A metal is know to form fluoride `MF_(2)`. When `10A` of electricity is passed through a molten sat for `330sec, 1.95g` of metal is deposited. Find the atomic weight of M. what will be the quantity electricity required to deposit the same mass of `Cu` form `CuSO_(4)`?

Explanation

To solve the problem step by step, we will break it down into two parts: finding the atomic weight of the metal (M) and calculating the quantity of electricity required to deposit the same mass of copper (Cu) from CuSO4.

### Part 1: Finding the Atomic Weight of Metal (M)

1. **Identify the Given Data:**
- Current (I) = 10 A
- Time (t) = 330 s
- Mass of metal deposited (W) = 1.95 g
- Valency factor of metal fluoride (MF2) = 2 (since it is bivalent)

2. **Use Faraday's Law of Electrolysis:**
The relationship can be expressed as:
\[ \frac{W}{\text{Equivalent Weight}} = \frac{I \cdot t}{96500} \]
where 96500 C is the Faraday constant.

3. **Calculate the Equivalent Weight:**
The equivalent weight (EW) is given by:
\[ \text{Equivalent Weight} = \frac{\text{Atomic Weight}}{\text{Valency Factor}} = \frac{A}{2} \]
where A is the atomic weight of metal M.

4. **Substitute Values into the Equation:**
\[ \frac{1.95}{\frac{A}{2}} = \frac{10 \cdot 330}{96500} \]

5. **Simplify the Equation:**
\[ \frac{1.95 \cdot 2}{A} = \frac{3300}{96500} \]
\[ \frac{3.90}{A} = \frac{3300}{96500} \]

6. **Cross Multiply to Solve for A:**
\[ 3.90 \cdot 96500 = 3300 \cdot A \]
\[ 376350 = 3300A \]
\[ A = \frac{376350}{3300} \approx 114 \text{ g/mol} \]

### Part 2: Quantity of Electricity Required to Deposit 1.95 g of Cu

1. **Identify the Given Data for Copper:**
- Mass of copper deposited (W) = 1.95 g
- Atomic weight of copper (A) = 63.5 g/mol
- Valency factor for Cu (from CuSO4) = 2 (since Cu²⁺ gains 2 electrons)

2. **Use Faraday's Law Again:**
\[ \frac{W}{\text{Equivalent Weight}} = \frac{Q}{96500} \]

3. **Calculate the Equivalent Weight for Copper:**
\[ \text{Equivalent Weight} = \frac{63.5}{2} = 31.75 \text{ g/equiv} \]

4. **Substitute Values into the Equation:**
\[ \frac{1.95}{31.75} = \frac{Q}{96500} \]

5. **Cross Multiply to Solve for Q:**
\[ 1.95 \cdot 96500 = 31.75 \cdot Q \]
\[ 188775 = 31.75Q \]
\[ Q = \frac{188775}{31.75} \approx 5917.45 \text{ C} \]

### Final Answers:
- The atomic weight of metal M is approximately **114 g/mol**.
- The quantity of electricity required to deposit 1.95 g of Cu from CuSO4 is approximately **5917.45 C**.