How many grams of solute would you use to prepare the following solutions? 1.08 L of 0.438 M glucose (C6H12O6) Express your answer with the appropriate units.

Answer 1

The definition of M (molarity) is moles of solute per liter of solution.

molar mass of glucose (C₆H₁₂O₆) = 180.2 g / mol

To find the grams of glucose needed to make 1.08 L of 0.438 M glucose, we calculate as follows:

0.438 mol / L x 1.08 L = 0.473 mols needed

0.473 mols x 180.2 g / mol = 85.24 g needed

rounding to 3 sig. figs. we have 85.2 g of glucose are needed