Calculate the molarity of each solution. 28.33 g c6h12o6 in 1.28 l of solution

Molarity is number of moles of solute in 1 L of solution

number of moles of glucose - 28.33 g / 180 g/mol = 0.1574 mol

volume of solution is 1.28 L

since molarity is number of moles in 1 L

the number of moles in 1.28 L - 0.1574 mol

therefore number of moles in 1 L - 0.1574 mol / 1.28 L = 0.123 M

molarity is 0.123 M