Certainly! Let's answer each question step-by-step.
#### Question 5: What amount of anhydrous sodium carbonate (Na₂CO₃) is required to prepare 250 mL of a decinormal solution?
1. Determine the Equivalent Weight of Na₂CO₃:
- The molecular weight of anhydrous sodium carbonate (Na₂CO₃) is 106 g/mol.
- Since Na₂CO₃ dissociates into 2 moles of sodium ions and 1 mole of carbonate ion, the equivalent weight is the molecular weight divided by 2.
- Equivalent Weight = 106 g/mol / 2 = 53 grams/equivalent.
2. Decinormal Solution (0.1 N):
- A decinormal (0.1 N) solution means it contains 0.1 equivalents per liter.
3. Calculate the Required Amount for 1 Liter:
- For 1 L of 0.1 N solution, the amount of Na₂CO₃ required is 0.1 equivalents.
- Required in 1 Liter = 0.1 53 grams = 5.3 grams.
4. Calculate the Required Amount for 250 mL:
- To find the amount needed for 250 mL (which is 0.25 L), use the proportion:
- Required for 250 mL = (250 mL / 1000 mL) 5.3 grams = 0.25 * 5.3 grams = 1.325 grams.
Answer: The amount of anhydrous sodium carbonate required to prepare 250 mL of a decinormal solution is 1.325 grams.
#### Question 6: What is the strength of a 1% (W/V) NaOH solution?
1. Understanding 1% (W/V) Solution:
- The notation 1% (Weight/Volume) means that there is 1 gram of the solute (NaOH) per 100 mL of solution.
Answer: The strength of a 1% (W/V) NaOH solution is 1 gram per 100 mL.
