The answers are in black.I just need help working it out. Concentration and the

Question

The answers are in black.I just need help working it out.

Concentration and the Importance of "Basis". A solution is prepared by dissolving 250 g of glucose, C_6H_12O_6, in water. The total volume of solution is one liter, and the specific gravity of the solution (with water as the reference) is 1.014. Calculate: the molarity (gmole glucose/liter) of the glucose solution the mass% of glucose and water in the mixture the mole% of glucose and water in the mixture the average molecular weight of the mixture

Explanation / Answer

a) molarity = (W/MW) (1000 / volume of the solution in mL)

molarity = (250 / 180) (1000 /1000 )

molarity = 1.39 x 1

molarity = 1.39 M

b) mass % = (mass of solute / total mass ) x 100

mass of solution = 1000 x 1.014 =1014 g

mass % = (250 / 1014) x 100

mass % = 24.65

c) mass of water = 1014 - 250 = 764

moles of glucose = 250 / 180 = 1.39

moles of water = 764 / 18 = 42.4

mole % glucose= (1.39 / 1.39 + 42.4) x 100 = (1.39 / 43.79)x 100 = 3.17

mole % of water = 42.4 / 43.79 x 100 = 96.83

d)

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