Question 32 of 50intermediate🔧 ApplyNumerical2 marks

At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?

Correct Answer

The concentration of the solution is 0.061 mol L⁻¹.

Exercise: EXERCISES | Q: 1.22 | (Page 28)
For More Understanding

Explanation

This question applies the osmotic pressure relationship π = CRT, where π is osmotic pressure, C is concentration, R is the gas constant, and T is temperature. Since temperature remains constant at 300 K for both solutions, the ratio of osmotic pressures equals the ratio of concentrations. The molar mass of glucose (C₆H₁₂O₆) is 180 g mol⁻¹, which is used to calculate the initial concentration from the given mass.

Solution Steps

  1. Step 1: Calculate the molar mass of glucose (C₆H₁₂O₆) = (12×612 \times 6) + (1×121 \times 12) + (16×616 \times 6) = 180 g mol⁻¹

  2. Step 2: Calculate initial concentration (C₁) = mass of glucose / (molar mass × volume) = 36 g / (180 g mol⁻¹ × 1 L) = 0.2 mol L⁻¹

  3. Step 3: Apply the relation π₁/C₁ = π₂/C₂ (since π = CRT and T is constant)

  4. Step 4: Rearrange to find C₂ = C₁ × π₂/π₁ = 0.2 mol L⁻¹ × (1.52 bar / 4.98 bar) = 0.061 mol L⁻¹

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