What is the molarity of an aqueous solution containing 31.7 g of glucose (C6H12O6) in 1.2 L of solution?

moles = grams/molar mass

solve for moles.
Then M = moles/L soln. Solve for M.

To calculate the molarity of a solution, you need to know the number of moles of solute and the volume of the solution in liters. The equation to calculate molarity is:

Molarity (M) = Moles of Solute / Volume of Solution (in liters)

First, let's calculate the number of moles of glucose (C6H12O6) in the given amount of grams.

The molar mass of glucose (C6H12O6) is:
(6 * atomic mass of carbon) + (12 * atomic mass of hydrogen) + (6 * atomic mass of oxygen)
= (6 * 12.01 g/mol) + (12 * 1.01 g/mol) + (6 * 16.00 g/mol) = 180.18 g/mol

Now, we can calculate the number of moles of glucose:
Moles of glucose (C6H12O6) = Mass of glucose / Molar mass of glucose
= 31.7 g / 180.18 g/mol
= 0.176 moles of glucose

Next, we need to convert the volume of the solution from milliliters to liters.
1.2 L = 1200 mL

Now, we can calculate the molarity:
Molarity (M) = Moles of Solute / Volume of Solution (in liters)
= 0.176 moles / 1.2 L
≈ 0.147 M

Therefore, the molarity of the aqueous glucose solution is approximately 0.147 M.