Correct answer Carries: 4.
Wrong Answer Carries: -1.
A solution of a non-volatile solute in 600 g of water freezes at -0.62°C. If the solute’s molar mass is 62 g/mol, what is the mass of the solute? (\( K_f = 1.86 \, \text{K kg mol}^{-1} \))
\( \Delta T_f = K_f \cdot m \).
\( 0.62 = 1.86 \times \frac{\text{moles}}{0.6} \).
Moles = \( \frac{0.62 \times 0.6}{1.86} \approx 0.2 \).
Mass = \( 0.2 \times 62 = 12.4 \, \text{g} \).
What is the mole fraction of water in a solution containing 18 g of water and 46 g of ethanol (molar mass of ethanol = 46 g/mol)?
Moles of water = \( \frac{18}{18} = 1 \, \text{mol} \).
Moles of ethanol = \( \frac{46}{46} = 1 \, \text{mol} \).
Total moles = 1 + 1 = 2.
Mole fraction of water = \( \frac{1}{2} = 0.5 \).
What is the molality of a solution containing 12 g of sucrose (molar mass = 342 g/mol) in 300 g of water?
Moles of sucrose = \( \frac{12}{342} \approx 0.0351 \, \text{mol} \).
Mass of solvent = 300 g = 0.3 kg.
Molality = \( \frac{0.0351}{0.3} \approx 0.117 \, \text{mol/kg} \).
What is the mass percentage of a solution if 25 g of a solute is dissolved in 225 g of water?
Total mass = \( 25 + 225 = 250 \, \text{g} \).
Mass % = \( \frac{25}{250} \times 100 = 10\% \).
A solution freezes at -0.558°C. What is the molality if \( K_f = 1.86 \, \text{K kg mol}^{-1} \)?
\( 0.558 = 1.86 \cdot m \).
\( m = \frac{0.558}{1.86} = 0.3 \, \text{mol/kg} \).
A gas follows Henry’s law with a constant of 200 bar. If its solubility in a solvent is 0.015 mol/kg at 3 bar, what is the mass of solvent required to dissolve 0.03 mole of the gas at the same pressure?
Molality = \( \frac{\text{moles}}{\text{mass of solvent in kg}} \).
\( 0.015 = \frac{n}{m} \), at \( p = 3 \, \text{bar} \), \( x = \frac{3}{200} = 0.015 \) (molality ≈ \( x \) for dilute solutions).
For 0.03 mole: \( 0.015 = \frac{0.03}{m} \), \( m = \frac{0.03}{0.015} = 2 \, \text{kg} = 2000 \, \text{g} \).
What is the freezing point depression of a solution containing 9 g of glucose (molar mass = 180 g/mol) in 250 g of water? (\( K_f = 1.86 \, \text{K kg mol}^{-1} \))
Moles of glucose = \( \frac{9}{180} = 0.05 \, \text{mol} \).
Molality = \( \frac{0.05}{0.25} = 0.2 \, \text{mol/kg} \).
\( \Delta T_f = 1.86 \times 0.2 = 0.372 \, \text{K} \).
What is the molality of a solution prepared by dissolving 15 g of urea (molar mass = 60 g/mol) in 250 g of water?
Moles of urea = \( \frac{15}{60} = 0.25 \, \text{mol} \).
Mass of solvent = 250 g = 0.25 kg.
Molality = \( \frac{0.25}{0.25} = 1 \, \text{mol/kg} \).
A 0.15 molal solution of a solute in water has a freezing point depression of 0.558°C. If the solute dissociates into 2 ions, what is the degree of dissociation? (\( K_f = 1.86 \, \text{K kg mol}^{-1} \))
\( \Delta T_f = i \cdot K_f \cdot m \).
\( 0.558 = i \times 1.86 \times 0.15 \).
\( i = \frac{0.558}{1.86 \times 0.15} \approx 2 \).
\( i = 1 + \alpha (n - 1) \), \( 2 = 1 + \alpha (2 - 1) \), \( \alpha = 1 \).
The van’t Hoff factor of a 0.08 m CaCl₂ solution is 2.6. What is the boiling point elevation? (\( K_b = 0.52 \, \text{K kg mol}^{-1} \))
\( \Delta T_b = i \cdot K_b \cdot m \).
\( \Delta T_b = 2.6 \times 0.52 \times 0.08 \approx 0.1082 \, \text{K} \).
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