Q.200 cm$^3$ of an aqueous solution of a protein contains 1.26 g of the protein. The osmotic pressure of such a solution at 300 K is found to be $2.57 \times 10^{-3}$ bar. Calculate the molar mass of the protein.

Showing the 12 most recent of 21 on this concept.

1. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.At 300 K, a decimolar solution of potassium ferrocyanide is 50% dissociated. The osmotic pressure (in atm) of the solution is ($R = 0.082\ L\ atm\ K^{-1}mol^{-1}$) (A) 6.83 (B) 3.87 (C) 7.38 (D) 7.83

   ›Reveal solutionSolution

   Potassium ferrocyanide gives 5 ions on complete dissociation; with $\alpha=0.5$, $i=3$, giving $\pi = iCRT = 7.38$ atm.

   Concept and Intuition

   Osmotic pressure of an electrolyte solution is corrected by the van't Hoff factor $i$, which accounts for the extra particles produced on dissociation: $\pi = iCRT$, where $i = 1+(n-1)\alpha$, $n$ being the number of ions per formula unit and $\alpha$ the degree of dissociation.

   $\mathrm{K_4[Fe(CN)_6]} \rightleftharpoons 4\mathrm{K^+} + [\mathrm{Fe(CN)_6}]^{4-}$ gives $n=5$ ions in total.

   Step-by-Step Solution

   1. $n = 5$ (4 K⁺ + 1 ferrocyanide ion), $\alpha = 0.5$.
   2. $i = 1+(5-1)(0.5) = 1+2 = 3$.
   3. $C = 0.1$ M (decimolar), $T=300$ K, $R=0.082$ L atm K⁻¹ mol⁻¹.
   4. $\pi = iCRT = 3\times0.1\times0.082\times300$. …

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2. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.At 27°C, x g of $C_6H_{12}O_6$ (molar mass = 180 g mol$^{-1}$) and y g of a non-volatile, non-electrolyte (molar mass = 92 g mol$^{-1}$) were present separately in 1.0 L solutions. The osmotic pressure of two solutions is equal. What is x/y? (A) 45/23 (B) 23/45 (C) 32/54 (D) 54/32

   ›Reveal solutionSolution

   Equal osmotic pressure at the same temperature means equal molar concentration; since both solutions occupy the same 1.0 L volume, this reduces to equal moles of solute. The answer is 45/23.

   Concept and Intuition

   Osmotic pressure of a dilute solution is given by van't Hoff's equation $\pi = CRT$, analogous to the ideal gas law, where $C$ is the molar concentration of solute particles. For two non-electrolyte solutes (no dissociation, so van't Hoff factor $i=1$) at the same temperature, equal osmotic pressures directly imply equal molar concentrations $C$.

   Step-by-Step Solution

   1. For glucose: $C_1 = \dfrac{x/180}{1.0}= \dfrac{x}{180}$ mol/L (moles divided by 1.0 L).
   2. For the other solute: $C_2 = \dfrac{y/92}{1.0} = \dfrac{y}{92}$ mol/L. …

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3. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.A centi molar solution of acetic acid is 50% dissociated at $27°C$. The osmotic pressure of the solution (in atm) is ($R = 0.083\ \mathrm{L\, atm\, K^{-1}\, mol^{-1}}$) (A) 0.37 (B) 3.7 (C) 0.037 (D) 0.73

   ›Reveal solutionSolution

   Applying the van't Hoff factor for a partially dissociated weak electrolyte to the osmotic-pressure formula gives π ≈ 0.37 atm. Answer: (A).

   Concept and Intuition

   Acetic acid is a weak electrolyte: $\mathrm{CH_3COOH \rightleftharpoons CH_3COO^- + H^+}$, dissociating into 2 particles per molecule. Because dissociation is only partial (here 50%), the effective number of particles in solution is more than 1 but less than 2 per formula unit — captured by the van't Hoff factor $i = 1+\alpha(n-1)$, which then scales the ideal (undissociated) osmotic pressure formula.

