Question

Glycerol. C3HgO3, is a substance used extensively in the manufacture of cosmetics, foodstuffs, antifreeze, and plastics. Glycerol is a water-soluble liquid

with a density of 1.2656 g/mL at 15 °C. Calculate the molarity of a solution of glycerol made by dissolving 50.000 mL glycerol at 15 °C in enough water to
make 250.00 mL of solution. The molecular weight of C3HgO3 is 92.094 amu.
O A 0.6871
O B. 3.600
O C. 63.28
O 0.92.10
O E. 2.749​

Answer 1

Answer: The correct option is E.) 2.749 M.

Step-by-step explanation:

Density is defined as the ratio of mass and volume of a substance.

`\text{Density}=\frac{\text{Mass}}{\text{Volume}}` ......(1)

Given values:

Volume of glycerol = 50.0 mL

Density of glycerol = 1.2656 g/mL

Putting values in equation 1, we get:

`\text{Mass of glycerol }=(1.2656g/mL* 50.0mL)=63.28g`

Molarity is defined as the amount of solute expressed in the number of moles present per liter of solution. The units of molarity are mol/L. The formula used to calculate molarity:

`\text{Molarity of solution}=\frac{\text{Given mass of solute}* 1000}{\text{Molar mass of solute}* \text{Volume of solution (mL)}}` .....(2)

Given values:

Given mass of glycerol = 63.28 g

Molar mass of glycerol = 92.094 g/mol

Volume of the solution = 250.00 mL

Putting values in equation 2, we get:

`\text{Molarity of solution}=(63.28* 1000)/(92.094* 250.00)\\\\\text{Molarity of solution}=2.749M`

Hence, the correct option is E.) 2.749 M.