What is the mass ratio of ethylene glycol (C₂H₆O₂, 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?

Solution

Chemistry ma’am · 4 steps

Molal is per kg of water, molar is per litre of solution: find each glycol mass, then compare.

1. Molal solution. Molality is moles per kg of solvent. Let the glycol be n mol. Its mass is 62n g, so the water is (500 - 62n) g.

   0.25 = n/((500 - 62n)/1000) ⇒ 1.0155 n = 0.125 ⇒ n = 0.123 mol

2. Mass needed. Mass of glycol = 0.123 × 62 ≈ 7.63 g. If you treat all 500 g as water, you get 0.125 × 62 = 7.75 g, which is very close.
3. Molar solution. Molarity is moles per litre of solution: 0.25 × 0.250 = 0.0625 mol. Mass = 0.0625 × 62 = 3.875 g.
4. So. The mass ratio is 7.63 : 3.875 ≈ 1.97, so it is about 2 : 1, option (c).

The idea to keep. Molality counts kg of solvent; molarity counts litres of solution. Turn each one into moles first, using its own basis, and only then compare masses.

Answer from the NTA final answer key. Solution by PrepWiser.