   Step-by-Step Solution

   1. "Centi molar" means $C = 0.01$ mol/L.
   2. Acetic acid dissociates into $n=2$ particles (the acetate ion and the proton).
   3. Degree of dissociation $\alpha = 0.5$ (50%).
   4. Van't Hoff factor: $i = 1 + \alpha(n-1) = 1 + 0.5(2-1) = 1 + 0.5 = 1.5$.
   5. Osmotic pressure with the van't Hoff correction: $\pi = iCRT$.
   6. $T = 27°C = 300$ K, $R = 0.083$ L atm K⁻¹ mol⁻¹. …

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4. AP EAPCET 2025Set ap-2025-05-19-AN1 markMCQ
   Q.A 1% (w/v) aqueous solution of a certain solute is isotonic with a 3% (w/v) solution of glucose (molar mass 180 g mol$^{-1}$). The molar mass of solute (in g mol$^{-1}$) is (A) 120 (B) 60 (C) 360 (D) 80

   ›Reveal solutionSolution

   Isotonic solutions share the same molar concentration; equating the glucose and solute molarities from their w/v percentages gives molar mass $= 60\ \text{g/mol}$.

   Concept and Intuition

   Two solutions are isotonic when they generate the same osmotic pressure at the same temperature, which (for non-electrolytes, same van't Hoff factor) means they have the same molar concentration ($\pi = CRT$, and $RT$ is identical for both solutions at the same $T$).

   Step-by-Step Solution

   1. Glucose solution: 3% w/v means 3 g of glucose per 100 mL, i.e. 30 g per litre. Molar concentration $= \dfrac{30}{180} = \dfrac16\ \text{mol/L} \approx 0.1667\ \text{M}$.
   2. Solute solution: 1% w/v means 1 g per 100 mL, i.e. 10 g per litre. Molar concentration $= \dfrac{10}{M}\ \text{mol/L}$, where $M$ is the unknown molar mass. …

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5. AP EAPCET 2024Set ap-2024-05-16-FN1 markMCQ
   Q.At 27 °C, $x$ g of $CaCl_2$ was dissolved in 2.5 L of water. The osmotic pressure of resultant solution is 0.82 atm. What is $x$ in g? ($i = 2.5$; $R = 0.082\ \mathrm{L\,atm\,mol^{-1}K^{-1}}$) (A) 37 (B) 1.85 (C) 3.7 (D) 18.5

   ›Reveal solutionSolution

   Using the van't Hoff osmotic-pressure equation with the given $i$, solve for moles of $CaCl_2$, then convert to mass using its molar mass (111 g/mol) → x = 3.7 g.

   Concept and Intuition

   For an electrolyte solution, the observed osmotic pressure is boosted by the van't Hoff factor $i$, which accounts for dissociation into multiple particles: $\pi = iCRT = i\dfrac{n}{V}RT$.

   Step-by-Step Solution

   1. $T = 27\,^\circ C = 300$ K.
   2. Rearranging: $n = \dfrac{\pi V}{iRT}$.
   3. Substitute: $n = \dfrac{0.82 \times 2.5}{2.5 \times 0.082 \times 300}$.
   4. The $2.5$ (volume) cancels: $n = \dfrac{0.82}{0.082\times300} = \dfrac{0.82}{24.6} = 0.03333$ mol. …

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6. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
   Q.[FIGURE] (an osmosis apparatus: a vessel divided by a semi-permeable membrane, SPM, into part-II (with a piston, left) and part-I (with a piston, right) containing sea water) The osmotic pressure of sea water is 1.05 atm. Four experiments were carried as shown in table. In which of the following experiments, pure water can be obtained in part-II of vessel? Table: Expt. No. | Pressure applied in part-I of Vessel | Pressure applied in part-II of Vessel I | 10 atm | - II | - | 10 atm III | 15 atm | - IV | - | 15 atm (A) I, III only (B) II, IV only (C) I, II, III, IV (D) IV only

   ›Reveal solutionSolution

   Reverse osmosis requires pressure greater than the solution's osmotic pressure applied on the solution (sea-water) side; only Experiments I and III do this — answer (A).

   Concept and Intuition

   In normal osmosis, pure solvent moves through a semi-permeable membrane INTO the solution side (to dilute it), driven by the difference in chemical potential. Reverse osmosis flips this: if you apply mechanical pressure exceeding the solution's osmotic pressure directly on the solution side, you force pure solvent to move OUT of the solution and into the pure-solvent side — this is exactly how sea water is desalinated. Applying pressure on the pure water side instead does not achieve this; it would, if anything, drive water from the pure side toward the solution side (enhancing normal osmosis or having no useful effect for extracting pure water into part-II).

