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
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.
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