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Chapter 10: Problem 58

The freezing point of a \(0.21 m\) aqueous solution of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is \(-0.796^{\circ} \mathrm{C}\) (a) What is \(i\) ? (b) Is the solution made up primarily of (i) \(\mathrm{H}_{2} \mathrm{SO}_{4}\) molecules only? (ii) \(\mathrm{H}^{+}\) and \(\mathrm{HSO}_{4}^{-}\) ions? (iii) \(2 \mathrm{H}^{+}\) and \(1 \mathrm{SO}_{4}^{2-}\) ions?

Short Answer

Expert verified
Answer: The van't Hoff factor (i) is approximately 2, and the solution is made up primarily of H+ and HSO4- ions.

Step by step solution

01

Calculate the Freezing Point Depression

The freezing point of water is 0°C. The freezing point depression (ΔTf) is the difference between the freezing point of the pure solvent (water) and the freezing point of the solution. So, ΔTf = Freezing point of water - Freezing point of solution ΔTf = 0 - (-0.796) ΔTf = 0.796°C
02

Use the Freezing Point Depression Formula to Find i

The freezing point depression formula is: ΔTf = i * Kf * m where: ΔTf = Freezing point depression (0.796°C, already calculated in Step 1) Kf = The cryoscopic constant for water (1.86°C·kg/mol) m = Molality of the solution (0.21 mol/kg) i = Van't Hoff factor (this is what we need to find) Substituting the known values, we get: 0.796 = i * 1.86 * 0.21 Now we will solve for i: i = 0.796 / (1.86 * 0.21) i ≈ 2
03

Determine the Primary Ions in the Solution

Since the calculated van't Hoff factor (i) is close to 2, we can conclude that the solution is made up primarily of H+ and HSO4- ions. Therefore, the solution is primarily composed of the ions mentioned in option (ii). #Answer#: (a) The van't Hoff factor (i) is approximately 2. (b) The solution is made up primarily of H+ and HSO4- ions (option (ii)).

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Most popular questions from this chapter

Pure benzene boils at \(80.10^{\circ} \mathrm{C}\) and has a boiling point constant, \(k_{\mathrm{b}}\), of \(2.53^{\circ} \mathrm{C} / \mathrm{m} .\) A sample of benzene is contaminated by naphthalene, \(\mathrm{C}_{10} \mathrm{H}_{8}\). The boiling point of the contaminated sample is \(81.20^{\circ} \mathrm{C}\). How pure is the sample? (Express your answer as mass percent of benzene.)

Explain why (a) the freezing point of \(0.10 \mathrm{~m} \mathrm{CaCl}_{2}\) is lower than the freezing point of \(0.10 \mathrm{~m} \mathrm{CaSO}_{4}\) (b) the solubility of solids in water usually increases as the temperature increases. (c) pressure must be applied to cause reverse osmosis to occur. (d) \(0.10 \mathrm{M} \mathrm{BaCl}_{2}\) has a higher osmotic pressure than \(0.10 \mathrm{M}\) glucose. (e) molarity and molality are nearly the same in dilute solutions.

Arrange \(0.30 \mathrm{~m}\) solutions of the following solutes in order of increasing freezing point and boiling point. (a) \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}\) (b) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) (c) \(\mathrm{Ba}(\mathrm{OH})_{2}\) (d) \(\mathrm{CaCr}_{2} \mathrm{O}_{7}\)

Solutions introduced directly into the bloodstream have to be "isotonic" with blood; that is, they must have the same osmotic pressure as blood. An aqueous \(\mathrm{NaCl}\) solution has to be \(0.90 \%\) by mass to be isotonic with blood. What is the molarity of sodium ions in solution? Take the density of the solution to be \(1.00 \mathrm{~g} / \mathrm{mL}\).

Show how 1 ppb (part per billion) is equivalent to 1 microgram \(/ \mathrm{kg}\). One microgram \(=10^{-6} \mathrm{~g}\).
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