Osmotic pressure of NaCl solution is 0.1 atm and glucose solution id 0.2 atm. if 1l of NaCl solution & 2L of glucose solution is mixed at same temperature, then osmotic pressure of resulting solution is `X xx 10^(-3)` atm, then value of X in nearest integer is

Answer

To solve the problem, we need to calculate the osmotic pressure of the resulting solution after mixing 1 liter of NaCl solution with 2 liters of glucose solution.

### Step-by-Step Solution:

1. **Identify the given values:**
- Osmotic pressure of NaCl solution (π₁) = 0.1 atm
- Volume of NaCl solution (V₁) = 1 L
- Osmotic pressure of glucose solution (π₂) = 0.2 atm
- Volume of glucose solution (V₂) = 2 L

2. **Use the formula for osmotic pressure of the resulting solution:**
The formula to calculate the osmotic pressure of the resulting solution when two solutions are mixed is:
\[ \pi_{result} = \frac{\pi_1 V_1 + \pi_2 V_2}{V_1 + V_2} \]

3. **Substitute the values into the formula:**
\[ \pi_{result} = \frac{(0.1 \, \text{atm} \times 1 \, \text{L}) + (0.2 \, \text{atm} \times 2 \, \text{L})}{1 \, \text{L} + 2 \, \text{L}} \]

4. **Calculate the numerator:**
- For NaCl: \(0.1 \times 1 = 0.1\)
- For glucose: \(0.2 \times 2 = 0.4\)
- Total: \(0.1 + 0.4 = 0.5\)

5. **Calculate the denominator:**
\[ V_1 + V_2 = 1 + 2 = 3 \, \text{L} \]

6. **Calculate the osmotic pressure of the resulting solution:**
\[ \pi_{result} = \frac{0.5}{3} \, \text{atm} \]

7. **Convert the result to the desired format:**
\[ \pi_{result} = \frac{500}{3} \times 10^{-3} \, \text{atm} \]

8. **Calculate the numerical value:**
\[ \frac{500}{3} \approx 166.67 \]
Thus, we can express it as:
\[ \pi_{result} \approx 166.67 \times 10^{-3} \, \text{atm} \]

9. **Determine the value of X:**
Since we need the nearest integer value of X:
\[ X \approx 167 \]

### Final Answer:
The value of X in the nearest integer is **167**.