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Chapter 13: Problem 63

Consider two solutions, one formed by adding \(150 \mathrm{~g}\) of glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) to $1 \mathrm{~L}\( of water and the other formed by adding \)150 \mathrm{~g}$ of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) to $1 \mathrm{~L}$ of water. Calculate the vapor pressure for each solution at \(25^{\circ} \mathrm{C} ;\) the vapor pressure of pure water at this temperature is \(3.17 \mathrm{kPa}\).

Short Answer

Expert verified
The vapor pressure of the glucose solution at 25°C is approximately 3.12 kPa, and the vapor pressure of the sucrose solution at 25°C is approximately 3.15 kPa.

Step by step solution

01

Calculate the moles of solute (glucose and sucrose)

To calculate the moles of solute, we use the formula: moles = mass (g) / molar mass (g/mol) For glucose, the mass is 150 g and the molar mass is: \(C_6H_{12}O_6 = 6(12.01) + 12(1.01) + 6(16.00) = 180.18 \mathrm{~g/mol}\) For sucrose, the mass is 150 g and the molar mass is: \(C_{12}H_{22}O_{11} = 12(12.01) + 22(1.01) + 11(16.00) = 342.30 \mathrm{~g/mol}\) Now, calculate the moles of both solutes: moles of glucose = 150 g / 180.18 g/mol ≈ 0.8329 mol moles of sucrose = 150 g / 342.30 g/mol ≈ 0.4382 mol
02

Calculate the moles of solvent (water)

To find the moles of water, we first need to convert the volume of water from liters to grams. The density of water is about 1 g/mL, so 1 L water = 1000 g water. The molar mass of water is 18.02 g/mol. moles of water = 1000 g / 18.02 g/mol ≈ 55.5 mol
03

Calculate the mole fraction of solvent (water) in each solution

Mole fraction is calculated as the ratio of moles of a particular component to the total moles in the solution. Calculate the mole fraction of water in both solutions using the moles of solutes and solvent calculated in steps 1 and 2: Mole fraction of water in glucose solution = moles of water / (moles of water + moles of glucose) = 55.5 / (55.5 + 0.8329) ≈ 0.9852 Mole fraction of water in sucrose solution = moles of water / (moles of water + moles of sucrose) = 55.5 / (55.5 + 0.4382) ≈ 0.9921
04

Calculate the vapor pressure of each solution

To find the vapor pressure (P_solution) of each solution, we will use Raoult's Law: P_solution = mole fraction of water × P_pure_water At 25°C, the vapor pressure of pure water is 3.17 kPa. For the glucose solution: P_solution_glucose = 0.9852 × 3.17 kPa ≈ 3.12 kPa For the sucrose solution: P_solution_sucrose = 0.9921 × 3.17 kPa ≈ 3.15 kPa #Conclusion# The vapor pressure of the glucose solution at 25°C is approximately 3.12 kPa, and the vapor pressure of the sucrose solution at 25°C is approximately 3.15 kPa.

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Most popular questions from this chapter

At \(35^{\circ} \mathrm{C}\) the vapor pressure of acetone, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO},\) is 47.9 \(\mathrm{kPa}\), and that of carbon disulfide, \(\mathrm{CS}_{2}\), is \(66.7 \mathrm{kPa}\). A solution composed of an equal number of moles of acetone and carbon disulfide has a vapor pressure of \(86.7 \mathrm{kPa}\) at $35^{\circ} \mathrm{C} .(\mathbf{a})$ What would be the vapor pressure of the solution if it exhibited ideal behavior? (b) Based on the behavior of the solution, predict whether the mixing of acetone and carbon disulfide is an exothermic \(\left(\Delta H_{\text {soln }}<0\right)\) or endothermic $\left(\Delta H_{\text {soln }}>0\right)$ process.

(a) Calculate the mass percentage of \(\mathrm{NaNO}_{3}\) in a solution containing \(13.6 \mathrm{~g}\) of \(\mathrm{NaNO}_{3}\) in \(834 \mathrm{~g}\) of water. (b) An alloy contains \(2.86 \mathrm{~g}\) of chromium per $100 \mathrm{~kg}$ of alloy. What is the concentration of chromium in ppm?

What is the molarity of each of the following solutions: (a) \(15.0 \mathrm{~g}\) of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) in \(0.250 \mathrm{~mL}\) solution, (b) \(5.25 \mathrm{~g}\) of $\mathrm{Mn}\left(\mathrm{NO}_{3}\right)_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O}\( in \)175 \mathrm{~mL}$ of solution, (c) \(35.0 \mathrm{~mL}\) of \(9.00 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) diluted to $0.500 \mathrm{~L} ?$

You make two solutions of a nonvolatile solute with a liquid solvent, $0.01 \mathrm{M}\( and \)1.00 \mathrm{M}$. Indicate whether each of the following statements is true or false. (a) The vapor pressure of the concentrated solution is higher than that of the diluted solution. (b) The osmotic pressure of the concentrated solution is higher than that of the diluted solution. (c) The boiling point of the concentrated solution is higher than that of the diluted solution. (d) The freezing point of the concentrated solution is higher than that of the diluted solution.

Calculate the number of moles of solute present in each of the following aqueous solutions: (a) \(750 \mathrm{~mL}\) of \(0.120 \mathrm{M}\) \(\operatorname{SrBr}_{2},(\mathbf{b}) 70.0 \mathrm{~g}\) of $0.200 \mathrm{~m} \mathrm{KCl},(\mathbf{c}) 150.0 \mathrm{~g}\( of a solution that is \)5.75 \%$ glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) by mass.
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