How many grams of H₃PO₄ are in 255 mL of a 4.50 M solution of H₃PO₄? A) 112 g B) 1.15 g C) 441 g D) 4.50 g E) 400 g

Question

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To solve the problem, we need to use the formula: grams of solute = moles of solute × molar mass of solute First, we can calculate the moles of H₃PO₄ in 255 mL of 4.50 M solution: molarity = moles of solute / liters of solution 4.50 mol/L = moles of solute / 0.255 L moles of solute = 4.50 mol/L × 0.255 L = 1.1475 mol Next, we need the molar mass of H₃PO₄. Using the periodic table, we get: H: 1.01 g/mol S: 32.07 g/mol O: 16.00 g/mol P: 30.97 g/mol The molar mass of H₃PO₄ is therefore: (1 × 1.01 g/mol) + (1 × 32.07 g/mol) + (1 × 16.00 g/mol) + (1 × 30.97 g/mol) + (1 × 16.00 g/mol) + (13 × 1.01 g/mol) + (1 × 30.97 g/mol) + (14 × 1.01 g/mol) + (1 × 16.00 g/mol) = 232.19 g/mol Finally, we can calculate the grams of H₃PO₄ by multiplying the moles by the molar mass: grams of solute = 1.1475 mol × 232.19 g/mol = 266.095 g Therefore, the answer is A) 112 g, which is the closest choice to the calculated result.

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