Correct answer Carries: 4.
Wrong Answer Carries: -1.
A solution is made by dissolving 12 g of a solute in water to make 300 mL of solution. If the osmotic pressure is 0.984 atm at 27°C, what is the molar mass of the solute? (\( R = 0.0821 \, \text{L atm mol}^{-1} \text{K}^{-1} \))
\( \Pi = \frac{w}{M V} RT \).
\( 0.984 = \frac{12}{M \times 0.3} \times 0.0821 \times 300 \).
\( M = \frac{12 \times 0.0821 \times 300}{0.984 \times 0.3} \approx 100 \, \text{g/mol} \).
The vapor pressure of pure water is 28 mm Hg at a certain temperature. A solution with a non-volatile solute has a vapor pressure of 26.6 mm Hg. If the solute’s molar mass is 50 g/mol, what is the mass of solute in 360 g of water?
\( \frac{p^0 - p}{p^0} = x_{\text{solute}} \).
\( \frac{28 - 26.6}{28} = \frac{1.4}{28} = 0.05 \).
Moles of water = \( \frac{360}{18} = 20 \).
\( x_{\text{solute}} = \frac{n_{\text{solute}}}{n_{\text{solute}} + 20} = 0.05 \).
\( n_{\text{solute}} = 0.05 (n_{\text{solute}} + 20) \), \( 0.95 n_{\text{solute}} = 1 \), \( n_{\text{solute}} \approx 1.0526 \).
Mass = \( 1.0526 \times 50 \approx 52.63 \, \text{g} \).
A gas has a Henry’s law constant of 500 bar at 25°C. If the partial pressure increases from 5 bar to 10 bar, what is the percentage increase in solubility?
Initial \( x = \frac{5}{500} = 0.01 \).
New \( x = \frac{10}{500} = 0.02 \).
Percentage increase = \( \frac{0.02 - 0.01}{0.01} \times 100 = 100\% \).
What is the molar mass of a solute if 6 g of it in 300 mL of solution produces an osmotic pressure of 0.821 atm at 27°C? (\( R = 0.0821 \, \text{L atm mol}^{-1} \text{K}^{-1} \))
\( 0.821 = \frac{6}{M \times 0.3} \times 0.0821 \times 300 \).
\( M = \frac{6 \times 0.0821 \times 300}{0.821 \times 0.3} \approx 60 \, \text{g/mol} \).
A solution of two volatile liquids with vapor pressures 300 mm Hg and 500 mm Hg has a total vapor pressure of 380 mm Hg. What is the mole fraction of the second component in the vapor phase?
Liquid phase: \( 380 = 300 x_1 + 500 (1 - x_1) \).
\( 380 = 300 x_1 + 500 - 500 x_1 \), \( 200 x_1 = 120 \), \( x_1 = 0.6 \), \( x_2 = 0.4 \).
Vapor phase: \( y_2 = \frac{P_2^0 \cdot x_2}{P_{\text{total}}} = \frac{500 \times 0.4}{380} \approx 0.5263 \).
A solution of NaCl (molar mass = 58.5 g/mol) in 500 g of water freezes at -1.116°C. If \( K_f = 1.86 \, \text{K kg mol}^{-1} \), what is the mass of NaCl? (Assume complete dissociation, i = 2)
\( \Delta T_f = i \cdot K_f \cdot m \).
\( 1.116 = 2 \times 1.86 \times \frac{\text{moles}}{0.5} \).
\( 1.116 = 3.72 \times \frac{\text{moles}}{0.5} \), moles = \( \frac{1.116 \times 0.5}{3.72} \approx 0.15 \).
Mass = \( 0.15 \times 58.5 = 8.775 \, \text{g} \).
What is the mole fraction of benzene (molar mass = 78 g/mol) in a solution containing 39 g of benzene and 92 g of toluene (molar mass = 92 g/mol)?
Moles of benzene = \( \frac{39}{78} = 0.5 \, \text{mol} \).
Moles of toluene = \( \frac{92}{92} = 1 \, \text{mol} \).
Total moles = \( 0.5 + 1 = 1.5 \).
Mole fraction = \( \frac{0.5}{1.5} \approx 0.333 \).
A solution contains 10 g of a solute in 90 g of water. What is the mass percentage of the solute?
Total mass = 10 g + 90 g = 100 g.
Mass % = \( \frac{10}{100} \times 100 = 10\% \).
A 0.5 M solution of Na₂SO₄ (molar mass = 142 g/mol) has a density of 1.2 g/mL. What is the mass of the solute in 200 mL of this solution?
Moles of Na₂SO₄ = \( 0.5 \times 0.2 = 0.1 \, \text{mol} \).
Mass of solute = \( 0.1 \times 142 = 14.2 \, \text{g} \).
(Density is extra info, not needed for molarity-based calculation.)
A mixture of ethanol (molar mass = 46 g/mol) and water has a total mass of 138 g. If the mole fraction of ethanol is 0.25, what is the mass of ethanol in the mixture?
Moles of ethanol = \( n_e \), moles of water = \( n_w \), total moles = \( n_e + n_w \).
\( \frac{n_e}{n_e + n_w} = 0.25 \).
Mass: \( 46 n_e + 18 n_w = 138 \).
From mole fraction: \( n_w = 3 n_e \).
\( 46 n_e + 18 (3 n_e) = 138 \), \( 46 n_e + 54 n_e = 138 \), \( 100 n_e = 138 \), \( n_e = 1.38 \).
Mass of ethanol = \( 1.38 \times 46 = 63.48 \, \text{g} \).
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