Step 1: Use the formula for osmotic pressure.
Osmotic pressure (\( \Pi \)) is given by the equation:
\[ \Pi = \frac{nRT}{V} \] Where: \( n \) is the number of moles of solute, \( R \) is the gas constant, \( T \) is the temperature in Kelvin, \( V \) is the volume of the solution.
Step 2: Relate the osmotic pressures.
Since the number of moles of solute and the volume remain constant, the osmotic pressure is directly proportional to the temperature. Hence, we can use the relation: \[ \frac{\Pi_2}{\Pi_1} = \frac{T_2}{T_1} \] Where:
\( \Pi_1 = 500 \, \text{mm of Hg} \), \( \Pi_2 = 105.3 \, \text{mm of Hg} \), \( T_1 = 10°C = 273 + 10 = 283 \, \text{K} \), \( T_2 = 25°C = 273 + 25 = 298 \, \text{K} \).
Step 3: Calculate the ratio of osmotic pressures.
Now, applying the formula: \[ \frac{105.3}{500} = \frac{298}{283} \] Simplifying: \[ \frac{105.3}{500} \approx 0.2106, \quad \frac{298}{283} \approx 1.053. \] So the degree of dilution is: \[ \text{Degree of dilution} = \frac{1}{4.5} = 4.5 \text{ times}. \]
The osmotic pressure of seawater is 1.05 atm. Four experiments were carried out as shown in the table. In which of the following experiments, pure water can be obtained in part-II of the vessel?
Calculate the EMF of the Galvanic cell: $ \text{Zn} | \text{Zn}^{2+}(1.0 M) \parallel \text{Cu}^{2+}(0.5 M) | \text{Cu} $ Given: $ E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.763 \, \text{V} $ and $ E^\circ_{\text{Cu}^{2+}/\text{Cu}} = +0.350 \, \text{V} $
Find the values of a, b, c, and d for the following redox equation: $ a\text{I}_2 + b\text{NO} + 4\text{H}_2\text{O} = c\text{HNO}_3 + d\text{HI} $
If equilibrium constant for the equation $ A_2 + B_2 \rightleftharpoons 2AB \quad \text{is} \, K_p, $ then find the equilibrium constant for the equation $ AB \rightleftharpoons \frac{1}{2} A_2 + \frac{1}{2} B_2. $