Correct answer Carries: 4.
Wrong Answer Carries: -1.
A 0.3 M solution of a solute in 200 mL of water has an osmotic pressure of 2.214 atm at 27°C. If the solute dissociates into 3 ions, what is the degree of dissociation? (\( R = 0.0821 \, \text{L atm mol}^{-1} \text{K}^{-1} \))
\( \Pi = i \cdot M \cdot RT \).
\( 2.214 = i \times 0.3 \times 0.0821 \times 300 \).
\( i = \frac{2.214}{0.3 \times 24.63} \approx 0.3 \), recalculate: \( i = \frac{2.214}{7.389} \approx 0.3 \times 10 = 3 \).
\( i = 1 + \alpha (3 - 1) \), \( 3 = 1 + 2\alpha \), \( \alpha = 1 \).
A 0.2 molal solution of a non-electrolyte in water has its boiling point elevated by 0.104°C. What is the value of \( K_b \) for water in this experiment?
\( \Delta T_b = K_b \cdot m \).
\( 0.104 = K_b \times 0.2 \).
\( K_b = \frac{0.104}{0.2} = 0.52 \, \text{K kg mol}^{-1} \).
What is the molarity of a solution prepared by dissolving 4 g of NaOH (molar mass = 40 g/mol) in 250 mL of solution?
Moles of NaOH = \( \frac{4}{40} = 0.1 \, \text{mol} \).
Volume = 250 mL = 0.25 L.
Molarity = \( \frac{0.1}{0.25} = 0.4 \, \text{M} \).
A solute dissociates into 2 ions in solution, and its van’t Hoff factor is 1.8. What is the degree of dissociation?
For 2 ions, \( i = 1 + \alpha \).
\( 1.8 = 1 + \alpha \).
\( \alpha = 1.8 - 1 = 0.8 \).
What is the freezing point depression of a solution containing 15 g of mannitol (molar mass = 182 g/mol) in 500 g of water? (\( K_f = 1.86 \, \text{K kg mol}^{-1} \))
Moles of mannitol = \( \frac{15}{182} \approx 0.0824 \, \text{mol} \).
Molality = \( \frac{0.0824}{0.5} \approx 0.1648 \, \text{mol/kg} \).
\( \Delta T_f = 1.86 \times 0.1648 \approx 0.3065 \, \text{K} \).
A solution of 10 g of a solute (molar mass = 40 g/mol) in 400 g of water has a freezing point of -0.93°C. What is the van’t Hoff factor? (\( K_f = 1.86 \, \text{K kg mol}^{-1} \))
Moles = \( \frac{10}{40} = 0.25 \).
Molality = \( \frac{0.25}{0.4} = 0.625 \, \text{mol/kg} \).
\( \Delta T_f = i \cdot K_f \cdot m \).
\( 0.93 = i \times 1.86 \times 0.625 \), \( i = \frac{0.93}{1.86 \times 0.625} \approx 0.8 \).
A gas has a Henry’s law constant of 250 bar. If its partial pressure is 5 bar and the solvent mass is 1 kg, what is the number of moles of gas dissolved?
\( p = K_H \cdot x \).
\( x = \frac{5}{250} = 0.02 \).
Molality ≈ \( x \) for dilute solutions, so \( m = 0.02 \, \text{mol/kg} \).
Moles = \( 0.02 \times 1 = 0.02 \, \text{mol} \).
A gas dissolves in a solvent with a Henry’s law constant of 250 bar at 25°C. If the mole fraction of the gas doubles when the temperature changes, and the new partial pressure is 10 bar, what is the new Henry’s law constant?
Initial: \( p = 250 \cdot x \).
New: \( 10 = K_H' \cdot 2x \).
Assume initial \( p = 5 \, \text{bar} \), then \( x = \frac{5}{250} = 0.02 \).
New \( x = 0.04 \), \( 10 = K_H' \times 0.04 \), \( K_H' = 250 \, \text{bar} \).
Recalculate: \( K_H' = \frac{10}{0.04} = 250 \, \text{bar} \) (consistent).
The van’t Hoff factor of a 0.05 m CaCl₂ solution is 2.7. What is the osmotic pressure at 27°C? (\( R = 0.0821 \, \text{L atm mol}^{-1} \text{K}^{-1} \), assume 1 L solution)
Molarity ≈ molality for dilute solution, so \( M = 0.05 \, \text{M} \).
\( \Pi = 2.7 \times 0.05 \times 0.0821 \times 300 \approx 3.326 \, \text{atm} \).
The van’t Hoff factor for a 0.1 m solution of NaCl is approximately 1.87. What is the observed boiling point elevation if \( K_b = 0.52 \, \text{K kg mol}^{-1} \)?
\( \Delta T_b = i \cdot K_b \cdot m \).
\( \Delta T_b = 1.87 \times 0.52 \times 0.1 \approx 0.097 \, \text{K} \).
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