Final answer:
To make a 2.00 M Na2CO3 solution with a volume of 750.0 mL, you would need 159.0 g of Na2CO3.
Step-by-step explanation:
To make a 2.00 M Na2CO3 solution with a volume of 750.0 mL, you would need to calculate the amount of solid Na2CO3 required. First, convert the volume from milliliters to liters by dividing by 1000: 750.0 mL ÷ 1000 = 0.750 L. Then, use the equation:Molarity (M) = moles of solute ÷ volume of solution (in liters)Rearrange the equation to solve for moles of solute: moles of solute = Molarity × volume of solution (in liters)Plugging in the given values:Moles of solute = 2.00 M × 0.750 L = 1.50 molesFinally, calculate the mass of Na2CO3 using its molar mass:Molar mass of Na2CO3 = 2(22.99 g/mol) + 12.01 g/mol + 3(16.00 g/mol) = 105.99 g/molMass of Na2CO3 = 1.50 moles × 105.99 g/mol = 159.0 g