Preparation Of Standard Solution
How much NaOH (molar mass = 40 g mol^-1) is required to prepare 250 mL of 0.5 mol L^-1 NaOH solution?
Select the correct option:
Solution
5.0 g
Use n = M x V. Here M = 0.5 mol L^-1 and V = 0.250 L, so moles needed = 0.5 x 0.250 = 0.125 mol. Required mass = 0.125 x 40 = 5.0 g. Unit check: g mol^-1 x mol gives grams.
🔒 Solution Hidden from View
Submit your answer to unlock the detailed step-by-step solution.
About This Question
- Subject
- chemistry
- Chapter
- principles related to practical chemistry
- Topic
- preparation of standard solution
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
5.0 g
Use n = M x V. Here M = 0.5 mol L^-1 and V = 0.250 L, so moles needed = 0.5 x 0.250 = 0.125 mol. Required mass = 0.125 x 40 = 5.0 g. Unit check: g mol^-1 x mol gives grams.
This medium difficulty chemistry question is from the chapter principles related to practical chemistry, covering the topic of preparation of standard solution. It appeared in the 2025 exam.
Looking for more practice? Explore all chemistry questions or browse principles related to practical chemistry questions on RankGuru.