Question

What mass of solid NaOH (97.0% NaOH by mass) is required to prepare 1.00 L of a 10.0% solution of NaOH by mass? The density of the 10.0% solution is 1.109 g/mL.

a) 77.24 g
b) 100 g
c) 108.01 g
d) 120.84 g

Answer 1

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.