MHT CET · Chemistry · Solutions
Calculate the osmotic pressure of 0.03 mole of non electrolyte solute dissolved in \(0.1 \mathrm{dm}^3\) of water at \(300 \mathrm{~K}\left[\mathrm{R}=0.0821 \mathrm{dm}^3 \mathrm{~atm} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right]\)
- A 7.4 atm
- B \(6 \cdot 4 \mathrm{~atm}\)
- C 8.0 atm
- D 5.6 atm
Answer & Solution
Correct Answer
(A) 7.4 atm
Step-by-step Solution
Detailed explanation
\(\Pi = \frac{n}{V} R T\) \(\Pi = \frac{0.03 \mathrm{~mol}}{0.1 \mathrm{~dm}^3} \times 0.0821 \mathrm{~dm}^3 \mathrm{~atm} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \times 300 \mathrm{~K}\)
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