Correct answer Carries: 4.
Wrong Answer Carries: -1.
What is the freezing point depression of a solution containing 18 g of sucrose (molar mass = 342 g/mol) in 400 g of water? (\( K_f = 1.86 \, \text{K kg mol}^{-1} \))
Moles of sucrose = \( \frac{18}{342} \approx 0.0526 \, \text{mol} \).
Molality = \( \frac{0.0526}{0.4} \approx 0.1315 \, \text{mol/kg} \).
\( \Delta T_f = 1.86 \times 0.1315 \approx 0.2446 \, \text{K} \).
The solubility of a gas in a liquid increases with an increase in:
According to Henry’s law, solubility of a gas is directly proportional to its partial pressure.
What is the osmotic pressure of a 0.025 M solution of a non-electrolyte at 27°C? (\( R = 0.0821 \, \text{L atm mol}^{-1} \text{K}^{-1} \))
\( \Pi = MRT \).
\( \Pi = 0.025 \times 0.0821 \times 300 \approx 0.6158 \, \text{atm} \).
The vapor pressure of pure water at a certain temperature is 50 mm Hg. What is the vapor pressure of a solution with a solute mole fraction of 0.1?
Mole fraction of water = \( 1 - 0.1 = 0.9 \).
Vapor pressure = \( 50 \times 0.9 = 45 \, \text{mm Hg} \).
The van’t Hoff factor of a 0.1 m K₂SO₄ solution is 2.4. What is the boiling point elevation? (\( K_b = 0.52 \, \text{K kg mol}^{-1} \))
\( \Delta T_b = i \cdot K_b \cdot m \).
\( \Delta T_b = 2.4 \times 0.52 \times 0.1 = 0.1248 \, \text{K} \).
What is the boiling point elevation of a solution containing 12 g of glucose (molar mass = 180 g/mol) in 600 g of water? (\( K_b = 0.52 \, \text{K kg mol}^{-1} \))
Moles of glucose = \( \frac{12}{180} \approx 0.0667 \, \text{mol} \).
Molality = \( \frac{0.0667}{0.6} \approx 0.111 \, \text{mol/kg} \).
\( \Delta T_b = 0.52 \times 0.111 \approx 0.0577 \, \text{K} \).
A solution contains 23 g of methanol (molar mass = 32 g/mol) and 72 g of water. If 64 g of water is added, what is the new mole fraction of methanol?
Moles of methanol = \( \frac{23}{32} \approx 0.7188 \).
Initial moles of water = \( \frac{72}{18} = 4 \).
New moles of water = \( \frac{72 + 64}{18} = \frac{136}{18} \approx 7.5556 \).
Total moles = \( 0.7188 + 7.5556 \approx 8.2744 \).
Mole fraction = \( \frac{0.7188}{8.2744} \approx 0.0869 \).
The vapor pressure of pure water is 25 mm Hg at a certain temperature. A solution with a non-volatile solute has a vapor pressure of 23 mm Hg. If the solute’s molar mass is 60 g/mol, what is the mass of solute in 180 g of water?
\( \frac{p^0 - p}{p^0} = x_{\text{solute}} \).
\( \frac{25 - 23}{25} = 0.08 \).
Moles of water = \( \frac{180}{18} = 10 \).
\( x_{\text{solute}} = \frac{n_{\text{solute}}}{n_{\text{solute}} + 10} = 0.08 \).
\( n_{\text{solute}} = 0.08 (n_{\text{solute}} + 10) \), \( n_{\text{solute}} - 0.08 n_{\text{solute}} = 0.8 \), \( 0.92 n_{\text{solute}} = 0.8 \), \( n_{\text{solute}} \approx 0.8696 \).
Mass = \( 0.8696 \times 60 \approx 52.18 \, \text{g} \).
A 0.25 M solution of a solute in 400 mL of water has an osmotic pressure of 1.845 atm at 27°C. If the solute dissociates into 2 ions, what is the degree of dissociation? (\( R = 0.0821 \, \text{L atm mol}^{-1} \text{K}^{-1} \))
\( \Pi = i \cdot M \cdot RT \).
\( 1.845 = i \times 0.25 \times 0.0821 \times 300 \).
\( i = \frac{1.845}{0.25 \times 24.63} \approx 0.3 \times 6 = 1.8 \).
\( i = 1 + \alpha \), \( 1.8 = 1 + \alpha \), \( \alpha = 0.8 \).
A 0.25 molal solution of a solute in water has a boiling point elevation of 0.26°C. If the solute dissociates into 2 ions, what is the degree of dissociation? (\( K_b = 0.52 \, \text{K kg mol}^{-1} \))
\( 0.26 = i \times 0.52 \times 0.25 \).
\( i = \frac{0.26}{0.52 \times 0.25} = 2 \).
\( i = 1 + \alpha (n - 1) \), \( 2 = 1 + \alpha (2 - 1) \), \( \alpha = 1 \).
Are you sure you want to submit your answers?