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

We need to first calculate the moles of glucose and water in the solution.

Number of moles of glucose = (18.0 g) / (180.16 g/mol) = 0.0999 mol

Number of moles of water = (125.9 g) / (18.01528 g/mol) = 6.986 mol

Molality of the solution = (0.0999 mol) / (0.1259 kg) = 0.794 mol/kg

Therefore, the molality of the glucose solution is 0.794 mol/kg, which is closest to option D (0.794 mol/kg).

What Is the Molality of a Glucose Solution Prepared by Dissolving