What is the mass ratio of ethylene glycol (, molar mass = 62 g/mol) required for making 500 g of 0.25 molal aqueous solution and 250 mL of 0.25 molal aqueous solution?
- A.3:1
- B.1:2
- C.2:1
- D.1:1
Show answer and solution
Correct option: C
Solution
For a molal solution, molality = moles of solute per kg of solvent.
Case 1: 500 g of 0.25 molal solution
Let mass of ethylene glycol = m₁ g, so mass of water = (500 − m₁) g.
0.25 = (m₁/62) / ((500 − m₁)/1000)
Since m₁ is small compared to 500: (500 − m₁) ≈ 500 g = 0.5 kg
moles of solute = 0.25 × 0.5 = 0.125 mol
m₁ = 0.125 × 62 = 7.75 g
Case 2: 250 mL of 0.25 molal solution
Here 250 mL refers to the volume of solvent (water), so mass of water = 250 g = 0.25 kg.
moles of solute = 0.25 × 0.25 = 0.0625 mol
m₂ = 0.0625 × 62 = 3.875 g
Mass ratio m₁ : m₂ = 7.75 : 3.875 = 2 : 1
Case 1: 500 g of 0.25 molal solution
Let mass of ethylene glycol = m₁ g, so mass of water = (500 − m₁) g.
0.25 = (m₁/62) / ((500 − m₁)/1000)
Since m₁ is small compared to 500: (500 − m₁) ≈ 500 g = 0.5 kg
moles of solute = 0.25 × 0.5 = 0.125 mol
m₁ = 0.125 × 62 = 7.75 g
Case 2: 250 mL of 0.25 molal solution
Here 250 mL refers to the volume of solvent (water), so mass of water = 250 g = 0.25 kg.
moles of solute = 0.25 × 0.25 = 0.0625 mol
m₂ = 0.0625 × 62 = 3.875 g
Mass ratio m₁ : m₂ = 7.75 : 3.875 = 2 : 1
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