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A 2.0 molal sugar solution has approximately the same freezing point as 1.0 molal solution of 1) CaCl2 2) CH3COOH 3) NaCl 4) C2H5OH

1 Answer

1 vote

Answer:

3) NaCl.

Step-by-step explanation:

∵ ΔTf = iKf.m

where, i is the van 't Hoff factor.

Kf is the molal depression freezing constant.

m is the molality of the solute.

The van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass.

  • For most non-electrolytes dissolved in water, the van 't Hoff factor is essentially 1.

So, for sugar: i = 1.

∴ ΔTf for sugar = iKf.m = (1)(Kf)(2.0 m) = 2 Kf.

  • For most ionic compounds dissolved in water, the van 't Hoff factor is equal to the number of discrete ions in a formula unit of the substance.

For NaCl, it is electrolyte compound which dissociates to Na⁺ and Cl⁻.

So, i for NaCl = 2.

∴ ΔTf for NaCl = iKf.m = (2)(Kf)(1.0 m) = 2 Kf.

So, the right choice is: 3) NaCl.

User Vinny M
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