Answer:
[tex]\boxed{\text{3.4 g}}[/tex]
Explanation:
Step 1. Calculate the moles of C₁₂H₂₂O₁₁
[tex]\text{n = 0.100 L} \times \dfrac{\text{1.0 mol}}{\text{1 L}}=\text{0.100 mol}[/tex]
Step 2. Calculate the mass of C₁₂H₂₂O₁₁
[tex]\text{m = 0.100 mol} \times \dfrac{\text{342.30 g}}{\text{1 mol}} = \text{3.4 g}\\\\\text{The mass of sucrose needed is }\boxed{\textbf{3.4 g}}[/tex]