19 g of molten `SnCl_2` is electrolysed for sometime using inert electrodes. 0.119g of Sn is deposited at the cathode. No substance is lost during the electrolysis. The ratio of the weights of `SnCl_2` : `SnCl_4` after electrolysis [Atomic weight of `Sn = 119`]

Options

A. `71.34 : 1`
B. `31.34 : 1`
C. `7.134 : 1`
D. None of these

Answer

To solve the problem, we need to determine the ratio of the weights of SnCl₂ to SnCl₄ after the electrolysis process. Let's break down the solution step by step.

### Step 1: Determine the molar masses of SnCl₂ and SnCl₄
- The molar mass of Sn (tin) = 119 g/mol
- The molar mass of Cl (chlorine) = 35.5 g/mol

**For SnCl₂:**
\[ \text{Molar mass of SnCl₂} = 119 + 2 \times 35.5 = 119 + 71 = 190 \text{ g/mol} \]

**For SnCl₄:**
\[ \text{Molar mass of SnCl₄} = 119 + 4 \times 35.5 = 119 + 142 = 261 \text{ g/mol} \]

### Step 2: Calculate the amount of Sn deposited
From the problem, we know that 0.119 g of Sn is deposited at the cathode.

### Step 3: Calculate the amount of SnCl₂ used to deposit 0.119 g of Sn
Using the molar mass of SnCl₂, we can find out how much SnCl₂ is needed to deposit 0.119 g of Sn.

From the stoichiometry of the reaction:
- 119 g of Sn is deposited from 380 g of SnCl₂ (as 1 mole of SnCl₂ produces 1 mole of Sn).

Using proportions:
\[ \text{If } 119 \text{ g Sn} \text{ is produced from } 380 \text{ g SnCl₂, then } 0.119 \text{ g Sn is produced from } x \text{ g SnCl₂} \]

Setting up the proportion:
\[ \frac{0.119 \text{ g Sn}}{119 \text{ g Sn}} = \frac{x \text{ g SnCl₂}}{380 \text{ g SnCl₂}} \]

Cross-multiplying gives:
\[ x = \frac{0.119 \times 380}{119} = 0.380 \text{ g SnCl₂} \]

### Step 4: Calculate the remaining SnCl₂ after electrolysis
Initially, we had 19 g of SnCl₂. After using 0.380 g for the deposition of Sn, the remaining amount of SnCl₂ is:
\[ 19 \text{ g} - 0.380 \text{ g} = 18.62 \text{ g SnCl₂} \]

### Step 5: Calculate the amount of SnCl₄ produced
Since 0.380 g of SnCl₂ is used, it will produce an equivalent amount of SnCl₄ based on the stoichiometry of the reaction.

Using the molar mass of SnCl₄:
- From 0.380 g of SnCl₂, we can find how much SnCl₄ is produced.

Using the ratio:
\[ \frac{0.380 \text{ g SnCl₂}}{380 \text{ g SnCl₂}} = \frac{y \text{ g SnCl₄}}{261 \text{ g SnCl₄}} \]

Cross-multiplying gives:
\[ y = \frac{0.380 \times 261}{380} = 0.261 \text{ g SnCl₄} \]

### Step 6: Find the ratio of weights of SnCl₂ to SnCl₄
Now we have:
- Weight of SnCl₂ = 18.62 g
- Weight of SnCl₄ = 0.261 g

The ratio of weights of SnCl₂ to SnCl₄ is:
\[ \text{Ratio} = \frac{18.62 \text{ g SnCl₂}}{0.261 \text{ g SnCl₄}} \]

Calculating this gives:
\[ \text{Ratio} = 71.34 \]

### Final Answer
The ratio of the weights of SnCl₂ to SnCl₄ after electrolysis is approximately **71.34:1**.

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