Answer :
Final answer:
a. The molarity of 23.0 g of ethanol (C₂H₅OH) in 4.80 × 102 mL of solution is 1.04 M.
b. The molarity of 20.0 g of sucrose (C₁₂H₂₂O₁₁) in 75.0 mL of solution is 0.773 M.
c. The molarity of 6.50 g of sodium chloride (NaCl) in 97.4 mL of solution is 1.13 M.
Explanation:
To calculate the molarity of the solution using the formula:
Molarity (M) = moles of solute / liters of solution
First, we need to convert grams of solute into moles, and milliliters of solvent into liters.
1. 23.0 g of ethanol (C₂H₅OH): The molar mass of ethanol is 46.068 g/mol. So, 23.0 g is 0.499 mol.
4.80 x 102 mL = 0.480 L
Thus, the molarity is 0.499 mol / 0.480 L = 1.04 M.
2. 20.0 g of sucrose (C₁₂H₂₂O₁₁): The molar mass of sucrose is 342.30 g/mol. So, 20.0 g is 0.058 mol.
75.0 mL = 0.075 L
Thus, the molarity is 0.058 mol / 0.075 L = 0.773 M.
3. 6.50 g of sodium chloride (NaCl): The molar mass of sodium chloride is 58.44 g/mol. So, 6.50 g is 0.11 mol.
97.4 mL = 0.0974 L
Thus, the molarity is 0.11 mol / 0.0974 L = 1.13 M.
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