Final answer:
The question seeks to determine the mass of NaOH in milligrams in a 500 mL of 0.2N sodium hydroxide solution. The correct calculation shows that there are 4000 mg of NaOH, but this option is not available in the answer choices, suggesting a potential error in the question or the answer choices provided.
Step-by-step explanation:
The question asks how many milligrams (mg) of sodium hydroxide (NaOH) are present in a 500 mL of 0.2N solution. Normality (N) is a measure of concentration equivalent to molarity (M) for monoprotic acids and bases. Since NaOH is a monoprotic base, 0.2N is also 0.2M.
To find the number of moles of NaOH in the solution, we use the formula:
moles of NaOH = Molarity (M) × Volume (L)
Here, Volume should be in liters, so we convert 500 mL to liters:
500 mL × (1 L / 1000 mL) = 0.500 L
Now we calculate the moles:
0.2M × 0.500 L = 0.1 moles of NaOH
Next, we convert moles of NaOH to mass in grams:
NaOH has a molar mass of 40.0 g/mol (from the periodic table).
0.1 mol NaOH × 40.0 g/mol = 4.0 grams
Finally, we convert grams to milligrams since the answer choices are in milligrams:
4.0 g × 1000 mg/g = 4000 mg
However, it seems there is a confusion because the answer suggests that we should have calculated for 500 mL of 0.2 M (which is identical to 0.2N for NaOH) and not for 0.02 M. If we go back and correct the molarity to 0.2 M, we have:
0.2 M NaOH × 0.500 L = 0.1 mol
and
0.1 mol × 40.0 g/mol = 4.0 g
4.0 g × 1000 mg/g = 4000 mg, which is not found in the answer choices. Therefore, there must be a typographical error in the question or answer choices.