   Step-by-Step Solution

   1. Sea water (with $\pi = 1.05\ \mathrm{atm}$) sits in part-I; part-II is the side where we want to collect pure water.
   2. To achieve reverse osmosis (pure water pushed from part-I into part-II across the SPM), the applied pressure must be on part-I (the solution side) and must exceed $\pi = 1.05\ \mathrm{atm}$.
   3. Expt I: 10 atm on part-I — exceeds 1.05 atm and is on the correct (solution) side. ✓ Works. …

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7. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
   Q.Given below are two statements. Statement I: Liquids A & B form a non-ideal solution with negative deviation. The interactions between A and B are weaker than A-A & B-B interactions. Statement II: In reverse osmosis, the applied pressure must be higher than the osmotic pressure of solution. The correct answer is (A) Both Statement-I and Statement-II are correct (B) Both Statement-I and Statement-II are not correct (C) Statement-I is correct but Statement-II is not correct (D) Statement-I is not correct but Statement-II is correct

   ›Reveal solutionSolution

   Negative deviation needs STRONGER unlike-molecule interactions (Statement I has this backwards), while the reverse osmosis condition (applied pressure > osmotic pressure) is correctly stated in Statement II.

   Concept and Intuition

   Raoult's law deviations are governed by the relative strength of A–B interactions compared to A–A and B–B interactions. If A–B attractions are stronger (e.g., due to hydrogen bonding, like acetone-chloroform), molecules are held back more than in the ideal case, lowering the escaping tendency and vapour pressure — this is negative deviation. Positive deviation happens when A–B interactions are weaker than the pure-component interactions. For reverse osmosis, in normal osmosis solvent flows from pure solvent into solution across a semipermeable membrane; to reverse this flow (purify the solution, e.g. desalination), you must apply external pressure exceeding the osmotic pressure of the solution.

   Step-by-Step Solution

   1. Evaluate Statement I: it claims A–B interactions are WEAKER than A–A/B–B interactions cause negative deviation. This is the reverse of the truth — weaker A–B interactions actually cause POSITIVE deviation (molecules escape more easily than in the ideal case). So Statement I is incorrect. …

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8. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
   Q.At 300 K, 6 g of urea was dissolved in 500 mL of water. What is the osmotic pressure (in atm) of resultant solution? ($R = 0.082\ \mathrm{L\,atm\,K^{-1}\,mol^{-1}}$) (C = 12; N = 14; O = 16; H = 1) (A) 0.492 (B) 2.46 (C) 4.92 (D) 49.2

   ›Reveal solutionSolution

   Osmotic pressure is computed from $\pi = CRT$ using urea's molar mass (60 g/mol); the answer is 4.92 atm, option (C).

   Concept and Intuition

   Osmotic pressure is a colligative property — it depends only on the number of solute particles per unit volume of solution, not on their identity. For a non-electrolyte like urea (which does not dissociate), van't Hoff's equation for dilute solutions mirrors the ideal gas law:

   $\pi V = nRT \quad\Rightarrow\quad \pi = CRT$

   where $C = n/V$ is the molar concentration.

   Step-by-Step Solution

   1. Molar mass of urea, $\mathrm{CO(NH_2)_2}$: $C(12) + O(16) + 2\times N(14) + 4\times H(1) = 12+16+28+4 = 60\ \text{g/mol}$.
   2. Moles of urea $= \dfrac{6}{60} = 0.1\ \text{mol}$.
   3. Volume of solution $= 500\ \text{mL} = 0.5\ \text{L}$. …

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9. AP EAPCET 2023Set ap-2023-05-22-AN1 markMCQ
   Q.Which of the following aqueous solution has highest osmotic pressure? (A) $5.85\%\left(\dfrac{W}{V}\right)\mathrm{NaCl}$ (B) $3.42\%\left(\dfrac{W}{V}\right)\mathrm{Al_2(SO_4)_3}$ (C) $6\%\left(\dfrac{W}{V}\right)\mathrm{Urea}$ (D) $18\%\left(\dfrac{W}{V}\right)\mathrm{Glucose}$

   ›Reveal solutionSolution

   Converting each percentage to molarity and multiplying by the van't Hoff factor shows NaCl's solution has the highest effective particle concentration, hence the highest osmotic pressure.

   Concept and Intuition

   Osmotic pressure is a colligative property: $\pi = iCRT$, where $i$ is the van't Hoff factor (number of particles the solute dissociates into) and $C$ is the molar concentration. For electrolytes, you must multiply the molarity by $i$ to get the effective particle concentration that actually determines $\pi$.

