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Chapter 14: Problem 90

A compound is known to be a potassium halide, KX. If \(4.00 \mathrm{g}\) of the salt is dissolved in exactly \(100 \mathrm{g}\) of water, the solution freezes at \(-1.28^{\circ} \mathrm{C} .\) Identify the halide ion in this formula.

Short Answer

Expert verified
The halide ion is bromide (Br\(^-\)).

Step by step solution

01

Determine the Molality of the Solution

The freezing point depression formula is given by \[\Delta T_f = i \cdot K_f \cdot m\]where \(\Delta T_f\) is the freezing point depression \(1.28^{\circ}C\), \(i\) is the van't Hoff factor (which is 2 for KX, since it dissociates into K\(^+\) and X\(^-\)), \(K_f\) is the freezing point depression constant for water (\(1.86^{\circ}C\cdot mol^{-1}\, kg\)), and \(m\) is the molality.\[m = \frac{\Delta T_f}{i \cdot K_f} = \frac{1.28}{2 \cdot 1.86} = 0.344 \text{ mol/kg}\]
02

Calculate the Number of Moles of Salt

Molality (\(m\)) is defined as the number of moles of solute per kilogram of solvent. Using the calculated molality:\[\text{molality} = \frac{\text{moles of salt}}{0.1 \text{ kg of water}}\]Thus,\[\text{moles of salt} = 0.344 \text{ mol/kg} \times 0.1 \text{ kg} = 0.0344 \text{ mol}\]
03

Determine the Molar Mass of the Salt

We know that \(4.00\) grams of the salt contains \(0.0344\) moles. Using the relationship between mass, moles, and molar mass:\[\text{Molar mass of KX} = \frac{4.00 \text{ g}}{0.0344 \text{ mol}} = 116.28 \text{ g/mol}\]
04

Identify the Halide Ion X

Knowing that K has a molar mass of \(39.10\) g/mol, we can find the mass of X:\[116.28 \text{ g/mol} - 39.10 \text{ g/mol} = 77.18 \text{ g/mol}\]The approximate molar mass of X (77.18 g/mol) matches the molar mass of bromide (Br).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Freezing Point Depression
Freezing point depression is a colligative property observed when a solute is added to a solvent, causing the solution's freezing point to be lower than that of the pure solvent. This phenomenon happens because the solute particles disrupt the forming of a solid phase, requiring a lower temperature to reach the freezing point. In our scenario, a potassium halide (KX) is dissolved in water, causing the solution to freeze at -1.28掳C.

The formula that describes freezing point depression is:- \[ \Delta T_f = i \cdot K_f \cdot m \] where: - \(\Delta T_f\) is the change in freezing point. - \(i\) is the van't Hoff factor, representing the number of particles the solute splits into in solution. For KX, \(i = 2\), because it dissociates into K\(^+\) and X\(^-\). - \(K_f\) is the freezing point depression constant for water, valued at 1.86掳C路mol鈦宦孤穔g. - \(m\) is the molality of the solution.

By applying the known values into the equation, we calculate the molality, an essential step in finding the molar mass of the unknown halide.
Molality Calculation
Molality (\(m\)) measures how many moles of solute are present per kilogram of solvent, which is a crucial step in determining the degree of freezing point depression. Unlike molarity, molality does not change with temperature or pressure as it involves mass, not volume. To find molality for our solution, use the derived formula:
  • \(m = \frac{\Delta T_f}{i \cdot K_f}\)
Given the change in freezing point (\(\Delta T_f\)) is 1.28掳C and the van't Hoff factor (\(i\)) is 2, along with \(K_f\) for water as 1.86掳C路mol鈦宦孤穔g, the molality can be calculated as:
  • \[m = \frac{1.28}{2 \cdot 1.86} = 0.344 \text{ mol/kg}\]
This value signifies 0.344 moles of potassium halide is present per 1 kg of water, providing a base for further calculations involving the mass and molar quantity of the halide.
Molar Mass Determination
The process of determining molar mass involves understanding the relationship between the mass of a substance, the number of moles, and the molecular composition. For the potassium halide (KX), knowing the sample's mass and its molality allows us to derive the moles of KX in solution.

From the calculated molality (0.344 mol/kg) and the fact that the solvent's mass is 0.1 kg, calculate the moles of KX:
  • \[\text{moles of KX} = 0.344 \times 0.1 = 0.0344 \text{ mol} \]
With 4.00 grams of KX provided, the molar mass can be calculated as:
  • \[\text{Molar mass of KX} = \frac{4.00}{0.0344} = 116.28 \text{ g/mol} \]
The discovered molar mass of KX allows identification of the halide ion X by subtracting the known mass of potassium(\(39.10\) g/mol). The remaining mass points to a halide, with 77.18 g/mol matching that of bromide (Br\(^-\)). This illustrates how freezing point depression and calculated molality lead to finding molar mass and identifying unknown ions in a compound.

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Most popular questions from this chapter

When solutions of \(\mathrm{BaCl}_{2}\) and \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) are mixed, the mixture becomes cloudy. After a few days, a white solid is observed on the bottom of the beaker with a clear liquid above it. (a) Write a balanced equation for the reaction that occurs. (b) Why is the solution cloudy at first? (c) What happens during the few days of waiting?

Pure iodine \((105 \mathrm{g})\) is dissolved in \(325 \mathrm{g}\) of \(\mathrm{CCl}_{4}\) at \(65^{\circ} \mathrm{C}\) Given that the vapor pressure of \(\mathrm{CCl}_{4}\) at this temperature is \(531 \mathrm{mm}\) Hg, what is the vapor pressure of the \(\mathrm{CCl}_{4}-\mathrm{I}_{2}\) solution at \(65^{\circ} \mathrm{C} ?\) (Assume that \(\mathrm{I}_{2}\) does not contribute to the vapor pressure.)

The Henry's law constant for \(\mathrm{O}_{2}\) in water at \(25^{\circ} \mathrm{C}\) is 1.66 \(\times 10^{-6} \mathrm{M} / \mathrm{mm}\) Hg. Which of the following is a reasonable constant when the temperature is \(50^{\circ} \mathrm{C}\) ? Explain the reason for your choice. (a) \(8.80 \times 10^{-7} \mathrm{M} / \mathrm{mm} \mathrm{Hg}\) (b) \(3.40 \times 10^{-6} \mathrm{M} / \mathrm{mm} \mathrm{Hg}\) (c) \(1.66 \times 10^{-6} \mathrm{M} / \mathrm{mm} \mathrm{Hg}\) (d) \(8.40 \times 10^{-5} \mathrm{M} / \mathrm{mm} \mathrm{Hg}\)

(a) Which aqueous solution is expected to have the higher boiling point: \(0.10 \mathrm{m} \mathrm{Na}_{2} \mathrm{SO}_{4}\) or \(0.15 \mathrm{m}\) sugar? (b) For which aqueous solution is the vapor pressure of water higher: \(0.30 \mathrm{m} \mathrm{NH}_{4} \mathrm{NO}_{3}\) or \(0.15 \mathrm{m} \mathrm{Na}_{2} \mathrm{SO}_{4} ?\)

Hydrochloric acid is sold as a concentrated aqueous solution. If the molarity of commercial HCl is 12.0 and its density is \(1.18 \mathrm{g} / \mathrm{cm}^{3},\) calculate the following: (a) the molality of the solution (b) the weight percent of HCl in the solution
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