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!
