Question 19 of 50intermediate🔧 ApplyNumerical3 marks

An antifreeze solution is prepared from 222.6 g of ethylene glycol (C2H6O2) and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g mL-1, then what shall be the molarity of the solution?

Correct Answer

Molality = 17.95 mol kg⁻¹, Molarity = 9.11 mol L⁻¹

Exercise: EXERCISES | Q: 1.8 | (Page 28)
For More Understanding

Explanation

This question tests understanding of molality and molarity calculations. Molality depends only on mass of solvent, while molarity requires the volume of solution calculated from total mass and density. The molar mass of ethylene glycol (C₂H₆O₂) is 62 g mol⁻¹ as established in Example 1.1 of the textbook.

Solution Steps

  1. Step 1: Calculate molar mass of ethylene glycol (C₂H₆O₂) Molar mass = (12×212 \times 2) + (1×61 \times 6) + (16×216 \times 2) = 62 g mol⁻¹

  2. Step 2: Calculate moles of ethylene glycol Moles = 222.6 g ÷ 62 g mol⁻¹ = 3.59 mol

  3. Step 3: Calculate molality Molality (m) = Moles of solute ÷ Mass of solvent in kg Molality = 3.59 mol ÷ (200 g ÷ 1000) = 3.59 mol ÷ 0.2 kg = 17.95 mol kg⁻¹

  4. Step 4: Calculate total mass of solution Total mass = Mass of solute + Mass of solvent = 222.6 g + 200 g = 422.6 g

  5. Step 5: Calculate volume of solution Volume = Mass ÷ Density = 422.6 g ÷ 1.072 g mL⁻¹ = 394.22 mL = 0.394 L

  6. Step 6: Calculate molarity Molarity (M) = Moles of solute ÷ Volume of solution in L Molarity = 3.59 mol ÷ 0.394 L = 9.11 mol L⁻¹

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