`COCl_(2)` gas dissociates according to the equation, `COCl_(2)hArrCO(g)+Cl_(2)(g)`. When heated to 700 K the density of the gas mixture at 1.16 atm and at equilibrium is `1.16 g//litre` The degree of dissociation of `COCl_(2)` at 700K is :
(a)`0.28`
(b)`0.50`
(c)`0.72`
(d)`0.42`

Options

A. `0.28`
B. `0.50`
C. `0.72`
D. `0.42`

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the observed molecular mass using the formula
The formula for observed molecular mass (M) is given by:
\[ M = \frac{dRT}{P} \]
Where:
- \( d \) = density of the gas mixture = 1.16 g/L
- \( R \) = gas constant = 0.0821 L·atm/(K·mol)
- \( T \) = temperature = 700 K
- \( P \) = pressure = 1.16 atm

Substituting the values into the formula:
\[ M = \frac{1.16 \times 0.0821 \times 700}{1.16} \]

### Step 2: Simplify the equation
Since the pressure \( P \) is in the denominator and also equals the density, we can simplify:
\[ M = 1.16 \times 0.0821 \times 700 \]

Calculating:
\[ M = 1.16 \times 57.47 \]
\[ M \approx 57.47 \text{ g/mol} \]

### Step 3: Calculate the theoretical molecular mass of \( COCl_2 \)
The molecular mass of \( COCl_2 \) can be calculated as follows:
- Carbon (C) = 12 g/mol
- Oxygen (O) = 16 g/mol
- Chlorine (Cl) = 35.5 g/mol (and there are 2 Cl atoms)

So,
\[ \text{Molecular mass of } COCl_2 = 12 + 16 + (2 \times 35.5) = 12 + 16 + 71 = 99 \text{ g/mol} \]

### Step 4: Determine the degree of dissociation (\( \alpha \))
The degree of dissociation can be calculated using the formula:
\[ \alpha = \frac{M_{theoretical} - M_{observed}}{n - 1} \times M_{observed} \]
Where:
- \( M_{theoretical} = 99 \text{ g/mol} \)
- \( M_{observed} = 57.47 \text{ g/mol} \)
- \( n \) = number of moles of gaseous products = 2 (1 mole of CO + 1 mole of Cl2)

Substituting the values:
\[ \alpha = \frac{99 - 57.47}{2 - 1} \times 57.47 \]

Calculating:
\[ \alpha = \frac{41.53}{1} \times 57.47 \]
\[ \alpha = 0.72 \]

### Conclusion
The degree of dissociation of \( COCl_2 \) at 700 K is \( 0.72 \).

### Answer
(c) 0.72