If `1` mole of `CH_(3)COOH` and `1` mole of `C_(2)H_(5)OH` are taken in `1` litre flask, `50%` of `CH_(3)COOH` is converted into ester as,
`CH_(3)COOH_((l))+C_(2)H_(5)OH_((l))hArrCH_(3)COOC_(2)H_(5)(l)+H_(2)O_((l))`
There is `33%` conversion of `CH_(3)COOH` into ester, if `CH_(3)COOH "and" C_(2)H_(5)OH` have been taken initially in molar ratio `x:1`, find `x`.

Explanation

To solve the problem, we need to analyze the reaction and the conversions given. Let's break it down step by step.

### Step 1: Understand the Reaction
The reaction given is:
\[ \text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \rightleftharpoons \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O} \]
This is the esterification reaction between acetic acid (CH₃COOH) and ethyl alcohol (C₂H₅OH).

### Step 2: Initial Moles and Concentrations
Initially, we have:
- 1 mole of acetic acid (CH₃COOH)
- 1 mole of ethyl alcohol (C₂H₅OH)

### Step 3: First Condition (50% Conversion)
From the problem, we know that if 1 mole of each is taken, 50% of CH₃COOH is converted into ester. This means:
- Moles of CH₃COOH at equilibrium = \(1 - 0.5 = 0.5\) moles
- Moles of C₂H₅OH at equilibrium = \(1 - 0.5 = 0.5\) moles
- Moles of CH₃COOC₂H₅ formed = 0.5 moles
- Moles of H₂O formed = 0.5 moles

### Step 4: Equilibrium Constant (Kc) Calculation
The equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[\text{CH}_3\text{COOC}_2\text{H}_5][\text{H}_2\text{O}]}{[\text{CH}_3\text{COOH}][\text{C}_2\text{H}_5\text{OH}]} \]
Substituting the equilibrium concentrations:
\[ K_c = \frac{(0.5)(0.5)}{(0.5)(0.5)} = 1 \]

### Step 5: Second Condition (Molar Ratio \( x:1 \))
Now, we consider the case where the initial moles of CH₃COOH and C₂H₅OH are in the ratio \( x:1 \). Let:
- Moles of CH₃COOH = \( x \)
- Moles of C₂H₅OH = \( 1 \)

Given that there is a 33% conversion of CH₃COOH, we have:
- Moles of CH₃COOH reacted = \( \frac{1}{3}x \)
- Moles of CH₃COOH remaining = \( x - \frac{1}{3}x = \frac{2}{3}x \)
- Moles of C₂H₅OH reacted = \( \frac{1}{3}x \)
- Moles of C₂H₅OH remaining = \( 1 - \frac{1}{3}x \)
- Moles of CH₃COOC₂H₅ formed = \( \frac{1}{3}x \)
- Moles of H₂O formed = \( \frac{1}{3}x \)

### Step 6: Equilibrium Constant for the Second Condition
Using the same formula for \( K_c \):
\[ K_c = \frac{[\text{CH}_3\text{COOC}_2\text{H}_5][\text{H}_2\text{O}]}{[\text{CH}_3\text{COOH}][\text{C}_2\text{H}_5\text{OH}]} \]
Substituting the equilibrium concentrations:
\[ K_c = \frac{\left(\frac{1}{3}x\right)\left(\frac{1}{3}x\right)}{\left(\frac{2}{3}x\right)\left(1 - \frac{1}{3}x\right)} \]
Since \( K_c \) remains the same (equal to 1 from the first condition), we set up the equation:
\[ 1 = \frac{\left(\frac{1}{3}x\right)^2}{\left(\frac{2}{3}x\right)\left(1 - \frac{1}{3}x\right)} \]

### Step 7: Solve for \( x \)
Cross-multiplying and simplifying:
\[ \left(\frac{1}{3}x\right)^2 = \left(\frac{2}{3}x\right)\left(1 - \frac{1}{3}x\right) \]
\[ \frac{1}{9}x^2 = \frac{2}{3}x - \frac{2}{9}x^2 \]
Combining like terms:
\[ \frac{1}{9}x^2 + \frac{2}{9}x^2 - \frac{2}{3}x = 0 \]
\[ \frac{3}{9}x^2 - \frac{2}{3}x = 0 \]
Factoring out \( x \):
\[ x\left(\frac{1}{3}x - \frac{2}{3}\right) = 0 \]
This gives us:
\[ x = 0 \quad \text{or} \quad x = 2 \]

### Final Answer
Thus, the value of \( x \) is:
\[ \boxed{2} \]