Question

Calculate the osmotic pressure of 0.03 mole of non electrolyte solute dissolved in 0.1 dm³ of water at 300 K. [R = 0.082 dm³ atm mol^-1 K^-1]

विकल्प

• 7.4 atm

• 6.4 atm

• 8.0 atm

• 5.6 atm

Answer 1

7.4 atm

Explanation:

π = MRT

\[\pi=\frac{\mathrm{n}_{2}\mathrm{RT}}{\mathrm{V}}\]

\[=\frac{0.03\mathrm{~mol}\times0.082\mathrm{~atm}\mathrm{dm}^3\mathrm{K}^{-1}\mathrm{mol}^{-1}\times300\mathrm{~K}}{0.1\mathrm{~dm}^3}\]

= 7.38 atm ≈ 7.4 atm