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Multiple Choice
What is the molality of a 10.5% by mass glucose (C₆H₁₂O₆) solution, given that the density of the solution is 1.03 g/mL?
A
0.75 mol/kg
B
0.45 mol/kg
C
0.90 mol/kg
D
0.61 mol/kg
Verified step by step guidance
1
Start by understanding that molality is defined as the number of moles of solute per kilogram of solvent. The formula for molality (m) is: m = \( \frac{\text{moles of solute}}{\text{kilograms of solvent}} \).
Calculate the mass of glucose in the solution. Since the solution is 10.5% by mass glucose, this means there are 10.5 grams of glucose in every 100 grams of solution.
Determine the mass of the solvent (water) in the solution. If the total mass of the solution is 100 grams, and 10.5 grams is glucose, then the mass of the solvent is 100 g - 10.5 g = 89.5 g.
Convert the mass of the solvent from grams to kilograms, as molality is expressed in moles per kilogram. So, 89.5 g = 0.0895 kg.
Calculate the moles of glucose. The molar mass of glucose (C₆H₁₂O₆) is approximately 180.18 g/mol. Use the formula: moles of glucose = \( \frac{\text{mass of glucose}}{\text{molar mass of glucose}} \) = \( \frac{10.5 \text{ g}}{180.18 \text{ g/mol}} \).