Osmotic pressure of solution formed by mixing \(50\, \ mL\) of \(0.2\, \ M\) urea and \(50\, \ mL\) of \(0.4\, \ m\) glucose aqueous solution at \(298\, \ K\) will be: \((R = 0.083\, \ L\, \ atm\, \ K^{-1}\, \ mol^{-1})\)

7.42 atm

The correct answer is option 2. 7.42 atm.

To calculate the osmotic pressure of the solution, we can use the formula:

\[ \pi = i \cdot M \cdot R \cdot T \]

where:

\( \pi \) is the osmotic pressure,

\( i \) is the van't Hoff factor,

\( M \) is the molarity of the solution,

\( R \) is the ideal gas constant, and

\( T \) is the temperature in Kelvin.

First, let's calculate the total number of moles of solute particles in the solution. Then, we'll determine the van't Hoff factor and plug in the values into the formula.

Given:

For urea (\( NH_2CONH_2 \)), \( i = 1 \) because it does not dissociate in solution.

For glucose (\( C_6H_{12}O_6 \)), \( i = 1 \) because it also does not dissociate in solution.

\( R = 0.083 \, L \, atm \, K^{-1} \, mol^{-1} \)

\( T = 298 \, K \)

First, calculate the total number of moles of solute particles:

For urea:
\[ \text{Moles of urea} = M \cdot V = 0.2 \, M \times 0.050 \, L = 0.010 \, moles \]

For glucose:
\[ \text{Moles of glucose} = M \cdot V = 0.4 \, M \times 0.050 \, L = 0.020 \, moles \]

Total moles of solute particles:
\[ \text{Total moles} = 0.010 \, moles \, (urea) + 0.020 \, moles \, (glucose) = 0.030 \, moles \]

Now, let's calculate the osmotic pressure using the formula:

\[ \pi = i \cdot M \cdot R \cdot T \]

\[ \pi = 1 \times 0.3 \, M \times 0.083 \, L \, atm \, K^{-1} \, mol^{-1} \times 298 \, K \]

\[ \pi = 7.434 \, atm \]

So, the osmotic pressure of the solution is approximately \( 7.434 \, atm \), which rounds to \( 7.42 \, atm \).

Therefore, the correct answer is option 2: \( 7.42 \, atm \).