1. Tardigrade
  2. Question
  3. Chemistry
  4. A solution is prepared by dissolving 0.6 g of urea (molar mass = 60 g mol-1) and 1.8 g of glucose (molar mass = 180 g mol-1) in 100 mL of water at 27°C. The osmotic pressure of the solution is: (R = 0.08206 L atm K-1 mol-1 )

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Q. A solution is prepared by dissolving $0.6\, g$ of urea (molar mass = $60\, g\, mol^{-1})$ and $1.8\, g$ of glucose (molar mass = $180\, g$ mol${-1}$) in $100\, mL$ of water at $27^{\circ}C$. The osmotic pressure of the solution is :
$(R = 0.08206 \; L \; atm \; K^{-1} \; mol^{-1} ) $

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Solution:

$\Pi = \frac{\left(\frac{0.6}{60} + \frac{1.8}{180}\right)}{0.1} \times0.08206 \times30$
$ \Pi =4.9236\, \text{atm}$