QuestionJuly 9, 2025

The osmotic pressure of a solution containing 2.63 mg of a sugar (342g/mole) in 13.0 mL of the solution at 25^circ C is __ torr. 3760 1.45times 10^-2 1.10times 10^-2 1.10times 10^4 11.0

The osmotic pressure of a solution containing 2.63 mg of a sugar (342g/mole) in 13.0 mL of the solution at 25^circ C is __ torr. 3760 1.45times 10^-2 1.10times 10^-2 1.10times 10^4 11.0
The osmotic pressure of a solution containing 2.63 mg of a sugar (342g/mole) in 13.0 mL of the solution at 25^circ C is __ torr.
3760
1.45times 10^-2
1.10times 10^-2
1.10times 10^4
11.0

Solution
4.1(197 votes)

Answer

1.10 \times 10^{-2} torr Explanation 1. Convert mass to moles Calculate moles of sugar using \text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} . Convert 2.63 mg to grams: 2.63 \, \text{mg} = 0.00263 \, \text{g}. Moles = \frac{0.00263}{342}. 2. Calculate molarity Convert 13.0 mL to liters: 13.0 \, \text{mL} = 0.013 \, \text{L}. Molarity M = \frac{\text{moles}}{\text{volume (L)}}. 3. Apply osmotic pressure formula Use **\Pi = MRT**, where R = 0.0821 \, \text{L atm/mol K} and T = 298 \, \text{K}. Convert atm to torr by multiplying by 760.

Explanation

1. Convert mass to moles<br /> Calculate moles of sugar using $ \text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} $. Convert 2.63 mg to grams: $2.63 \, \text{mg} = 0.00263 \, \text{g}$. Moles = $\frac{0.00263}{342}$.<br /><br />2. Calculate molarity<br /> Convert 13.0 mL to liters: $13.0 \, \text{mL} = 0.013 \, \text{L}$. Molarity $M = \frac{\text{moles}}{\text{volume (L)}}$.<br /><br />3. Apply osmotic pressure formula<br /> Use **$\Pi = MRT$**, where $R = 0.0821 \, \text{L atm/mol K}$ and $T = 298 \, \text{K}$. Convert atm to torr by multiplying by 760.
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