JEE Main · 2023 · Shift-IImediumSOL-007

What is the mass ratio of ethylene glycol (C2H6O2, molar mass = 62 g/mol) required for making 500 g of 0.25 molal…

Solutions · Class 12 · JEE Main Previous Year Question

Question

What is the mass ratio of ethylene glycol (C2H6O2\mathrm{C_2H_6O_2}, molar mass = 62 g/mol) required for making 500 g of 0.25 molal aqueous solution and 250 mL of 0.25 molar aqueous solution?

Options
  1. a

    1 : 1

  2. b

    3 : 1

  3. c

    2 : 1

  4. d

    1 : 2

Correct Answerc

2 : 1

Detailed Solution

Strategy:\n> Calculate the moles of ethylene glycol required for each solution and compare them.\n\nStep 1: Calculate moles for 500 g of 0.25 molal solution\nMolality (mm) =0.25= 0.25. Solvent mass 500textg=0.5textkg\approx 500\\text{ g} = 0.5\\text{ kg} (assuming dilute solution).\nn1=mtimes0.5=0.25times0.5=0.125textmoln_1 = m \\times 0.5 = 0.25 \\times 0.5 = 0.125\\text{ mol}\n\nStep 2: Calculate moles for 250 mL of 0.25 molar solution\nMolarity (MM) =0.25= 0.25. Volume =0.25textL= 0.25\\text{ L}.\nn2=Mtimes0.25=0.25times0.25=0.0625textmoln_2 = M \\times 0.25 = 0.25 \\times 0.25 = 0.0625\\text{ mol}\n\nStep 3: Determine mass ratio\nRatio =n1:n2=0.125:0.0625=2:1= n_1 : n_2 = 0.125 : 0.0625 = 2 : 1.\n\ntextAnswer:(3)\boxed{\\text{Answer: (3)}}

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