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 decreasing order of osmotic pressure for the given solutions, we must understand the concept of osmotic pressure. Osmotic pressure is a colligative property, meaning it depends on the number of solute particles in a solution and not on the nature of the solute itself. The formula for osmotic pressure \((\pi)\) is given by: 

\(\pi = iCRT\)

• \(i\) is the van't Hoff factor (number of particles the solute dissociates into)
• \(C\) is the molar concentration of the solution
• \(R\) is the universal gas constant
• \(T\) is the temperature in Kelvin

In the given problem, the solute and solvent amounts are the same for each solution, so the concentration \((C)\) can be calculated using:

Molar mass of glucose \((C_{6}H_{12}O_{6})\) = 180 g/mol

Molar mass of urea \((CH_{4}N_{2}O)\) = 60 g/mol

Molar mass of sucrose \((C_{12}H_{22}O_{11})\) = 342 g/mol

For glucose \((P_1)\), \(10\, g\) of glucose is dissolved. The number of moles of glucose is:

\(\text{Moles of glucose} = \frac{10}{180} = 0.0556 \text{ mol}\)

For urea \((P_2)\), \(10\, g\) of urea is dissolved. The number of moles of urea is:

\(\text{Moles of urea} = \frac{10}{60} = 0.1667 \text{ mol}\)

For sucrose \((P_3)\), \(10\, g\) of sucrose is dissolved. The number of moles of sucrose is:

\(\text{Moles of sucrose} = \frac{10}{342} = 0.0292 \text{ mol}\)

Since all the solutes are non-ionic compounds without dissociation in water, the van’t Hoff factor \((i)\) for all is 1. Thus, the osmotic pressure is directly proportional to the number of moles of solute present in the solution.

Comparing the moles of the solute, we have:

• Urea has the greatest number of moles (0.1667 mol), so \(P_2\) will have the highest osmotic pressure.
• Glucose has the next greatest number of moles (0.0556 mol), so \(P_1\) will have an intermediate osmotic pressure.
• Sucrose has the least number of moles (0.0292 mol), so \(P_3\) will have the lowest osmotic pressure.

Thus, the decreasing order of osmotic pressure is:

\(P_2 > P_1 > P_3\)

Therefore, the correct answer is:

\(P_2 > P_1 > P_3\)

Answer 2

$\pi=\frac{W}{M} S T$ $\pi \propto \frac{1}{\text { Mol.wt }}$ Glucose $(180)$, Urea $(60)$, Sucrose $(342)$ $\pi_{\text {urea }}>\pi_{\text {glucose }}>\pi_{\text {sucrose }}$ $P_{2}>P_{1}>P_{3}$