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Calculate the Molarity of 15.4 g of sucrose (C12H22O11) in 74.0 mL of solution

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samueladesida43

The molarity of 15.4 g of sucrose (C12H22O11) in 74.0 mL of solution is 0.61M

HOW TO CALCULATE MOLARITY:

  • The molarity of a solution can be calculated by dividing the number of moles by the volume as follows:

Molarity (M) = no. of moles (mol) ÷ volume (L)

  • According to this question, 15.4 g of sucrose (C12H22O11) is said to be in 74.0 mL of solution. The mass of sucrose must be converted to moles by dividing by its molar mass:

Molar mass of C12H22O11 = 12(12) + 1(22) + 16(11)

= 144 + 22 + 176

= 342g/mol

  • no. of moles = 15.4g ÷ 342g/mol

  • no. of moles = 0.045mol

Volume = 74mL = 74/1000 = 0.074L

Molarity = 0.045mol ÷ 0.074L

Molarity = 0.61M

  • Therefore, the molarity of 15.4 g of sucrose (C12H22O11) in 74.0 mL of solution is 0.61M.

Learn more: https://brainly.com/question/15948514?referrer=searchResults

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