Correct answer Carries: 4.
Wrong Answer Carries: -1.
A solution boils at 100.13°C at 1 atm. What is the molality if \( K_b = 0.52 \, \text{K kg mol}^{-1} \)?
\( \Delta T_b = K_b \cdot m \).
\( 100.13 - 100 = 0.52 \cdot m \).
\( m = \frac{0.13}{0.52} = 0.25 \, \text{mol/kg} \).
What is the mole fraction of chloroform (molar mass = 119.5 g/mol) in a solution containing 59.75 g of chloroform and 36 g of water?
Moles of chloroform = \( \frac{59.75}{119.5} = 0.5 \, \text{mol} \).
Moles of water = \( \frac{36}{18} = 2 \, \text{mol} \).
Total moles = \( 0.5 + 2 = 2.5 \).
Mole fraction = \( \frac{0.5}{2.5} = 0.2 \).
What is the mass percentage of a solution prepared by dissolving 25 g of sugar in 75 g of water?
Total mass of solution = 25 g + 75 g = 100 g.
Mass % = \( \frac{\text{Mass of solute}}{\text{Total mass}} \times 100 = \frac{25}{100} \times 100 = 25\% \).
How many grams of NaCl (molar mass = 58.5 g/mol) are required to prepare 400 mL of a 0.25 M solution?
Moles of NaCl = \( 0.25 \times 0.4 = 0.1 \, \text{mol} \).
Mass = \( 0.1 \times 58.5 = 5.85 \, \text{g} \).
A gas has a Henry’s law constant of 200 bar. If its partial pressure is 4 bar, what is the mole fraction in the solution?
\( p = K_H \cdot x \).
\( x = \frac{4}{200} = 0.02 \).
The vapor pressure of a solvent decreases from 50 mm Hg to 47 mm Hg when a non-volatile solute is added. If an additional amount of the same solute is added to double the molality, what will be the new vapor pressure?
\( \frac{p^0 - p}{p^0} = x_{\text{solute}} \).
Initial: \( \frac{50 - 47}{50} = 0.06 \).
Doubling molality doubles \( x_{\text{solute}} \) (approximately for dilute solutions), so new \( x_{\text{solute}} = 0.12 \).
New \( p = p^0 (1 - x_{\text{solute}}) = 50 (1 - 0.12) = 44 \, \text{mm Hg} \).
A solution boils at 100.208°C at 1 atm. What is the molality if \( K_b = 0.52 \, \text{K kg mol}^{-1} \)?
\( 100.208 - 100 = 0.52 \cdot m \).
\( m = \frac{0.208}{0.52} = 0.4 \, \text{mol/kg} \).
A gas has a partial pressure of 3 bar above a solution, and its Henry’s law constant is 150 bar. What is the mole fraction of the gas in the solution?
Henry’s law: \( p = K_H \cdot x \).
\( x = \frac{3}{150} = 0.02 \).
A solution of two volatile liquids has vapor pressures of 500 mm Hg and 700 mm Hg. If the total vapor pressure is 580 mm Hg, what is the mole fraction of the first component in the vapor phase?
Liquid phase: \( 580 = 500 x_1 + 700 (1 - x_1) \).
\( 580 = 500 x_1 + 700 - 700 x_1 \), \( 200 x_1 = 120 \), \( x_1 = 0.6 \), \( x_2 = 0.4 \).
Vapor phase: \( y_1 = \frac{500 \times 0.6}{580} \approx 0.5172 \).
How many grams of MgCl₂ (molar mass = 95 g/mol) are needed to prepare 500 mL of a 0.3 M solution?
Moles of MgCl₂ = \( 0.3 \times 0.5 = 0.15 \, \text{mol} \).
Mass = \( 0.15 \times 95 = 14.25 \, \text{g} \).
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