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?

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!

the answer is the first option

explanation:

just googled it m8. not that hard

lol

every chemical equation adheres to the law of conservation of mass, which states that matter cannot be created or destroyed. therefore, there must be the same number of atoms of each element on each side of a chemical equation.