A solution is prepared by dissolving 78.6 g of glucose (C6H12O6 180.16 grams per

Question

A solution is prepared by dissolving 78.6 g of glucose (C6H12O6 180.16 grams per mole) in 0.402 megagrams of water. The final volume of the solution is 621 mL. Calculate the part by million of the solution

Calculate the volume of water, in Liters, that can be vaporized at its normal boiling point with 9.28 megajoules of heat. The heat of vaporization of water at its normal noiling point of 100 Celsius is +40.7 likojoules per mole

Determine the solubility of oxygen in water at 25 degrees Celsius exposed to air at one atmosphere. Assume partial pressure of 2.85 atmosphere.

Explanation / Answer

1) 78.6 g of glucose = 78.6 / 180.16 = 0.436 gm = 436 mg.

part by million of the solution = 436 / 0.621 = 702 ppm

2) 9.28 megajoules of heat = 9.28 * 10^6 J

moles of water vaporized =  9.28 * 10^6 / (40.7 * 1000) = 228 mole water = 228 * 18 = 4104 gm water

as the density of water = 1 gm / ml.

so volume of water, in Liters = 4.104 L

3)

we know,

Henrys law :

KH = concentration / partial pressure.

we need KH (Henrys constant) to get concentration that is solubility of O2.

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