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(C) The osmotic pressure $\pi$ is given by the formula $\pi = CRT$,where $C$ is the total molarity of the solution.First,calculate the moles of urea:

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$4.92$

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(C) The osmotic pressure $\pi$ is given by the formula $\pi = CRT$,where $C$ is the total molarity of the solution.First,calculate the moles of urea: $n_{	ext{urea}} = \frac{0.6 \ g}{60 \ g \ mol^{-1}} = 0.01 \ mol$.Next,calculate the moles of glucose: $n_{	ext{glucose}} = \frac{1.8 \ g}{180 \ g \ mol^{-1}} = 0.01 \ mol$.Total moles of solute $= 0.01 + 0.01 = 0.02 \ mol$.The volume of the solution is $100 \ mL = 0.1 \ L$.Total molarity $C = \frac{0.02 \ mol}{0.1 \ L} = 0.2 \ M$.Temperature $T = 27 \ ^oC = 27 + 273 = 300 \ K$.Now,calculate osmotic pressure: $\pi = 0.2 \ mol \ L^{-1} 	imes 0.08206 \ L \ atm \ K^{-1} \ mol^{-1} 	imes 300 \ K$.$\pi = 0.2 	imes 0.08206 	imes 300 = 4.9236 \ atm$.

$A$ 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 $27 \ ^oC$. The osmotic pressure of the solution is : ............. $atm$ $(R = 0.08206 \ L \ atm \ K^{-1} \ mol^{-1})$

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