Question

The following solutions were prepared by dissolving $10 g$ of glucose $\left( C _{6} H _{12} O _{6}\right)$ in $250 \,ml$ of water $\left( P _{1}\right)$ $10 \,g$ of urea $\left( CH _{4} N _{2} O \right)$ in $250 \,ml$ of water $\left( P _{2}\right)$ and $10 g$ of sucrose $\left( C _{12} H _{22} O _{11}\right)$ in $250\, ml$ of water $\left( P _{3}\right)$. The right option for the decreasing order of osmotic pressure of these solutions is:

• A. $P _{2}> P _{1}> P _{3}$
• B. $P _{1}> P _{2}> P _{3}$
• C. $P _{2}> P _{3}> P _{1}$
• D. $P _{3}> P _{1}> P _{2}$

Answer 1

The Correct Option is A

To determine the order of osmotic pressure for the given solutions, we need to understand the concept of osmotic pressure and how it is related to the concentration of solute particles. Osmotic pressure (\(\pi\)) is given by the formula:

\(\pi = iCRT\)

where:

• \(i\) is the van't Hoff factor (which is 1 for non-electrolytes like glucose, urea, and sucrose as they do not dissociate in solution).
• \(C\) is the molarity of the solution.
• \(R\) is the ideal gas constant.
• \(T\) is the absolute temperature in Kelvin.

Since all solutions are prepared in the same volume of water, the temperature and \(R\) are constant for all, and \(i\) is the same (1), osmotic pressure mainly depends on the molarity of the solute.

1. Calculate the molar mass of each solute:
   • Glucose (\(C_6H_{12}O_6\)): Molar mass = \(6(12) + 12(1) + 6(16) = 180 \, \text{g/mol}\)
   • Urea (\(CH_4N_2O\)): Molar mass = \(1(12) + 4(1) + 2(14) + 1(16) = 60 \, \text{g/mol}\)
   • Sucrose (\(C_{12}H_{22}O_{11}\)): Molar mass = \(12(12) + 22(1) + 11(16) = 342 \, \text{g/mol}\)
2. Calculate the molarity for each solution using the formula:

\(\text{Molarity (M)} = \frac{\text{mass of solute (g)}}{\text{molar mass (g/mol)}} \times \frac{1000}{\text{volume of solution (ml)}}\)

• For glucose: \(\text{M} = \frac{10}{180} \times \frac{1000}{250} = \frac{1}{45} \times 4 = \frac{4}{45} \, \text{mol/L}\)
• For urea: \(\text{M} = \frac{10}{60} \times \frac{1000}{250} = \frac{1}{6} \times 4 = \frac{4}{6} = \frac{2}{3} \, \text{mol/L}\)
• For sucrose: \(\text{M} = \frac{10}{342} \times \frac{1000}{250} = \frac{1}{34.2} \times 4 \approx \frac{4}{34.2} \, \text{mol/L}\)

1. Compare the molarities:
   • Urea: \(\frac{2}{3} \, \text{mol/L} \approx 0.666 \, \text{mol/L}\)
   • Glucose: \(\frac{4}{45} \, \text{mol/L} \approx 0.089 \, \text{mol/L}\)
   • Sucrose: \(\frac{4}{34.2} \, \text{mol/L} \approx 0.117 \, \text{mol/L}\)
2. Order the osmotic pressures based on molarity:
   • \(P_2\) (urea) has the highest molarity, hence the highest osmotic pressure.
   • \(P_1\) (glucose) has a higher molarity than \(P_3\) (sucrose).
   • Thus, the decreasing order of osmotic pressure is: \(P_2 > P_1 > P_3\)

The correct option is \(P _{2}> P _{1}> P _{3}\).