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andreev551
7 days ago
10

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 solutio

n in the following units?
Chemistry
1 answer:
KiRa [2.7K]7 days ago
6 0

Response:

Mole fraction: 0.0157.

Molality: 0.889m

Mass%: 16%

Clarification:

The measurement units are mole fraction, molality, and mass percentage.

Mole fraction is calculated using moles of glucose relative to the total moles present.

The glucose moles for 1L is:

0.944 moles.

The moles of water in 1L are:

1L × (1.0624kg/L) × (1000g / 1kg) × (1mol / 18.02g) = 59.0 moles of water

Mole fraction is given by: 0.944 moles / (59.0 mol + 0.944mol) = 0.0157

Molality is defined as moles of solute (0.944) per kilogram of solution (1.0624kg):0.944mol / 1.0624kg =

0.889m

For mass percentage, the total mass equals 1062.4g, and the mass of 0.944 moles of glucose is:

0.944mol×(180.156g/1mol) = 170g of glucose. Mass%:

170g / 1062.4g ×100 = 16%

I trust this information is useful!
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