Question

A 0.944 M solution of glucose, C6H12O6, in water has a density of 1.0624 g/mL at 20∘C. What is the concentration of this solution in the following units?

Answer 1

Answer:

Mole fraction: 0,0157.

Molality: 0,889m

Mass%: 16%

Explanation:

The units are mole fraction, molality and mass%

Units of mole fraction are moles of glucose per total moles.

Moles of glucose assuming 1L are:

0,944 moles.

Moles of water in 1L are:

1L × (1,0624kg/L) × (1000g / 1kg) × (1mol / 18,02g) = 59,0 moles of water

Mole fraction is: 0,944 moles / (59,0 mol + 0,944mol) = 0,0157

Molality is mole of solute (0,944) per kg of solution (1,0624kg):

0,944mol / 1,0624kg = 0,889m

In mass percent total mass is 1062,4g and mass of 0,944 moles of glucose is:

0,944mol×(180,156g/1mol) = 170g of glucose. Mass%:

170g / 1062,4g ×100 = 16%

I hope it helps!