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Chapter 10: Problem 36

Consider an aqueous solution of urea, \(\left(\mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2}\right)\) at \(26^{\circ} \mathrm{C} .\) Its vapor pressure is \(21.15 \mathrm{~mm} \mathrm{Hg}\) (vapor pressure of pure \(\left.\mathrm{H}_{2} \mathrm{O}=25.21 \mathrm{~mm} \mathrm{Hg}\right)\). How would you prepare \(250.0 \mathrm{~mL}\) of this solution \((d=1.06 \mathrm{~g} / \mathrm{mL}) ?\)

Short Answer

Expert verified
Answer: To prepare this solution, first determine the mole fraction of urea using Raoult's law (\(\chi_\text{urea} = \frac{P_\text{solution}}{P_\text{water}}\)). Next, calculate the molar mass of urea, then find the mass of urea needed by multiplying the mole fraction of urea by the total mass of the solution. Subtract the mass of urea from the total mass of the solution to determine the mass of water needed. Finally, dissolve the calculated mass of urea in distilled water and adjust the final volume to 250.0 mL.

Step by step solution

01

Calculate the mole fraction of urea in the solution

Using Raoult's law, we can determine the mole fraction of urea in the solution. Raoult's law states that the vapor pressure of a solution is equal to the product of the mole fraction of the solute and the vapor pressure of the pure solvent. In this problem, we are given the vapor pressure of the urea solution (\(P_\text{solution} = 21.15 \mathrm{~mm} \mathrm{Hg}\)) and the vapor pressure of pure water (\(P_\text{water} = 25.21 \mathrm{~mm} \mathrm{Hg}\)). The mole fraction of urea can be calculated using the following formula: \(\chi_\text{urea} = \frac{P_\text{solution}}{P_\text{water}}\)
02

Calculate the molar mass of urea

To find the mass of urea needed, we first need to find its molar mass. The molecular formula for urea is \(\left(\mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2}\right)\). The molar mass is the sum of the atomic masses of all the elements in the molecule: Molar mass of urea = M(C) + M(O) + 2 x [M(N) + 2 x M(H)]
03

Calculate the mass of urea needed

Using the mole fraction of urea and the molar mass, we can calculate the mass of urea needed for the 250.0 mL solution. First, we need to find the total mass of the solution using the given density: Mass of solution = Volume of solution x Density = 250.0 mL x 1.06 g/mL Next, we can find the mass of urea in the solution by multiplying the mole fraction of urea by the total mass of the solution: Mass of urea = Mole fraction of urea x Mass of solution
04

Calculate the mass of water needed

Since we now know the mass of urea needed, we can find the mass of water needed to prepare the solution. The mass of water is the difference between the total mass of the solution and the mass of urea: Mass of water = Mass of solution - Mass of urea
05

Prepare the 250.0 mL solution

Weigh out the calculated mass of urea and dissolve it in a small amount of distilled water. Then, add enough distilled water to bring the final volume of the solution to 250.0 mL. Mix the solution well to ensure that the urea is completely dissolved. In conclusion, by calculating the mole fraction of urea and the required masses of urea and water, we can prepare the desired 250.0 mL aqueous solution of urea.

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Most popular questions from this chapter

Consider three test tubes. Tube A has pure water. Tube \(B\) has an aqueous \(1.0 \mathrm{~m}\) solution of ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} .\) Tube \(\mathrm{C}\) has an aqueous \(1.0 \mathrm{~m}\) solution of \(\mathrm{NaCl}\). Which of the following statements are true? (Assume that for these solutions \(1.0 m=1.0 \mathrm{M} .)\) (a) The vapor pressure of the solvent over tube \(\mathrm{A}\) is greater than the solvent pressure over tube \(\mathrm{B}\). (b) The freezing point of the solution in tube \(\mathrm{B}\) is higher than the freezing point of the solution in tube \(\mathrm{A}\). (c) The freezing point of the solution in tube \(\mathrm{B}\) is higher than the freezing point of the solution in tube \(\mathrm{C}\). (d) The boiling point of the solution in tube \(\mathrm{B}\) is higher than the boiling point of the solution in tube \(\mathrm{C}\). (e) The osmotic pressure of the solution in tube \(\mathrm{B}\) is greater than the osmotic pressure of the solution in tube C.

The Rast method uses camphor \(\left(\mathrm{C}_{10} \mathrm{H}_{16} \mathrm{O}\right)\) as a solvent for determining the molar mass of a compound. When \(2.50 \mathrm{~g}\) of cortisone acetate are dissolved in \(50.00 \mathrm{~g}\) of camphor \(\left(k_{\mathrm{f}}=40.0^{\circ} \mathrm{C} / \mathrm{m}\right)\), the freezing point of the mixture is \(173.44^{\circ} \mathrm{C}\) that of pure camphor is \(178.40^{\circ} \mathrm{C}\). What is the molar mass of cortisone acetate?

Consider two nonelectrolytes \(\mathrm{X}\) and \(\mathrm{Y} . \mathrm{X}\) has a higher molar mass than Y. Twenty-five grams of \(\mathrm{X}\) are dissolved in \(100 \mathrm{~g}\) of solvent \(\mathrm{C}\) and labeled solution 1. Twenty-five grams of \(Y\) are dissolved in \(100 \mathrm{~g}\) of solvent \(\mathrm{C}\) and labeled solution 2. Both solutions have the same density. Which solution has (a) a higher molarity? (b) a higher mass percent? (c) a higher molality? (d) a larger multiplier i? (e) a larger mole fraction of solvent?

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Aqueous solutions introduced into the bloodstream by injection must have the same osmotic pressure as blood; that is, they must be "isotonic" with blood. At \(25^{\circ} \mathrm{C},\) the average osmotic pressure of blood is 7.7 atm. What is the molarity of an isotonic saline solution \(\left(\mathrm{NaCl}\right.\) in \(\left.\mathrm{H}_{2} \mathrm{O}\right)\) ? Recall that \(\mathrm{NaCl}\) is an electrolyte; assume complete conversion to \(\mathrm{Na}^{+}\) and \(C]^{-}\) inns
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