrm{mole}\end{aligned}$
If we assume graph between $\frac{\pi}{\mathrm{C}}$ and $\mathrm{C}$, then right graph should be like this in the question,
$$ \therefore $$ Question's graph is wrong.
Assuming $$\pi $$ vs C graph,
$\begin{aligned} & \pi=C R T \\\\ & \Rightarrow \pi=\frac{\text { mole }}{\text { volume }} \times R T \\\\ & \Rightarrow \pi=\frac{\text { mole }}{\text { volume }} \times \frac{\mathrm{M}}{\mathrm{M}} \times \mathrm{RT} \\\\ & \Rightarrow \pi=\frac{\text { mass }}{\text { volume }} \times \frac{\mathrm{RT}}{\mathrm{M}} \\\\ & \Rightarrow \pi(\mathrm{atm})=\frac{R T}{M} \times C\left(\mathrm{g} \mathrm{lit}^{-1}\right)\end{aligned}$
$\begin{aligned} & \text {So, Slope }=\frac{\mathrm{RT}}{\mathrm{M}}=\frac{0.083 \times 300}{\mathrm{M}}=6 \times 10^{-4} \\\\ & \therefore \mathrm{M}=\frac{0.083 \times 300}{6 \times 10^{-4}}=\frac{830 \times 300}{6} \\\\ & =41,500 \mathrm{g} / \mathrm{mole}\end{aligned}$
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