3) Here is a problem about equilibrium constants and standard states involving t

Question

3) Here is a problem about equilibrium constants and standard states involving the solubility of KCl in water. We said in class that the activity of water solvent is a(H20) Y(H20)x(H20) such that Y- 1 as x 1. Note that we use the mole fraction convention for the solvent. Because most solutions are dilute and x is close to one, a(H2O) is also close to one. Let's now consider a very concentrated solution: 4.8 m KCl, where b molality F moles solute in 1 kg of solvent. In this case, a(H2O) is not close to one

Explanation / Answer

Water activity is usually measured as equilibrium relative humidity (ERH). The water activity(aw) represents the ratio of the water vapor pressure of the food to the water vapor pressure of pure water under the same conditions and it is expressed as a fraction.

This actually means that 4.8* 58.5 grams = 280.8 gms of KCl present in 1000 grams of solvent
Thus it means that 1000- 280.8 = 719.2 grams of H2O are present.
Number of moles of H2O are thus 719.2/18 I.e., 39.9
Number of moles of KCl are 4.8
Thus mole fraction of water is 39.9 / (39.9 + 4.8)=0.89

water activity coefficient is 20.0 * 0.89 = 17.8

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