Question

C6H1206+5KIO4=5KIO3+5HCO2H+H2CO

Answer 1

Answer:

5 % by mass

Explanation:

1. Calculate the moles of KIO₄.

$\text{Moles of KIO}_{4} = \text{30 mL} \times \frac{\text{0.5 mmol}}{\text{1 mL}} = \text{15 mmol KIO}_{4}$

2. Calculate the moles of sugar

$\text{Moles of sugar} = \text{15 mmol KIO}_{4} \times \frac{\text{ 1 mmol sugar}}{\text{5 mmol KIO}_{4}} = \text{3.0 mmol sugar}$

3. Calculate the mass of sugar

$\text{Mass of sugar} = \text{3.0 mmol sugar} \times \frac{\text{180.16 mg sugar}}{\text{1 mmol sugar}} = \text{540 mg sugar} = \text{0.54 g sugar}$

4. Calculate the mass of the sugar solution

$\text{Mass of solution} = \text{ 10 mL solution} \times \frac{\text{1.10 g solution}}{\text{1 mL solution}} = \text{11.0 g solution}$  

5. Calculate the percent by mass of sugar

$\text{\% by mass} = \frac{\text{mass of sugar}}{\text{mass of solution}} \times 100 \%$

$\text{\% by mass} = \frac{\text{0.54 g sugar}}{\text{11.0 g solution}} \times 100 \% = 5 \%$

Note: The answer can have only one significant figure because that is all you gave for the concentration of KIO₄.