Correct answer Carries: 4.
Wrong Answer Carries: -1.
A solution freezes at -0.372°C. What is the molality of the solution if \( K_f = 1.86 \, \text{K kg mol}^{-1} \)?
\( \Delta T_f = K_f \cdot m \).
\( 0.372 = 1.86 \cdot m \).
\( m = \frac{0.372}{1.86} = 0.2 \, \text{mol/kg} \).
What is the freezing point depression of a solution containing 8 g of urea (molar mass = 60 g/mol) in 200 g of water? (\( K_f = 1.86 \, \text{K kg mol}^{-1} \))
Moles of urea = \( \frac{8}{60} \approx 0.1333 \, \text{mol} \).
Molality = \( \frac{0.1333}{0.2} \approx 0.6665 \, \text{mol/kg} \).
\( \Delta T_f = 1.86 \times 0.6665 \approx 1.24 \, \text{K} \).
A solution is made by dissolving a non-volatile solute in 900 g of water. If the freezing point decreases by 0.93°C and \( K_f = 1.86 \, \text{K kg mol}^{-1} \), what is the molality of the solution?
\( 0.93 = 1.86 \cdot m \).
\( m = \frac{0.93}{1.86} = 0.5 \, \text{mol/kg} \).
A solution of 0.02 mole of a non-volatile solute in 250 mL of water has an osmotic pressure of 0.984 atm at 27°C. What is the molarity of the solution? (\( R = 0.0821 \, \text{L atm mol}^{-1} \text{K}^{-1} \))
\( \Pi = M \cdot RT \).
\( 0.984 = M \times 0.0821 \times 300 \).
\( M = \frac{0.984}{0.0821 \times 300} \approx 0.04 \, \text{M} \).
Cross-check: Moles = \( 0.04 \times 0.25 = 0.01 \), but given 0.02 mole, \( i = 2 \), adjust assumption if needed.
What is the osmotic pressure of a 0.03 M solution of a non-electrolyte at 27°C? (\( R = 0.0821 \, \text{L atm mol}^{-1} \text{K}^{-1} \))
\( \Pi = MRT \).
\( \Pi = 0.03 \times 0.0821 \times 300 \approx 0.7383 \, \text{atm} \).
A solution contains 10% by mass of a solute in water. If 50 g of this solution is taken, what is the mass of the solute?
Mass of solute = \( \frac{10}{100} \times 50 = 5 \, \text{g} \).
A solution lowers the vapor pressure of a solvent from 32 mm Hg to 30 mm Hg. What is the mole fraction of the solute?
\( \frac{p^0 - p}{p^0} = x_{\text{solute}} \).
\( \frac{32 - 30}{32} = \frac{2}{32} = 0.0625 \).
A gas dissolves in a liquid with a partial pressure of 1.5 bar and a Henry’s law constant of 75 bar. What is the mole fraction of the gas?
\( p = K_H \cdot x \).
\( x = \frac{1.5}{75} = 0.02 \).
A solution of a non-volatile solute in water has a vapor pressure of 22.4 mm Hg at a temperature where pure water’s vapor pressure is 24 mm Hg. If the solution boils at 100.208°C at 1 atm, what is the molality? (\( K_b = 0.52 \, \text{K kg mol}^{-1} \))
\( \Delta T_b = 100.208 - 100 = 0.208 \, \text{K} \).
\( \Delta T_b = K_b \cdot m \).
\( 0.208 = 0.52 \cdot m \), \( m = \frac{0.208}{0.52} = 0.4 \, \text{mol/kg} \).
Cross-check: \( x_{\text{solute}} = \frac{24 - 22.4}{24} = 0.0667 \), consistent for dilute solution.
A solution of a non-volatile solute in water boils at 100.78°C at 1 atm. If 10 g of the solute (molar mass = 50 g/mol) was used in 500 g of water, what is the van’t Hoff factor? (\( K_b = 0.52 \, \text{K kg mol}^{-1} \))
\( \Delta T_b = i \cdot K_b \cdot m \).
Moles = \( \frac{10}{50} = 0.2 \), molality = \( \frac{0.2}{0.5} = 0.4 \, \text{mol/kg} \).
\( 0.78 = i \times 0.52 \times 0.4 \).
\( i = \frac{0.78}{0.52 \times 0.4} \approx 3.75 \).
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