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A solution containing 3.00gm of calcium ...

A solution containing `3.00gm` of calcium nitrate in `100 c.c.` of solution had an osmotic pressure of `11.2` atmosphere at `12^(@)C`. Calculate the degree of ionisation of calcium nitrate at this dilution and temperature.

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To solve the problem of calculating the degree of ionization of calcium nitrate in the given solution, we will follow these steps: ### Step 1: Calculate the Molar Mass of Calcium Nitrate The formula for calcium nitrate is \( \text{Ca(NO}_3\text{)}_2 \). - Molar mass of Calcium (Ca) = 40 g/mol - Molar mass of Nitrogen (N) = 14 g/mol (2 Nitrogen atoms = 28 g/mol) - Molar mass of Oxygen (O) = 16 g/mol (6 Oxygen atoms = 96 g/mol) So, the molar mass of calcium nitrate is: \[ \text{Molar mass} = 40 + 28 + 96 = 164 \text{ g/mol} \] ### Step 2: Calculate the Number of Moles of Calcium Nitrate Given mass of calcium nitrate = 3 g. Using the formula: \[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{3 \text{ g}}{164 \text{ g/mol}} \approx 0.0183 \text{ moles} \] ### Step 3: Calculate the Molarity of the Solution The volume of the solution is given as 100 cm³, which is equivalent to 0.1 L. Molarity (C) is given by: \[ C = \frac{\text{Number of moles}}{\text{Volume in L}} = \frac{0.0183 \text{ moles}}{0.1 \text{ L}} = 0.183 \text{ mol/L} \] ### Step 4: Use the Osmotic Pressure Equation The osmotic pressure (\( \Pi \)) is given by the formula: \[ \Pi = iCRT \] Where: - \( \Pi = 11.2 \text{ atm} \) - \( C = 0.183 \text{ mol/L} \) - \( R = 0.0821 \text{ L atm/(K mol)} \) - \( T = 12^\circ C = 285 \text{ K} \) Rearranging the formula to find the van't Hoff factor \( i \): \[ i = \frac{\Pi}{CRT} \] Substituting the values: \[ i = \frac{11.2}{0.183 \times 0.0821 \times 285} \] Calculating the denominator: \[ 0.183 \times 0.0821 \times 285 \approx 4.29 \] Thus, \[ i \approx \frac{11.2}{4.29} \approx 2.61 \] ### Step 5: Determine the Degree of Ionization Calcium nitrate dissociates in solution as follows: \[ \text{Ca(NO}_3\text{)}_2 \rightarrow \text{Ca}^{2+} + 2\text{NO}_3^{-} \] This means that for every 1 mole of calcium nitrate, we get 3 moles of ions (1 calcium ion and 2 nitrate ions). Let \( \alpha \) be the degree of ionization. Then: \[ i = 1 + 2\alpha \] Substituting the value of \( i \): \[ 2.61 = 1 + 2\alpha \] Solving for \( \alpha \): \[ 2\alpha = 2.61 - 1 = 1.61 \implies \alpha = \frac{1.61}{2} \approx 0.805 \] ### Final Result The degree of ionization of calcium nitrate at this dilution and temperature is approximately \( 0.805 \) or \( 80.5\% \). ---

To solve the problem of calculating the degree of ionization of calcium nitrate in the given solution, we will follow these steps: ### Step 1: Calculate the Molar Mass of Calcium Nitrate The formula for calcium nitrate is \( \text{Ca(NO}_3\text{)}_2 \). - Molar mass of Calcium (Ca) = 40 g/mol - Molar mass of Nitrogen (N) = 14 g/mol (2 Nitrogen atoms = 28 g/mol) - Molar mass of Oxygen (O) = 16 g/mol (6 Oxygen atoms = 96 g/mol) ...
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