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 find the osmotic pressure of the resulting solution when 1 L of NaCl solution with an osmotic pressure of 0.1 atm is mixed with 2 L of glucose solution with an osmotic pressure of 0.2 atm.

### 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 the osmotic pressure of the resulting solution:**
The formula for the osmotic pressure (π) of the resulting solution when two solutions are mixed is given by:
\[ \pi = \frac{\pi_1 V_1 + \pi_2 V_2}{V_1 + V_2} \]

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

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

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

6. **Calculate the osmotic pressure:**
\[ \pi = \frac{0.5 \, \text{atm L}}{3 \, \text{L}} = \frac{0.5}{3} \, \text{atm} \approx 0.16667 \, \text{atm} \]

7. **Convert to the required format:**
\[ 0.16667 \, \text{atm} = 166.67 \times 10^{-3} \, \text{atm} \]

8. **Round to the nearest integer:**
The nearest integer to 166.67 is 167.

### Final Answer:
The value of \(X\) in the expression \(X \times 10^{-3}\) atm is **167**.