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Movement of Solvent Molecules and Osmotic Pressure Calculation

Given below are two statements :

Chamber 118 g glucose in100 mL aqueous solutionSemi-permeablemembraneChamber 230 g glucose in250 mL aqueous solution\begin{array}{|c|c|c|} \hline \begin{array}{c} \text{Chamber 1} \\ \hline 18\text{ g glucose in} \\ 100\text{ mL aqueous solution} \end{array} & \begin{array}{c} \text{Semi-permeable} \\ \text{membrane} \end{array} & \begin{array}{c} \text{Chamber 2} \\ \hline 30\text{ g glucose in} \\ 250\text{ mL aqueous solution} \end{array} \\ \hline \end{array}

Statement I : H2O\text{H}_2\text{O} molecules move from the chamber 1 to chamber 2.

Statement II : The osmotic pressure of a solution prepared by dissolving 50 mg50\text{ mg} of potassium sulphate (molar mass =174 g/mol= 174\text{ g/mol}) in 2 L2\text{ L} of water (at 27 ∘C27\,^\circ\text{C}) is 0.0107 bar0.0107\text{ bar}. (Given: R=0.083 dm3 bar K−1 mol−1\text{R} = 0.083\text{ dm}^3\text{ bar K}^{-1}\text{ mol}^{-1} and assume complete dissociation of electrolyte)

In the light of the above statements, choose the correct answer from the options given below :

Options

A

Both Statement I and Statement II are true

B

Both Statement I and Statement II are false

C

Statement I is true but Statement II is false

D

Statement I is false but Statement II is true

Correct

Topics & Concepts

SolutionsOsmotic PressureConcentration Terms

Step-by-Step Solution

To determine the correctness of Statement I and Statement II, let us analyze both statements step-by-step.

Analysis of Statement I:

Osmosis is the net flow of solvent (H2O\text{H}_2\text{O}) molecules through a semi-permeable membrane from a solution of lower solute concentration (hypotonic) to a solution of higher solute concentration (hypertonic).

  1. Concentration of Chamber 1 (C1C_1):

    • Mass of glucose =18 g= 18\text{ g}
    • Molar mass of glucose (C6H12O6\text{C}_6\text{H}_{12}\text{O}_6) =180 g mol−1= 180\text{ g mol}^{-1}
    • Moles of glucose =18180=0.1 mol= \frac{18}{180} = 0.1\text{ mol}
    • Volume of solution =100 mL=0.1 L= 100\text{ mL} = 0.1\text{ L} Molarity (C1)=0.1 mol0.1 L=1.0 M\text{Molarity } (C_1) = \frac{0.1\text{ mol}}{0.1\text{ L}} = 1.0\text{ M}

  2. Concentration of Chamber 2 (C2C_2):

    • Mass of glucose =30 g= 30\text{ g}
    • Moles of glucose =30180=16 mol≈0.1667 mol= \frac{30}{180} = \frac{1}{6}\text{ mol} \approx 0.1667\text{ mol}
    • Volume of solution =250 mL=0.25 L= 250\text{ mL} = 0.25\text{ L} Molarity (C2)=1/6 mol0.25 L=46 M=0.667 M\text{Molarity } (C_2) = \frac{1/6\text{ mol}}{0.25\text{ L}} = \frac{4}{6}\text{ M} = 0.667\text{ M}

Since C1(1.0 M)>C2(0.667 M)C_1 (1.0\text{ M}) > C_2 (0.667\text{ M}), Chamber 2 has a lower solute concentration compared to Chamber 1. Therefore, H2O\text{H}_2\text{O} molecules will move from Chamber 2 to Chamber 1.

Thus, Statement I is false.

Analysis of Statement II:

Potassium sulphate (K2SO4\text{K}_2\text{SO}_4) dissociates completely in water as: K2SO4 (aq)→2K+ (aq)+SO42− (aq)\text{K}_2\text{SO}_4\text{ (aq)} \rightarrow 2\text{K}^+\text{ (aq)} + \text{SO}_4^{2-}\text{ (aq)}

Therefore, the van 't Hoff factor (ii) is: i=3i = 3

The osmotic pressure (π\pi) of an electrolyte solution is given by: π=i⋅C⋅R⋅T=i⋅(wM⋅V)⋅R⋅T\pi = i \cdot C \cdot R \cdot T = i \cdot \left(\frac{w}{M \cdot V}\right) \cdot R \cdot T

Given parameters:

  • Mass of solute (ww) =50 mg=0.05 g= 50\text{ mg} = 0.05\text{ g}
  • Molar mass of K2SO4\text{K}_2\text{SO}_4 (MM) =174 g mol−1= 174\text{ g mol}^{-1}
  • Volume of solution (VV) =2 L=2 dm3= 2\text{ L} = 2\text{ dm}^3
  • Temperature (TT) =27 ∘C=(27+273.15) K=300 K= 27\,^\circ\text{C} = (27 + 273.15)\text{ K} = 300\text{ K}
  • Universal gas constant (RR) =0.083 dm3 bar K−1 mol−1= 0.083\text{ dm}^3\text{ bar K}^{-1}\text{ mol}^{-1}

Substituting the values into the formula: π=3×0.05 g174 g mol−1×2 dm3×0.083 dm3 bar K−1 mol−1×300 K\pi = 3 \times \frac{0.05\text{ g}}{174\text{ g mol}^{-1} \times 2\text{ dm}^3} \times 0.083\text{ dm}^3\text{ bar K}^{-1}\text{ mol}^{-1} \times 300\text{ K}

π=3×0.05×0.083×300348=3.735348≈0.01073 bar≈0.0107 bar\pi = \frac{3 \times 0.05 \times 0.083 \times 300}{348} = \frac{3.735}{348} \approx 0.01073\text{ bar} \approx 0.0107\text{ bar}

Thus, Statement II is true.

Conclusion:

  • Statement I is false.
  • Statement II is true.

Correct Option: D (Statement I is false but Statement II is true)

Movement of Solvent Molecules and Osmotic Pressure Calculation | Chemistry PYQ Solution - JEE Challenger