What mass of iodine is liberated from a solution of potassium iodide when 1 litre of chlorine gas at `10^(@)C` and 750 mm pressure is passed through it?

Explanation

To find the mass of iodine liberated from a solution of potassium iodide when 1 liter of chlorine gas at 10°C and 750 mm pressure is passed through it, we can follow these steps:

### Step 1: Convert Temperature to Kelvin
The temperature in Kelvin (K) is calculated by adding 273.15 to the temperature in Celsius (°C).
\[ T = 10°C + 273.15 = 283.15 \, K \]

### Step 2: Convert Pressure to Atmospheres
To convert the pressure from mm of Hg to atmospheres (ATM), we use the conversion factor where 1 ATM = 760 mm Hg.
\[ P = \frac{750 \, \text{mm Hg}}{760 \, \text{mm Hg/ATM}} = 0.9868 \, \text{ATM} \]

### Step 3: Use the Ideal Gas Law to Find Moles of Chlorine
The ideal gas equation is given by:
\[ PV = nRT \]
Where:
- \( P \) = pressure in ATM
- \( V \) = volume in liters (1 L)
- \( n \) = number of moles
- \( R \) = ideal gas constant (0.0821 L·ATM/(K·mol))
- \( T \) = temperature in Kelvin

Rearranging the equation to solve for \( n \):
\[ n = \frac{PV}{RT} \]
Substituting the values:
\[ n = \frac{(0.9868 \, \text{ATM}) \times (1 \, \text{L})}{(0.0821 \, \text{L·ATM/(K·mol)}) \times (283.15 \, K)} \]
Calculating \( n \):
\[ n = \frac{0.9868}{23.247} \approx 0.04245 \, \text{moles of Cl}_2 \]

### Step 4: Determine Moles of Iodine Produced
From the balanced chemical equation:
\[ Cl_2 + 2KI \rightarrow 2KCl + I_2 \]
It can be seen that 1 mole of \( Cl_2 \) produces 1 mole of \( I_2 \). Thus, the moles of iodine produced will also be:
\[ \text{Moles of } I_2 = 0.04245 \, \text{moles} \]

### Step 5: Calculate the Mass of Iodine
The molar mass of iodine (\( I_2 \)) is calculated as follows:
\[ \text{Molar mass of } I_2 = 2 \times 127 \, \text{g/mol} = 254 \, \text{g/mol} \]
Now, to find the mass of iodine:
\[ \text{Mass of } I_2 = \text{Moles} \times \text{Molar mass} = 0.04245 \, \text{moles} \times 254 \, \text{g/mol} \]
Calculating the mass:
\[ \text{Mass of } I_2 \approx 10.78 \, \text{grams} \]

### Final Answer
The mass of iodine liberated is approximately **10.78 grams**.

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