Osmotic pressure of solution formed by mixing \(50\, \ mL\) of \(0.2\, \ M\) urea and \(50\, \ mL\) of \(0.4\, \ m\) glucose aqueous solution at \(298\, \ K\) will be: \((R = 0.083\, \ L\, \ atm\, \ K^{-1}\, \ mol^{-1})\)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 7.42 atm
Osmotic pressure depends only on the total number of solute particles: π = CRT. Urea and glucose are non-electrolytes (i = 1), so their moles add up. Taking the 0.4 m glucose solution as 0.4 mol L−1 (dilute solution):
\( \displaystyle n_{\text{urea}} = 0.2 \times \frac{50}{1000} = 0.010\ \text{mol} \)
\( \displaystyle n_{\text{glucose}} = 0.4 \times \frac{50}{1000} = 0.020\ \text{mol} \)
\( \displaystyle C = \frac{0.010 + 0.020}{0.1\ \text{L}} = 0.3\ \text{mol L}^{-1} \)
\( \displaystyle \pi = CRT = 0.3 \times 0.083 \times 298 = 7.42\ \text{atm} \)
NCERT (2026-27 edition):
Osmosis and Osmotic Pressure – …
