MHT CET · Chemistry · Solutions
Calculate osmotic pressure of solution of 0.025 mole glucose in \(100 \mathrm{~mL}\) water at \(300 \mathrm{~K}\). \(\left[\mathrm{R}=0.082 \mathrm{~atm} \mathrm{dm}^3 \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right]\)
- A \(1.54 \mathrm{~atm}\)
- B \(2.05 \mathrm{~atm}\)
- C \(6.15 \mathrm{~atm}\)
- D \(3.08 \mathrm{~atm}\)
Answer & Solution
Correct Answer
(C) \(6.15 \mathrm{~atm}\)
Step-by-step Solution
Detailed explanation
\(\pi =\mathrm{MRT}=\frac{\mathrm{n}_2 \mathrm{RT}}{\mathrm{V}} \)
\( =\frac{0.025 \mathrm{~mol} \times 0.082 \mathrm{dm}^3 \mathrm{~atm} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \times 300 \mathrm{~K}}{0.1 \mathrm{dm}^3} \)
\( =6.15 \mathrm{~atm}\)
\( =\frac{0.025 \mathrm{~mol} \times 0.082 \mathrm{dm}^3 \mathrm{~atm} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \times 300 \mathrm{~K}}{0.1 \mathrm{dm}^3} \)
\( =6.15 \mathrm{~atm}\)
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