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Solutions - Osmotic pressure

Grade 12ICSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Osmosis is the spontaneous flow of solvent molecules through a semi-permeable membrane (SPM) from a region of lower solute concentration (pure solvent) to a region of higher solute concentration.

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Osmotic Pressure (π\pi) is defined as the minimum excess pressure that must be applied to the solution side to prevent the entry of the solvent through the semi-permeable membrane.

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Osmotic pressure is a colligative property because it depends on the number of solute particles and not on their identity.

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Van't Hoff Equation: For dilute solutions, the osmotic pressure is directly proportional to the molarity (CC) and the absolute temperature (TT).

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Isotonic solutions are two solutions having the same osmotic pressure at a given temperature. If one solution has lower osmotic pressure, it is 'Hypotonic', and if it has higher, it is 'Hypertonic'.

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Reverse Osmosis (RO) occurs when the pressure applied to the solution side is greater than the osmotic pressure, causing solvent molecules to move from the solution to the pure solvent side.

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For electrolytes that undergo association or dissociation, the Van't Hoff factor (ii) must be included: π=iCRT\pi = iCRT.

📐Formulae

π=CRT\pi = CRT

π=nBVRT\pi = \frac{n_B}{V}RT

π=wBRTMBV\pi = \frac{w_B RT}{M_B V}

MB=wBRTπVM_B = \frac{w_B RT}{\pi V}

π=iCRT\pi = iCRT

💡Examples

Problem 1:

Calculate the osmotic pressure of a 5%5\% (w/v) solution of urea (NH2CONH2NH_2CONH_2) at 273 K273\ K. (Given R=0.0821 L atm K−1 mol−1R = 0.0821\ L\ atm\ K^{-1}\ mol^{-1}, Molar mass of urea = 60 g/mol60\ g/mol)

Solution:

  1. Concentration (CC) in mol/Lmol/L: 5%5\% w/v means 5 g5\ g of urea in 100 mL100\ mL of solution. Therefore, wB=5 gw_B = 5\ g, V=100 mL=0.1 LV = 100\ mL = 0.1\ L.
  2. Moles of urea (nBn_B) = wBMB=560=0.0833 mol\frac{w_B}{M_B} = \frac{5}{60} = 0.0833\ mol.
  3. Molarity (CC) = nBV=0.08330.1=0.833 M\frac{n_B}{V} = \frac{0.0833}{0.1} = 0.833\ M.
  4. Using π=CRT\pi = CRT: π=0.833×0.0821×273\pi = 0.833 \times 0.0821 \times 273.
  5. π≈18.67 atm\pi \approx 18.67\ atm.

Explanation:

The osmotic pressure is calculated using the Van't Hoff equation by first converting the weight/volume percentage into molarity (mol/Lmol/L) and then applying the absolute temperature.

Problem 2:

200 cm3200\ cm^3 of an aqueous solution of a protein contains 1.26 g1.26\ g of the protein. The osmotic pressure of such a solution at 300 K300\ K is found to be 2.57×10−3 bar2.57 \times 10^{-3}\ bar. Calculate the molar mass of the protein. (R=0.083 L bar K−1 mol−1R = 0.083\ L\ bar\ K^{-1}\ mol^{-1})

Solution:

  1. Given: π=2.57×10−3 bar\pi = 2.57 \times 10^{-3}\ bar, V=200 cm3=0.2 LV = 200\ cm^3 = 0.2\ L, T=300 KT = 300\ K, wB=1.26 gw_B = 1.26\ g.
  2. Formula: MB=wBRTπVM_B = \frac{w_B RT}{\pi V}.
  3. Calculation: MB=1.26×0.083×3002.57×10−3×0.2M_B = \frac{1.26 \times 0.083 \times 300}{2.57 \times 10^{-3} \times 0.2}.
  4. MB=31.3740.000514≈61038 g/molM_B = \frac{31.374}{0.000514} \approx 61038\ g/mol.

Explanation:

Osmotic pressure is the preferred method for determining the molar masses of macromolecules like proteins and polymers because the pressure changes are large enough to be measured accurately even at very low concentrations.

Osmotic pressure - Revision Notes & Key Formulas | ICSE Class 12 Chemistry