Question

How to prepare 200 mL of a 0.15 M sodium hydroxide solution?

• A. 1.0 g
• B. 1.2 g
• C. 1.5 g
• D. 1.8 g

Hint:

Use the formula \( \text{mass} = M \times V \times \text{molar mass} \) to prepare molar solutions accurately.

Answer 1

The Correct Option is B

- Given: Volume \( V = 200 \, \text{mL} = 0.2 \, \text{L} \), Molarity \( M = 0.15 \, \text{mol/L} \)
- Molar mass of sodium hydroxide (NaOH) \( = 23 + 16 + 1 = 40 \, \text{g/mol} \)
- Number of moles required: \[ n = M \times V = 0.15 \times 0.2 = 0.03 \, \text{mol} \]
- Mass of NaOH needed: \[ m = n \times \text{Molar mass} = 0.03 \times 40 = 1.2 \, \text{g} \]
- Therefore, to prepare 200 mL of 0.15 M NaOH solution, dissolve 1.2 g of sodium hydroxide in water and make up the volume to 200 mL.

Answer 2

The question asks how much sodium hydroxide is needed to make 200 mL of a 0.15 M solution. Instead of deriving the mass from scratch, let's check each option by working backward: if that mass of NaOH were dissolved in 200 mL of water, what molarity would result?

1. 1.0 g: Dividing by the molar mass of NaOH (40 g/mol) gives 0.025 mol. Over 0.2 L, that is 0.125 M, which is lower than the required 0.15 M.
2. 1.2 g: Dividing by 40 g/mol gives exactly 0.03 mol. Over 0.2 L, that works out to 0.03 / 0.2 = 0.15 M, matching the required concentration exactly.
3. 1.5 g: Dividing by 40 g/mol gives 0.0375 mol, which over 0.2 L gives 0.1875 M, higher than needed.
4. 1.8 g: Dividing by 40 g/mol gives 0.045 mol, which over 0.2 L gives 0.225 M, well above the target concentration.

Checking each mass this way shows that only 1.2 g of NaOH dissolved in 200 mL of water gives exactly 0.15 M.

Therefore, the correct answer is 1.2 g.