Same quantity of charge is being used to liberate iodine (at anode) and a metal `M` (at cathode). The mass of metal `M` liberated is `0.617 g` and the liberated iodine is completely reduced by `46.3 mL` of `0.124 M` sodium thio-sulphate. Calculate equivalent weight of metal. Also calculate the total time to bring this change if `10` ampere current passed through solution of metal iodide.

Explanation

To solve the given problem step by step, we will first calculate the equivalent weight of the metal \( M \) and then determine the total time taken for the electrochemical reaction when a current of 10 amperes is passed through the solution.

### Step 1: Calculate the mass of sodium thiosulfate used
The formula for sodium thiosulfate is \( \text{Na}_2\text{S}_2\text{O}_3 \cdot 5\text{H}_2\text{O} \) and its molar mass is approximately 248.17 g/mol.

Using the formula:
\[ \text{Mass of sodium thiosulfate} = \text{Molarity} \times \text{Volume (L)} \times \text{Molar mass} \]
Given:
- Molarity = 0.124 M
- Volume = 46.3 mL = 0.0463 L

Calculating the mass:
\[ \text{Mass} = 0.124 \, \text{mol/L} \times 0.0463 \, \text{L} \times 248.17 \, \text{g/mol} \]
\[ \text{Mass} = 0.124 \times 0.0463 \times 248.17 \approx 1.42479 \, \text{g} \]

### Step 2: Calculate the equivalent of sodium thiosulfate
The equivalent weight of sodium thiosulfate can be calculated as follows:
\[ \text{Equivalent of sodium thiosulfate} = \frac{\text{Mass}}{\text{Equivalent weight}} \]
The equivalent weight of sodium thiosulfate is its molar mass divided by the number of electrons transferred in the reaction (which is typically 2 for thiosulfate):
\[ \text{Equivalent weight} = \frac{248.17 \, \text{g/mol}}{2} = 124.085 \, \text{g/equiv} \]
Thus:
\[ \text{Equivalent of sodium thiosulfate} = \frac{1.42479 \, \text{g}}{124.085 \, \text{g/equiv}} \approx 0.01148 \, \text{equiv} \]

### Step 3: Relate the equivalent of metal and sodium thiosulfate
Since the same quantity of charge is used to liberate iodine and metal \( M \):
\[ \text{Equivalent of metal} = \text{Equivalent of sodium thiosulfate} \]
Let \( E \) be the equivalent weight of metal \( M \):
\[ \frac{0.617 \, \text{g}}{E} = 0.01148 \, \text{equiv} \]
Rearranging gives:
\[ E = \frac{0.617 \, \text{g}}{0.01148 \, \text{equiv}} \approx 53.9 \, \text{g/equiv} \]

### Step 4: Calculate the time taken using Faraday's law
Using Faraday's law:
\[ \text{Equivalent} = \frac{I \cdot t}{96500} \]
Where:
- \( I = 10 \, \text{A} \)
- \( t \) is the time in seconds.

We can express the equivalent of sodium thiosulfate as:
\[ 0.01148 = \frac{10 \cdot t}{96500} \]
Rearranging gives:
\[ t = \frac{0.01148 \cdot 96500}{10} \]
Calculating \( t \):
\[ t \approx \frac{1108.54}{10} \approx 110.85 \, \text{s} \]

### Final Answers
- Equivalent weight of metal \( M \) is approximately \( 53.9 \, \text{g/equiv} \).
- Total time taken is approximately \( 110.85 \, \text{s} \).