What is the molality of a glucose solution prepared by dissolving 18.0 g of glucose, C₆H₁₂O₆, in 125.9 g of water? A)7.94 × 10^-4 mol kg^-1 B)0.143 mol kg^-1 C)0.695 mol kg^-1 D)0.794 mol kg^-1

Question

Quizplus · Accepted Answer

First, we need to calculate the number of moles of glucose in the solution:

mass of glucose / molar mass of glucose = 18.0 g / 180.16 g/mol = 0.0999 mol

Next, we need to calculate the mass of water in kilograms:

mass of water = 125.9 g = 0.1259 kg

Now we can calculate the molality:

molality = moles of solute / mass of solvent in kg

molality = 0.0999 mol / 0.1259 kg = 0.794 mol kg^-1

Therefore, the answer is D, 0.794 mol kg^-1.

What Is the Molality of a Glucose Solution Prepared by Dissolving