Correct answer Carries: 4.
Wrong Answer Carries: -1.
What is the mass percentage of a solution if 20 g of a solute is dissolved in 180 g of water?
Total mass = \( 20 + 180 = 200 \, \text{g} \).
Mass % = \( \frac{20}{200} \times 100 = 10\% \).
What is the boiling point elevation of a solution containing 5 g of glucose (molar mass = 180 g/mol) in 250 g of water? (\( K_b = 0.52 \, \text{K kg mol}^{-1} \))
Moles of glucose = \( \frac{5}{180} \approx 0.0278 \, \text{mol} \).
Molality = \( \frac{0.0278}{0.25} \approx 0.1112 \, \text{mol/kg} \).
\( \Delta T_b = 0.52 \times 0.1112 \approx 0.0578 \, \text{K} \).
A solute dissociates into 2 ions with a van’t Hoff factor of 1.4. What is the degree of dissociation?
\( i = 1 + \alpha (n - 1) \), where \( n = 2 \).
\( 1.4 = 1 + \alpha \).
\( \alpha = 1.4 - 1 = 0.4 \).
A solution is prepared by dissolving 18 g of a solute in water to make 500 mL of solution with a density of 1.2 g/mL. If the molality is 0.2 mol/kg, what is the molar mass of the solute?
Total mass = \( 500 \times 1.2 = 600 \, \text{g} \).
Mass of water = \( 600 - 18 = 582 \, \text{g} = 0.582 \, \text{kg} \).
Molality = \( \frac{\text{moles}}{0.582} = 0.2 \).
Moles = \( 0.2 \times 0.582 = 0.1164 \).
Molar mass = \( \frac{18}{0.1164} \approx 154.64 \, \text{g/mol} \).
A solution is made by dissolving 15 g of a solute (molar mass = 30 g/mol) in water. If the boiling point elevation is 0.26°C and \( K_b = 0.52 \, \text{K kg mol}^{-1} \), what is the mass of water?
Moles = \( \frac{15}{30} = 0.5 \).
\( \Delta T_b = K_b \cdot m \).
\( 0.26 = 0.52 \times \frac{0.5}{w} \), \( w = \frac{0.5 \times 0.52}{0.26} = 1 \, \text{kg} = 1000 \, \text{g} \).
What is the molar mass of a solute if 5 g of it in 250 mL of solution produces an osmotic pressure of 0.984 atm at 27°C? (\( R = 0.0821 \, \text{L atm mol}^{-1} \text{K}^{-1} \))
\( \Pi = \frac{w}{M V} RT \).
\( 0.984 = \frac{5}{M \times 0.25} \times 0.0821 \times 300 \).
\( M = \frac{5 \times 0.0821 \times 300}{0.984 \times 0.25} \approx 50 \, \text{g/mol} \).
A solution is prepared by mixing 50 g of a solute with 150 g of water. If the solute’s mass percentage in a second solution (made with the same solute and water) is 40%, what mass of water is required for the second solution?
Mass % = \( \frac{\text{Mass of solute}}{\text{Total mass}} \times 100 \).
For 40%: \( 40 = \frac{50}{50 + w} \times 100 \).
\( 0.4 (50 + w) = 50 \), \( 50 + w = 125 \), \( w = 75 \, \text{g} \).
The van’t Hoff factor of a 0.05 m NaCl solution is 1.9. What is the osmotic pressure at 27°C? (\( R = 0.0821 \, \text{L atm mol}^{-1} \text{K}^{-1} \), assume 1 L solution)
\( \Pi = i \cdot M \cdot RT \).
Molarity ≈ molality for dilute solution, so \( M = 0.05 \, \text{M} \).
\( \Pi = 1.9 \times 0.05 \times 0.0821 \times 300 \approx 2.339 \, \text{atm} \).
What is the mole fraction of a solute if the vapor pressure of a solution is 28 mm Hg and that of pure solvent is 30 mm Hg?
\( \frac{p^0 - p}{p^0} = x_{\text{solute}} \).
\( \frac{30 - 28}{30} = \frac{2}{30} \approx 0.0667 \).
A solution of two volatile liquids has vapor pressures of 250 mm Hg and 350 mm Hg for pure components. If the total vapor pressure is 310 mm Hg, what is the mole fraction of the first component in the vapor phase?
Liquid phase: \( 310 = 250 x_1 + 350 (1 - x_1) \).
\( 310 = 250 x_1 + 350 - 350 x_1 \), \( 100 x_1 = 40 \), \( x_1 = 0.4 \), \( x_2 = 0.6 \).
Vapor phase: \( y_1 = \frac{P_1^0 \cdot x_1}{P_{\text{total}}} = \frac{250 \times 0.4}{310} \approx 0.3226 \).
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