The temperature at which 10% aqueous solution of glucose will show the osmotic pressure of 14 atm is:
Answer & explanation
Correct answer: option 3
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.