Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
Osmotic pressure of one molar solution at $27^{\circ} \mathrm{C}$ is $(\mathrm{R}=0.082)$
ChemistrySolutionsMHT CETMHT CET 2020 (14 Oct Shift 1)
Options:
  • A $2 \cdot 46 \mathrm{~atm}$
  • B $12 \cdot 1$ atm
  • C $24 \cdot 6 \mathrm{~atm}$
  • D $1 \cdot 21$ atm
Solution:
2547 Upvotes Verified Answer
The correct answer is: $24 \cdot 6 \mathrm{~atm}$
$\begin{aligned} \pi &=\operatorname{MRT}(M=\text { Molarity }) \\ &=1 \mathrm{~mol} \mathrm{dm}^{-3} \times 0.082 \mathrm{dm}^{3} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \times 300 \mathrm{~K}=24.6 \mathrm{~atm} \end{aligned}$

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.