Final answer:
To prepare 1.00 L of a 10.0% NaOH solution from solid NaOH that is 97.0% pure, one would need 114.33 g of the impure solid, which means the given options do not include the correct answer.
Step-by-step explanation:
To determine the mass of solid NaOH required to prepare 1.00 L of a 10.0% solution by mass, we first need to calculate the total mass of the solution. Since the density of the 10.0% solution is 1.109 g/mL, we use that to find the total mass:
Mass of the solution = Density × Volume
= 1.109 g/mL × 1000 mL
= 1109 g
A 10.0% NaOH solution by mass means that 10.0% of the solution's mass is NaOH, so:
Mass of NaOH in the solution = 0.100 × Mass of the solution
= 0.100 × 1109 g
= 110.9 g
However, the solid NaOH we have is 97.0% NaOH by mass. To get the correct mass of this impure solid that will give us the 110.9 g of pure NaOH, we divide by the percentage purity:
Mass of impure solid NaOH = Mass of pure NaOH / Purity
= 110.9 g / 0.970
= 114.33 g
So, you would need 114.33 g of the solid NaOH with 97.0% purity to prepare 1.00 L of a 10.0% NaOH solution, which means option (c) 108.01 g is closest to the calculated value but not quite correct.