To determine how many liters of a solution of C₁₂H₂₂O₁₁ (sucrose) will be prepared using 1.50 moles to make a 0.300 M solution, we can use the formula for molarity:

[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}} ]

Rearranging this formula to find the volume in liters gives us:

[ \text{liters of solution} = \frac{\text{moles of solute}}{\text{Molarity (M)}} ]

Using the provided information:

• Moles of solute = 1.50 moles
• Molarity = 0.300 M

Substituting the values into the rearranged formula results in:

[ \text{liters of solution} = \frac{1.50 \text{ moles}}{0.300 \text{ M}} = 5.00 \text{ liters} ]

Therefore, you would prepare 5.00 liters of the sucrose solution.

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For calculating the number of grams of sodium sulfate (Na₂SO₄) needed to make 200 mL of a 0.80 M solution, we will follow a similar calculation process.

First, calculate the number of moles of Na₂SO₄ required. We can use the formula:

[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}} ]

Rearranging gives us:

[ \text{moles of solute} = \text{Molarity (M)} \times \text{liters of solution} ]

For this calculation:

• Molarity = 0.80 M
• Volume = 200 mL = 0.200 L

Calculating the moles needed:

[ \text{moles of Na}_2\text{SO}_4 = 0.80 \text{ M} \times 0.200 \text{ L} = 0.16 \text{ moles} ]

Next, we need to convert moles to grams. The molar mass of sodium sulfate (Na₂SO₄) is approximately:

• Sodium (Na) = 23 g/mol × 2 = 46 g/mol
• Sulfur (S) = 32 g/mol
• Oxygen (O) = 16 g/mol × 4 = 64 g/mol

Total molar mass of Na₂SO₄:

[ \text{Molar mass} = 46 + 32 + 64 = 142 \text{ g/mol} ]

Now, convert the moles of Na₂SO₄ to grams:

[ \text{mass (g)} = \text{moles} \times \text{molar mass} = 0.16 \text{ moles} \times 142 \text{ g/mol} = 22.72 \text{ g} ]

Therefore, to prepare 200 mL of a 0.80 M sodium sulfate solution, you need 22.72 grams of Na₂SO₄.To prepare a solution of C12H22O11 using 1.50 moles to achieve a concentration of 0.300 M, you can use the molarity formula:

[ M = \frac{\text{moles of solute}}{\text{liters of solution}} ]

Rearranging the formula to solve for liters gives:

[ \text{liters of solution} = \frac{\text{moles of solute}}{M} ]

Inserting the values:

[ \text{liters of solution} = \frac{1.50 \text{ moles}}{0.300 \text{ M}} = 5.00 \text{ liters} ]

Thus, 5.00 liters of the solution will be prepared.

For the calculation of grams of sodium sulfate (Na2SO4) required to make a 200 mL (0.200 L) solution at a concentration of 0.80 M, the formula to determine the mass needed is:

[ \text{mass (g)} = \text{molarity (M)} \times \text{volume (L)} \times \text{molar mass (g/mol)} ]

First, the volume in liters (0.200 L) is multiplied by the molarity (0.80 M):

[ \text{moles of Na2SO4} = 0.80 \text{ M} \times 0.200 \text{ L} = 0.160 \text{ moles} ]

The molar mass of Na2SO4 is approximately 142.04 g/mol.

Now, calculate the mass:

[ \text{mass (g)} = 0.160 \text{ moles} \times 142.04 \text{ g/mol} \approx 22.73 \text{ g} ]

Therefore, you will need approximately 22.73 grams of sodium sulfate to prepare the required solution.