Question

The molarity (m) of an aqueous solution containing 22.5 g of glucose (c6h12o6) in 35.5 ml of solution is ________.

Answer 1

Molarity is found by dividing moles of solute/liters of solution. First, find the molar mass of glucose:

22.5 grams C6H12O6 x(1 mole/180 grams) = 0.125 moles glucose

Then, divide moles over liters:
M=0.125 moles glucose/0.035 L sol'n
M=3.57 

Answer 2

Explanation:

Molarity is the number of moles present in a liter of solution.

Mathematically,     Molarity = $\frac{\text{no. of moles}}{\text{volume in liter}}$

And, no. of moles = $\frac{mass}{\text{molar mass}}$

Molar mass of $C_{6}H_{12}O_{6}$ is 180.15 g/mol. Therefore, number of moles present will be as follows.

   No. of moles = $\frac{mass}{\text{molar mass}}$

                         = $\frac{22.5 g}{180.15 g/mol}$

                          = 0.124 mol

Hence, calculate the molarity as follows.

                Molarity = $\frac{\text{no. of moles}}{\text{volume in liter}}$

                       = $\frac{0.124 mol}{0.0355 L}$      (as 1 L = 1000 mL)

                       = 3.51 M

Thus, we can conclude that molarity of the solution is 3.51 M.