What is the molarity of an aqueous solution containing 43.7 g of glucose (C6H12O6) in 1.1 L of solution?

moles glucose = grams/molar mass glucose

M = moles/L soln.

To find the molarity of the solution, we need to first calculate the number of moles of glucose present in the solution using the given mass of glucose.

The molar mass of glucose (C6H12O6) can be calculated by adding up the atomic masses of its constituent elements:
6(12.01 g/mol) + 12(1.01 g/mol) + 6(16.00 g/mol) = 180.18 g/mol

Now, we can use the formula:

Molarity (M) = moles of solute / volume of solution (in liters)

To find the number of moles of glucose, we divide the given mass of glucose by its molar mass:

moles of glucose = mass of glucose / molar mass of glucose
= 43.7 g / 180.18 g/mol
≈ 0.2423 mol

Finally, we can calculate the molarity using the given volume of solution:

Molarity (M) = moles of solute / volume of solution
= 0.2423 mol / 1.1 L
≈ 0.22 M

Therefore, the molarity of the aqueous solution containing 43.7 g of glucose in 1.1 L of solution is approximately 0.22 M.