Question:medium

Calculate the osmotic pressure of solution containing 0.822 gm of sucrose in 300 mL of water at 298 K. [ Given : Molar mass of sucrose 342 g/mol, R = 0.08205 $dm^3$ atm $K^{-1}$ $mol^{-1}$ ]

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Always ensure the units of Volume ($V$) match the units of the Gas Constant ($R$). If $R$ is in $dm^3$, $V$ must be in Liters.
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Solution and Explanation

Step 1: Understanding the Concept:
Osmotic pressure (\(\pi\)) is a colligative property that depends on the molar concentration of the solute.
Step 2: Key Formula or Approach:
\[ \pi = CRT = \frac{n}{V} RT = \frac{w_2}{M_2 V} RT \]
Where:
\(w_2 = 0.822 \text{ g}\) (Mass of solute)
\(M_2 = 342 \text{ g/mol}\) (Molar mass of solute)
\(V = 300 \text{ mL} = 0.3 \text{ L} = 0.3 \text{ dm}^3\)
\(R = 0.08205 \text{ dm}^3 \text{ atm K}^{-1} \text{ mol}^{-1}\)
\(T = 298 \text{ K}\)
Step 3: Detailed Explanation:
First, calculate the number of moles of sucrose:
\[ n_2 = \frac{0.822}{342} \approx 0.002403 \text{ mol} \]
Now, substitute into the osmotic pressure formula:
\[ \pi = \frac{0.002403}{0.3} \times 0.08205 \times 298 \]
\[ \pi = 0.00801 \times 0.08205 \times 298 \]
\[ \pi \approx 0.196 \text{ atm} \]
Step 4: Final Answer: The osmotic pressure of the solution is \(0.196 \text{ atm}\).
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