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Answers:5. Mole Fraction of NaCl and Water:Moles of NaCl = 12.50 g ÷ 58.44 g/mol = 0.214 molesMoles of H₂O = 115.0 g ÷ 18.02 g/mol = 6.386 molesTotal moles = 0.214 + 6.386 = 6.600Mole fraction of NaCl = 0.0324Mole fraction of H₂O = 0.96766. Mass of Water for 0.750 Molal Solution:Moles of NaCl = 22.5 g ÷ 58.44 g/mol = 0.385 molesMass of water = 0.385 ÷ 0.750 = 0.513 kg = 513 g7. Mass of KCl for 0.250 M Solution:Moles of KCl = 0.250 × 0.500 = 0.125 molesMass of KCl = 0.125 × 74.55 g/mol = 9.32 g8. Molarity of NaCl Solution:Moles of NaCl = 8.00 g ÷ 58.44 g/mol = 0.137 molesMolarity = 0.137 ÷ 0.250 = 0.548 MThe given problems involve calculations related to mole fractions, molality, molarity, and mass of solutes in solutions. In Question 5, the mole fraction of NaCl and water was determined by first converting their given masses into moles and then dividing each by the total moles in the solution, resulting in 0.0324 for NaCl and 0.9676 for water. In Question 6, the mass of water required to prepare a 0.750 molal solution of NaCl was found by dividing the moles of NaCl (0.385 moles) by the given molality and converting the result into grams, giving 513 g of water. Question 7 required calculating the mass of KCl needed for a 0.250 M solution in 500 mL of water, which was determined by multiplying molarity by volume to get 0.125 moles, then converting it into mass, resulting in 9.32 g of KCl. Lastly, in Question 8, the molarity of an 8.00 g NaCl solution in 250.0 mL of water was calculated by finding the moles of NaCl and dividing by the volume in liters, yielding a final molarity of 0.548 M. These calculations demonstrate the importance of using proper formulas and unit conversions when working with solutions in chemistry.
