After electrlysis of `NaCI` solution with inert electrodes for a certain period of time `600mL` of the solution was left. Which was found to be `1N` in `NaOH`. During the same time, `31.75 g` of `Cu` deposited in the copper voltameter in series the electrolytic cell. calculate the percentage yield of `NaOH` obtained.

Explanation

To solve the problem, we need to calculate the percentage yield of NaOH obtained after the electrolysis of NaCl solution. Here are the steps to arrive at the solution:

### Step 1: Calculate the number of moles of Cu deposited
Given that 31.75 g of Cu is deposited, we can calculate the number of moles of Cu using its molar mass (63.5 g/mol).

\[ \text{Number of moles of Cu} = \frac{\text{mass of Cu}}{\text{molar mass of Cu}} = \frac{31.75 \, \text{g}}{63.5 \, \text{g/mol}} \approx 0.500 \, \text{mol} \]

### Step 2: Calculate the number of moles of electrons used
The reduction of Cu²⁺ to Cu involves the transfer of 2 moles of electrons for every mole of Cu deposited.

\[ \text{Number of moles of electrons} = 2 \times \text{Number of moles of Cu} = 2 \times 0.500 \, \text{mol} = 1.000 \, \text{mol} \]

### Step 3: Relate the moles of electrons to moles of OH⁻ produced
During the electrolysis of water, 2 moles of electrons produce 1 mole of OH⁻. Thus, the number of moles of OH⁻ produced is equal to the number of moles of electrons divided by 2.

\[ \text{Number of moles of OH⁻} = \frac{\text{Number of moles of electrons}}{2} = \frac{1.000 \, \text{mol}}{2} = 0.500 \, \text{mol} \]

### Step 4: Calculate the normality of the NaOH solution
We know that the volume of the NaOH solution left after electrolysis is 600 mL, which is equivalent to 0.600 L. The normality (N) of the NaOH solution is given as 1N.

\[ \text{Normality} = \frac{\text{Number of equivalents}}{\text{Volume in L}} \]

Since NaOH is a strong base and dissociates completely, 1 mole of NaOH provides 1 equivalent of OH⁻. Therefore, the number of equivalents of NaOH produced is equal to the number of moles of OH⁻ produced.

\[ \text{Number of equivalents of NaOH} = 0.500 \, \text{mol} \]

Calculating the normality:

\[ \text{Normality} = \frac{0.500 \, \text{equivalents}}{0.600 \, \text{L}} = \frac{0.500}{0.600} \approx 0.833 \, \text{N} \]

### Step 5: Calculate the percentage yield of NaOH
The theoretical yield of NaOH based on the moles of OH⁻ produced is 1N (as given in the problem). The actual yield is 0.833N.

\[ \text{Percentage yield} = \left( \frac{\text{Actual yield}}{\text{Theoretical yield}} \right) \times 100 = \left( \frac{0.833}{1} \right) \times 100 \approx 83.3\% \]

### Final Answer
The percentage yield of NaOH obtained is approximately **83.3%**.

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