The temperature at which 10% aqueous solution of glucose will show the osmotic pressure of 14 atm is:

307.3 K

The correct answer is option 3. 307.3 K.

To find the temperature at which a 10% aqueous solution of glucose will show an osmotic pressure of 14 atm, we can use the formula for osmotic pressure:

\(\Pi = iMRT \)

where:

\(\Pi\) is the osmotic pressure,

\(i\) is the van 't Hoff factor (for glucose, which does not dissociate in solution, \(i = 1\)),

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

\(R\) is the universal gas constant (0.0821 L·atm·K⁻¹·mol⁻¹),

\(T\) is the temperature in Kelvin.

First, we need to determine the molarity (\(M\)) of the glucose solution.

A 10% by weight aqueous solution of glucose means there are 10 grams of glucose in 100 grams of solution.

The molar mass of glucose \((C_6H_{12}O_{6})\) is approximately 180 g/mol.

Number of moles of glucose \( = \frac{10 \text{ g}}{180 \text{ g/mol}} \approx 0.0556 \text{ mol} \).

The density of water is approximately 1 g/mL, so 100 grams of the solution is roughly 100 mL or 0.1 L (since the solution is dilute, we can assume the volume is close to that of water).

Molarity \(M = \frac{\text{moles of solute}}{\text{volume of solution in liters}} = \frac{0.0556 \text{ mol}}{0.1 \text{ L}} = 0.556 \text{ M} \).

\( \Pi = iMRT \)

Given:

\(\Pi = 14 \text{ atm}\),

\(i = 1\),

\(M = 0.556 \text{ M}\),

\(R = 0.0821 \text{ L·atm·K⁻¹·mol⁻¹}\).

Rearrange the formula to solve for \(T\):

\(T = \frac{\Pi}{iMR} \)

Substitute the known values:

\(T = \frac{14 \text{ atm}}{(1)(0.556 \text{ mol/L})(0.0821 \text{ L atm K}^{-1}mol^{-1})} \)

\(T = \frac{14}{0.04566} \)

\(T \approx 306.8 \text{ K} \)

The closest answer to this value is: 307.3 K.