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Chapter 11: Problem 141

An aqueous solution containing 0.250 mole of \(\mathrm{Q},\) a strong electrolyte, in \(5.00 \times 10^{2} \mathrm{g}\) water freezes at \(-2.79^{\circ} \mathrm{C} .\) What is the van't Hoff factor for \(\mathrm{Q} ?\) The molal freezing-point depression constant for water is $1.86^{\circ} \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol}$ . What is the formula of \(\mathrm{Q}\) if it is 38.68\(\%\) chlorine by mass and there are twice as many anions as cations in one formula unit of \(\mathrm{Q} ?\)

Short Answer

Expert verified
The van't Hoff factor for Q is approximately 3. The formula of Q is MgCl₂, considering it has 38.68% chlorine by mass, and there are twice as many anions as cations in one formula unit of Q.

Step by step solution

01

Calculate the molality of the solution

To calculate the molality (mol/kg) of the solution, we need to divide the number of moles of the solute by the mass of the solvent (in kg). Molality = (moles of solute) / (mass of solvent in kg) Molality = \(0.250 \, mol / (5.00 \times 10^2 \, g \times (1 \, kg / 1000 \, g))\) Molality = 0.500 mol/kg
02

Calculate the van't Hoff factor

We'll use the freezing-point depression formula: ΔT = K_f * molality * i Where ΔT is the change in freezing point, K_f is the molal freezing-point depression constant, and i is the van't Hoff factor. We're given ΔT = -2.79°C (since the freezing point has decreased) and K_f = 1.86 °C (kg/mol) Rearrange the formula to solve for i: i = ΔT / (K_f * molality) i = \(-2.79 \, °C / (1.86 \, °C \cdot kg/mol) \times 0.500\, mol/kg\) i = \(-2.79 \, °C / 0.930 \times °C\) i ≈ 3
03

Finding the formula of Q

We know that Q is 38.68% chlorine by mass and there are twice as many anions as cations in one formula unit of Q. Since the van't Hoff factor is 3, we can infer that Q dissociates into 3 ions in the solution: one cation and two anions. Cation: Chlorine's most common charge is -1, so the cation should be +2 to balance the charge. The element with a +2 charge commonly found with chlorine is Magnesium (Mg). Anion: Chlorine (Cl) with a -1 charge We double the number of anions per cation, so the formula of Q is MgCl₂.

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Most popular questions from this chapter

A solution of phosphoric acid was made by dissolving 10.0 \(\mathrm{g}\) \(\mathrm{H}_{3} \mathrm{PO}_{4}\) in 100.0 \(\mathrm{mL}\) water. The resulting volume was 104 \(\mathrm{mL}\) . Calculate the density, mole fraction, molarity, and molality of the solution. Assume water has a density of 1.00 $\mathrm{g} / \mathrm{cm}^{3}$ .

For an acid or a base, when is the normality of a solution equal to the molarity of the solution and when are the two concentration units different?

Explain the terms isotonic solution, crenation, and hemolysis.

Which of the following statements is(are) true? Correct the false statements. a. The vapor pressure of a solution is directly related to the mole fraction of solute. b. When a solute is added to water, the water in solution has a lower vapor pressure than that of pure ice at \(0^{\circ} \mathrm{C} .\) c. Colligative properties depend only on the identity of the solute and not on the number of solute particles present. d. When sugar is added to water, the boiling point of the solution increases above \(100^{\circ} \mathrm{C}\) because sugar has a higher boiling point than water.

Plants that thrive in salt water must have internal solutions (inside the plant cells) that are isotonic with (have the same osmotic pressure as) the surrounding solution. A leaf of a saltwater plant is able to thrive in an aqueous salt solution (at \(25^{\circ} \mathrm{C} )\) that has a freezing point equal to \(-0.621^{\circ} \mathrm{C} .\) You would like to use this information to calculate the osmotic pressure of the solution in the cell. a. In order to use the freezing-point depression to calculate osmotic pressure, what assumption must you make (in addition to ideal behavior of the solutions, which we will assume)? b. Under what conditions is the assumption (in part a) reasonable? c. Solve for the osmotic pressure (at \(25^{\circ} \mathrm{C} )\) of the solution in the plant cell. d. The plant leaf is placed in an aqueous salt solution (at $25^{\circ} \mathrm{C}\( ) that has a boiling point of \)102.0^{\circ} \mathrm{C} .$ What will happen to the plant cells in the leaf?
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