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)

The correct option is 7.42 atm.

Step-by-Step Explanation:

To find the osmotic pressure of the resulting mixture, we need to determine the total concentration of solute particles in the final solution. Both urea and glucose are non-electrolytes, meaning they do not dissociate or associate in aqueous solutions. Thus, their van 't Hoff factor ($i$) is equal to 1.

Step 1: Calculate the moles of each solute.
The number of moles ($n$) of a solute is given by the formula:
$n=\text{Molarity}\times \text{Volume in Liters}$

For urea:
Volume, $V_1=50\text{mL}=0.050\text{L}$
Molarity, $M_1=0.2\text{M}$
Moles of urea, $n_{\text{urea}}=0.2\text{mol/L}\times 0.050\text{L}=0.01\text{mol}$

For glucose (assuming the concentration is approximately 0.4 M for the dilute aqueous solution):
Volume, $V_2=50\text{mL}=0.050\text{L}$
Molarity, $M_2=0.4\text{M}$
Moles of glucose, $n_{\text{glucose}}=0.4\text{mol/L}\times 0.050\text{L}=0.02\text{mol}$

Step 2: Calculate the total volume of the mixture.
$V_{\text{total}}=V_1+V_2=50\text{mL}+50\text{mL}=100\text{mL}=0.1\text{L}$

Step 3: Calculate the total molar concentration ($C$) of the final solution.
Since both solutes contribute to the osmotic pressure, we sum their moles over the total volume:
$C=\frac{n_{\text{urea}}+n_{\text{glucose}}}{V_{\text{total}}}$

$C=\frac{0.01\text{mol}+0.02\text{mol}}{0.1\text{L}}=\frac{0.03}{0.1}=0.3\text{M}$

Step 4: Calculate the osmotic pressure ($\pi$) using the osmotic pressure formula.
The osmotic pressure is given by:
$\pi =CRT$

Given constants:
$R=0.083{\text{L atm K}}^{-1}{\text{mol}}^{-1}$
$T=298\text{K}$

Substitute the values into the formula:
$\pi =0.3\times 0.083\times 298$

$\pi =0.0249\times 298=7.4202\text{atm}$

Thus, the osmotic pressure of the solution is approximately 7.42 atm.