JEE MainChemistry

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Solutions: JEE Main Chemistry Question with Solution

Q: Why does the slope equal RT/M instead of just RT?

The x-axis represents mass concentration in g/L ( ), not molar concentration ( ). Since , we have , meaning the slope of vs is .

The osmotic pressure of solutions of PVC in cyclohexanone at

300\mathrm{~K}

are plotted on the graph. The molar mass of PVC is

\underline{\quad\quad\quad}\mathrm{gmol}^{-1}

(Nearest integer) (Given :

\mathrm{R}=0.083\mathrm{~L}\mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}

)

A graph with y-axis as \frac{\pi}{C} (\text{atm}/\text{g L}^{-1}) and x-axis as C (\text{g L}^{-1}), showing a straight line with Slope = 6 \times 10^{-4}.

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Correct answer

The molar mass of PVC is determined to be $41500\text{ g mol}^{-1}$ using the relation between the slope of the $\Pi$ versus $C$ plot and $\frac{RT}{M}$ .

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01given

Temperature $T = 300\text{ K}$ , gas constant $R = 0.083\text{ L atm K}^{-1}\text{mol}^{-1}$ , and a linear plot of osmotic pressure $\pi$ (in atm) versus mass concentration $C = w/V$ (in $\text{g L}^{-1}$ ) with slope $m = 6 \times 10^{-4}$ .

02find

The molar mass $M$ of PVC in $\text{g mol}^{-1}$ .

03strategise

Osmotic pressure is given by van 't Hoff's law: $\pi = CRT/M$ , where $C = w/V$ is mass concentration. Plotting $\pi$ against $C$ gives a straight line through the origin with slope $\text{slope} = \frac{RT}{M}$ . Thus, $M = \frac{RT}{\text{slope}}$ .

04execute

Substitute $R = 0.083$ , $T = 300$ , and $\text{slope} = 6 \times 10^{-4}$ into $M = \frac{RT}{\text{slope}}$ : $M = \frac{0.083 \times 300}{6 \times 10^{-4}} = \frac{24.9}{6 \times 10^{-4}} = 41500\text{ g mol}^{-1}$

✓verify

A molar mass around $4.15 \times 10^4\text{ g mol}^{-1}$ is typical for a synthetic polymer like PVC.

Hints that build this answer step by step

What relates the osmotic pressure $\Pi$ (in $\text{atm}$ ) to polymer concentration $C$ (in $\text{g L}^{-1}$ ) and molar mass $M$ (in $\text{g mol}^{-1}$ )?
$\Pi = \frac{C}{M}RT$ , meaning the slope of $\Pi$ versus $C$ equals $\frac{RT}{M}$
Given the plot passes through the origin and $(C = 50\text{ g L}^{-1},\, \Pi = 3.0 \times 10^{-2}\text{ atm})$ , what is the slope of the line?
$6.0 \times 10^{-4}\text{ atm L g}^{-1}$
Using $\text{slope} = \frac{RT}{M}$ with $R = 0.083\text{ L atm K}^{-1}\text{mol}^{-1}$ and $T = 300\text{ K}$ , what is the value of $M$ ?
$41500\text{ g mol}^{-1}$

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Chemistry
Concepts assessed: Chemistry
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 8 September 2026

Quick checks

Students also ask

Why does the slope equal RT/M instead of just RT?

The x-axis represents mass concentration in g/L ( $C = w/V$ ), not molar concentration ( $n/V$ ). Since $n = w/M$ , we have $\pi = (w/V)(RT/M) = C(RT/M)$ , meaning the slope of $\pi$ vs $C$ is $RT/M$ .

Answer

The molar mass of PVC is determined to be $41500\text{ g mol}^{-1}$ using the relation between the slope of the $\Pi$ versus $C$ plot and $\frac{RT}{M}$ .

Why each option works or fails

Step-by-step solution

given: Temperature $300\mathrm{~K}$ , gas constant $R = 0.083\text{ L atm K}^{-1}\text{mol}^{-1}$ , and a linear plot of osmotic pressure $\pi$ (in atm) versus mass concentration $C = w/V$ (in $\text{g L}^{-1}$ ) with slope $m = 6 \times 10^{-4}$ .
find: The molar mass $M$ of PVC in $\text{g mol}^{-1}$ .
strategise: Osmotic pressure is given by van 't Hoff's law: $\pi = CRT/M$ , where $C = w/V$ is mass concentration. Plotting $\pi$ against $C$ gives a straight line through the origin with slope $\text{slope} = \frac{RT}{M}$ . Thus, $M = \frac{RT}{\text{slope}}$ .
execute: Substitute $R = 0.083$ , $T = 300$ , and $\text{slope} = 6 \times 10^{-4}$ into $M = \frac{RT}{\text{slope}}$ : $M = \frac{0.083 \times 300}{6 \times 10^{-4}} = \frac{24.9}{6 \times 10^{-4}} = 41500\text{ g mol}^{-1}$
verify: A molar mass around $4.15 \times 10^4\text{ g mol}^{-1}$ is typical for a synthetic polymer like PVC.

Solutions: JEE Main Chemistry Question with Solution

The molar mass of PVC is determined to be 41500 g mol−141500\text{ g mol}^{-1}41500 g mol−1 using the relation between the slope of the Π\PiΠ versus CCC plot and RTM\frac{RT}{M}MRT​.

01given

02find

03strategise

04execute

✓verify

Students also ask

The molar mass of PVC is determined to be $41500\text{ g mol}^{-1}$ using the relation between the slope of the $\Pi$ versus $C$ plot and $\frac{RT}{M}$ .