The mass percent of a solution is given by:

In a 10% by mass glucose (C₆H₁₂O₆) solution, this means there are 10 g of glucose for every 100 g solution. From this, we can calculate the mass of water (the solvent):

We need to find the molality of the solution since we'll use it to calculate the freezing point depression and boiling point elevation. Recall that molality is:

The molar mass of glucose is 6(12.01 g/mol C) + 12(1.01 g/mol H) + 6(16.00 g/mol O) = 180.18 g/mol. Calculating the molality of the glucose solution: