A .845 M solution of glucose in water has a density of 1.0624 g/mL at 20...

A .845 M solution of glucose in water has a density of 1.0624 g/mL at 20 degree C what is the concentration of this solution in the following units. Assume 1 L of solution

a. mole fraction

b. mass percent

c Molality

Expert Solution

To solve this problem we need an assumtion. So we assume that we have 1000 mL of glucose solution. The mass of the glucose can be found by multiplying density and its volume. If the mass of glucose has been found we can calculate the mass of both glucose and water. Changing into moles we can calculate mole fraction, molality, and the masss percent.

mass of 1000 mL glucose solution
= density x volume
= 1.0624g/mL x 1000 mL
= 1062.4 g

moles of glucose in 1000 mL of 0.845 glucose solution
= MxV
= 0.845 x 1
= 0.845 mole

mass of glucose
= mole x Mr
= 0.845 x 180
= 152.1 g

Mass of water
= mass of solution - mass of glucose
= 1062.4 g - 152.1 g
= 910.3

moles of water
= mass/Mr
= 910.3/18
= 50.57 mole

a. mole fraction
= mole glucosa / (mole total)
= 0.845 / ( 0.845 + 51.415)
= 0.845/51.415
= 0.016

b. mass percent
= mass of glucose / (mass total) x 100%
= 152.1 g / (152.1 g + 910.3 ) x 100%
= 152.1 / 1062.40 x 100%
= 14.31%

c. molality
= mole / Kg solvent
= 0.845/0.9103
= 0.9282 molal

queen_honey_blossom answered 1 hour ago

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