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:

( $R = 0.08206\ \mathrm{L\ atm\ K^{-1}\ mol^{-1}}$ )

Question

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:

( $R = 0.08206\ \mathrm{L\ atm\ K^{-1}\ mol^{-1}}$ )

Paaras Thakur · Accepted Answer

4.92 atm. Strategy: For a mixture of non-volatile, non-electrolytic solutes, the total osmotic pressure is the sum of the partial pressures, which corresponds to the total molarity of particles (C = C1 + C2).Step 1: Calculate moles for each component- n = 0.6 / 60 = 0.01\ mol- n = 1.8 / 180 = 0.01\ mol- n = 0.02\ molStep 2: Calculate total molarity (C) = 100 mL = 0.1 L. = \{0.02\{ mol}}{0.1\{ L}} = 0.2\{ M}Step 3: Solve for Osmotic Pressure ()- T = 27 + 273 = 300\ K- R = 0.08206 = CRT = 0.2 \ 0.08206 \ 300 = 60 \ 0.08206 = 4.9236\{ atm}{\{Answer: (3)}}

Answer

8.2 atm

Answer

2.46 atm

Answer

4.92 atm

Answer

1.64 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…

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