How many moles of solute are present in 50.0 grams of solution that is 2.50% by mass glucose (C₆H₁₂O₆) ?
MM(C₆H₁₂O₆) = 180.16 g/mol

Question

How many moles of solute are present in 50.0 grams of solution that is 2.50% by mass glucose (C₆H₁₂O₆)? MM(C₆H₁₂O₆) = 180.16 g/mol

Quizplus · Accepted Answer

The mass of glucose in the solution is 2.50% of 50.0 grams, which equals 1.25 grams. The molar mass of glucose (C6H12O6) is 180.16 g/mol. Therefore, the number of moles of glucose is 1.25 grams / 180.16 g/mol = 6.94×10^-3 moles.

How Many Moles of Solute Are Present in 50