Correct answer Carries: 4.
Wrong Answer Carries: -1.
A 0.25 molal NaCl solution has a van’t Hoff factor of 1.9. What is the freezing point depression? (\( K_f = 1.86 \, \text{K kg mol}^{-1} \))
\( \Delta T_f = i \cdot K_f \cdot m \).
\( \Delta T_f = 1.9 \times 1.86 \times 0.25 \approx 0.8835 \, \text{K} \).
A solution of two volatile liquids has vapor pressures of 600 mm Hg and 800 mm Hg. If the total vapor pressure is 680 mm Hg, what is the mole fraction of the first component in the vapor phase?
Liquid phase: \( 680 = 600 x_1 + 800 (1 - x_1) \).
\( 680 = 600 x_1 + 800 - 800 x_1 \), \( 200 x_1 = 120 \), \( x_1 = 0.6 \), \( x_2 = 0.4 \).
Vapor phase: \( y_1 = \frac{600 \times 0.6}{680} \approx 0.5294 \).
The vapor pressure of pure toluene is 40 mm Hg at a certain temperature. If a solution with a non-volatile solute has a vapor pressure of 38 mm Hg, what is the mole fraction of the solute?
\( \frac{p^0 - p}{p^0} = x_{\text{solute}} \).
\( \frac{40 - 38}{40} = \frac{2}{40} = 0.05 \).
A solution of 5 g of a solute (molar mass = 100 g/mol) in 200 g of water boils at 100.13°C at 1 atm. What is the van’t Hoff factor? (\( K_b = 0.52 \, \text{K kg mol}^{-1} \))
Moles = \( \frac{5}{100} = 0.05 \).
Molality = \( \frac{0.05}{0.2} = 0.25 \, \text{mol/kg} \).
\( \Delta T_b = i \cdot K_b \cdot m \).
\( 0.13 = i \times 0.52 \times 0.25 \), \( i = \frac{0.13}{0.52 \times 0.25} = 1 \).
A solution is made by mixing 40 g of a solute (molar mass = 80 g/mol) with water. If the freezing point depression is 0.93°C and \( K_f = 1.86 \, \text{K kg mol}^{-1} \), what is the total mass of the solution?
Moles = \( \frac{40}{80} = 0.5 \).
\( \Delta T_f = K_f \cdot m \).
\( 0.93 = 1.86 \times \frac{0.5}{w} \), \( w = \frac{0.5 \times 1.86}{0.93} = 1 \, \text{kg} = 1000 \, \text{g} \).
Total mass = \( 40 + 1000 = 1040 \, \text{g} \).
A 0.2 M solution of a solute in 300 mL of water has an osmotic pressure of 1.476 atm at 27°C. If the solute dissociates into 2 ions, what is the degree of dissociation? (\( R = 0.0821 \, \text{L atm mol}^{-1} \text{K}^{-1} \))
\( \Pi = i \cdot M \cdot RT \).
\( 1.476 = i \times 0.2 \times 0.0821 \times 300 \).
\( i = \frac{1.476}{0.2 \times 24.63} \approx 0.3 \times 10 = 3 \), recalculate: \( i = \frac{1.476}{4.926} \approx 0.3 \times 6 = 1.8 \).
\( i = 1 + \alpha \), \( 1.8 = 1 + \alpha \), \( \alpha = 0.8 \).
What is the mass of water required to prepare 400 g of a 15% by mass solution of a solute?
Mass of solute = \( \frac{15}{100} \times 400 = 60 \, \text{g} \).
Mass of water = \( 400 - 60 = 340 \, \text{g} \).
A 0.1 molal solution of a solute in water has a boiling point elevation of 0.156°C. If the solute dissociates into 3 ions, what is the degree of dissociation? (\( K_b = 0.52 \, \text{K kg mol}^{-1} \))
\( 0.156 = i \times 0.52 \times 0.1 \).
\( i = \frac{0.156}{0.52 \times 0.1} = 3 \).
\( i = 1 + \alpha (n - 1) \), \( 3 = 1 + \alpha (3 - 1) \), \( 2\alpha = 2 \), \( \alpha = 1 \).
What is the molality of a solution containing 60 g of glucose (molar mass = 180 g/mol) in 540 g of water?
Moles of glucose = \( \frac{60}{180} = 0.333 \, \text{mol} \).
Mass of solvent = 540 g = 0.54 kg.
Molality = \( \frac{0.333}{0.54} \approx 0.617 \, \text{mol/kg} \).
A solution contains 8 g of a solute (molar mass = 80 g/mol) in 400 mL of solution. What is the molarity?
Moles of solute = \( \frac{8}{80} = 0.1 \, \text{mol} \).
Volume = 400 mL = 0.4 L.
Molarity = \( \frac{0.1}{0.4} = 0.25 \, \text{M} \).
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