How many milliliters of a 0.205 M solution of glucose, C6H12O6, are required to obtain 150.1 g of glucose?

4050 mL

Explanation:

Given dа ta:

Mass of glucose = 150.1 g

Molarity of solution = 0.205 M

Volume of solution = ?

Solution:

Molarity = number of moles of solute / L of solution.

Now we will calculate the moles of sugar first.

Number of moles = mass / molar mass

Number of moles = 150.1 g / 180.156 g/mol

Number of moles = 0.83 mol

Now we will determine the volume:

Molarity = number of moles of solute / L of solution.

0.205 M = 0.83 mol / L of solution.

L of solution = 0.83 mol / 0.205 M

L of solution = 4.05 L

L to mL conversion:

4.05 L * 1000 mL / 1 L = 4050 mL