Question:
An aqueous solution containing 6.5 g of NaCl of 90% purity was subjected to electrolysis. After the complete electrolysis, the solution was evaporated to get solid NaOH. The volume of 1 M acetic acid required to neutralise NaOH obtained above is
An aqueous solution containing 6.5 g of NaCl of 90% purity was subjected to electrolysis. After the complete electrolysis, the solution was evaporated to get solid NaOH. The volume of 1 M acetic acid required to neutralise NaOH obtained above is
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Moles of $NaCl$ = Moles of $NaOH$ produced = Moles of acid required for neutralization.
Updated On: Apr 10, 2026
- $1000~cm^{3}$
- $2000~cm^{3}$
- $100~cm^{3}$
- $200~cm^{3}$
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The Correct Option is C
Solution and Explanation
Step 1: Find Pure NaCl
Weight of pure $NaCl = 6.5 \times 0.9 = 5.85~g$.
Step 2: Determine Equivalents
Equivalents of $NaCl = 5.85 / 58.5 = 0.1$. This yields 0.1 equivalents of $NaOH$.
Step 3: Neutralization Calculation
Volume of 1 M acetic acid $= \frac{0.1 \times 1000}{1} = 100~cm^{3}$.
Final Answer: (c)
Weight of pure $NaCl = 6.5 \times 0.9 = 5.85~g$.
Step 2: Determine Equivalents
Equivalents of $NaCl = 5.85 / 58.5 = 0.1$. This yields 0.1 equivalents of $NaOH$.
Step 3: Neutralization Calculation
Volume of 1 M acetic acid $= \frac{0.1 \times 1000}{1} = 100~cm^{3}$.
Final Answer: (c)
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