   Step-by-Step Solution

   1. NaCl: $5.85\%\ (w/v)$ means $5.85$ g per 100 mL $= 58.5$ g/L. Molar mass of NaCl $=58.5$ g/mol, so $C=1$ M. NaCl dissociates into 2 ions ($Na^+ + Cl^-$), so $i=2$. Effective concentration $=1\times2=2$ M.
   2. $Al_2(SO_4)_3$: $3.42\%\ (w/v) = 34.2$ g/L. Molar mass $=2(27)+3(96)=342$ g/mol, so $C=0.1$ M. It dissociates into $2\,Al^{3+}+3\,SO_4^{2-}=5$ ions, so $i=5$. Effective concentration $=0.1\times5=0.5$ M.
   3. Urea: $6\%\ (w/v)=60$ g/L. Molar mass $=60$ g/mol, so $C=1$ M. Urea is a non-electrolyte, $i=1$. Effective concentration $=1$ M. …

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10. AP EAPCET 2023Set ap-2023-05-23-FN1 markMCQ
    Q.34.2% $\left(\dfrac{W}{V}\right)$ sucrose $(\mathrm{C_{12}H_{22}O_{11}})$ solution is isotonic with 18% $\left(\dfrac{W}{V}\right)$ of unknown solution X. Formula of X is (A) $\mathrm{C_6H_{10}O_4}$ (B) $\mathrm{C_6H_{12}O_2}$ (C) $\mathrm{C_6H_{12}O_6}$ (D) $\mathrm{C_5H_{12}O_5}$

    ›Reveal solutionSolution

    Isotonic solutions must have equal molar concentrations; matching the given w/V percentages to molarities pins the unknown's molar mass at 180 g/mol, i.e. glucose.

    Concept and Intuition

    Osmotic pressure (and hence isotonicity) depends on the molar concentration of solute particles, not on mass concentration. Two solutions are isotonic (produce the same osmotic pressure at the same temperature) exactly when they have the same molarity (assuming both are non-electrolytes, van't Hoff factor = 1).

    Step-by-Step Solution

    1. Sucrose solution: 34.2% w/V means 34.2 g of sucrose per 100 mL, i.e. 342 g/L.
    2. Sucrose's molar mass, $\mathrm{C_{12}H_{22}O_{11}}$, is $12(12)+22(1)+11(16)=144+22+176=342$ g/mol.
    3. Molarity of sucrose solution $= \dfrac{342\text{ g/L}}{342\text{ g/mol}} = 1\ \text{mol/L}$.
    4. For X at 18% w/V $=180$ g/L to be isotonic (same molarity, 1 mol/L), its molar mass must be $\dfrac{180\text{ g/L}}{1\text{ mol/L}} = 180$ g/mol. …

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11. AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQ
    Q.What is the osmotic pressure (in atm) of 0.02M aqueous glucose solution at 300 K? (R=0.082 L atm mol$^{-1}$K$^{-1}$) (A) $\frac{1}{0.492}$ (B) 0.492 (C) 0.988 (D) $\frac{1}{0.988}$

    ›Reveal solutionSolution

    Osmotic pressure follows the van't Hoff equation $\pi=CRT$ — a direct plug-in gives 0.492 atm.

    Concept and Intuition

    For a dilute non-electrolyte solution, osmotic pressure behaves like an ideal-gas-like law: $\pi = CRT$, where $C$ is molar concentration. Glucose is a non-electrolyte (van't Hoff factor $i=1$), so no correction is needed.

    Step-by-Step Solution

    1. $\pi = CRT = 0.02\times0.082\times300$.
    2. $0.02\times0.082 = 0.00164$.
    3. $0.00164\times300 = 0.492\ \text{atm}$.

    Common Mistakes …

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12. AP EAPCET 2023Set eng-2023-05-17-AN1 markMCQ
    Q.If the osmotic pressure of cane sugar solution is 2.46 atm at 27 °C, then what is the concentration (in mol $\mathrm{L^{-1}}$) of the solution $(R = 0.0821\ \mathrm{L\ atm\ mol^{-1}\ K^{-1}})$ (A) 0.1 (B) 0.2 (C) 0.01 (D) 0.02

    ›Reveal solutionSolution

    This tests the van't Hoff osmotic pressure equation for a non-electrolyte solution. Answer: C ≈ 0.1 mol/L.

    Concept and Intuition

    Osmotic pressure of a dilute non-electrolyte solution follows the van't Hoff equation $\pi = CRT$, analogous to the ideal gas law, where C is molar concentration.

    Step-by-Step Solution

    1. Convert temperature: $27°C = 300$ K.
    2. Apply $\pi = CRT \Rightarrow C = \dfrac{\pi}{RT} = \dfrac{2.46}{0.0821 \times 300}$.
    3. Compute denominator: $0.0821 \times 300 = 24.63$.
    4. $C = 2.46/24.63 \approx 0.0999 \approx 0.1$ mol/L. …

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