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 m B) 0.143 m C) 0.695 m D) 0.794 m

Question

Quizplus · Accepted Answer

First step is to convert the mass of glucose to moles:

n(glucose) = 18.0 g / 180.16 g/mol = 0.0999 mol

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

m(water) = 125.9 g / 1000 = 0.1259 kg

Now we can calculate the molality (moles of solute per kilogram of solvent):

molality = n(glucose) / m(water) = 0.0999 mol / 0.1259 kg = 0.794 m

Therefore, the molality of the glucose solution is 0.794 m.

What Is the Molality of a Glucose Solution Prepared by Dissolving