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 data:

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

the number of valence electrons for each molecule or ion is shown beneath the structure. remember hydrogen will not have more than two electrons. this hydrogen is part of a covalent bond (sharing two electrons). a double bond here would cause hydrogen to share four electrons with phosphorus.