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QJEEM23Ch2e1e: JEE Main 2023 Chemistry - Class 12 JEEM23Ch2e1e Solution: The osmotic pressure of solutions of PVC in cyclohexanone at…

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^{-1} (Nearest integer)

JEE Main 2023 (Online) 25th January Morning Shift Chemistry - Solutions Question 39 English

(Given : R = 0.083 L atm K −1^{-1} mol −1^{-1} )

Step-by-Step Verified Solution

amp;π=M′RT=(W/MV)RTamp;⇒ π=(WV)(1M)RT=C(RTM)amp;⇒ πC=RTM=Constant\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 πC\frac{\pi}{\mathrm{C}} and C\mathrm{C} , then right graph should be like this in the question,

JEE Main 2023 (Online) 25th January Morning Shift Chemistry - Solutions Question 39 English Explanation

∴\therefore Question's graph is wrong.

Assuming π\pi vs C graph,

amp;π=CRTamp;⇒π= mole  volume ×RTamp;⇒π= mole  volume ×MM×RTamp;⇒π= mass  volume ×RTMamp;⇒π(atm)=RTM×C(glit−1)\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}

amp;So, Slope =RTM=0.083×300M=6×10−4amp;∴M=0.083×3006×10−4=830×3006amp;=41,500g/mole\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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