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

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​

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Hải Đăng 4 years 2021-07-20T17:28:50+00:00 1 Answers 173 views 0

Answers ( )

  1. MichaelMet
    0
    2021-07-20T17:30:24+00:00
    Reply

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

    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\times 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}\times 1000}{\text{Molar mass of solute}\times \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}=\frac{63.28\times 1000}{92.094\times 250.00}\\\\\text{Molarity of solution}=2.749M

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

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