Final answer:
To prepare 100 mL of 0.13 M C₁₂H₂₂O₁₁ solution, approximately 4.452 grams of sucrose would be needed.
Step-by-step explanation:
To calculate the mass of sucrose needed to prepare a 0.13 M solution, we need to use the formula:
moles of solute = molarity x volume (in L)
First, we need to convert the volume given in mL to L:
100 mL = 100/1000
= 0.1 L
Now, we can use the formula:
moles of sucrose = 0.13 M x 0.1 L
= 0.013 moles
To calculate the mass, we can use the formula:
mass = moles x molar mass
The molar mass of sucrose (C₁₂H₂₂O₁₁) is 342.3 g/mol.
mass = 0.013 moles x 342.3 g/mol
= 4.452 g
Therefore, you would need approximately 4.452 grams of sucrose to prepare 100 mL of 0.13 M C₁₂H₂₂O₁₁ solution.