When HNO 2 is dissolved in water, it partially dissociates according to the equation HNO 2 ⇌ H + + NO 2 - . A solution is prepared that contains 7.050 g of HNO 2 in 1.000 kg of water. Its freezing point is -0.2929 °C. Calculate the fraction of HNO 2 that has dissociated.

Calculate the molality (m) of the solution using the formula: \( m = \frac{\text{moles of solute}}{\text{kilograms of solvent}} \). First, find the moles of HNO2 by dividing the mass of HNO2 (7.050 g) by its molar mass (47.013 g/mol). Then, divide the moles of HNO2 by the mass of water in kilograms (1.000 kg).. The freezing point depression (\( \Delta T_f \)) is given by the formula: \( \Delta T_f = i \cdot K_f \cdot m \), where \( i \) is the van't Hoff factor, \( K_f \) is the freezing point depression constant for water (1.86 °C kg/mol), and \( m \) is the molality. Since the freezing point is -0.2929 °C, \( \Delta T_f = 0.2929 \) °C.. Rearrange the freezing point depression formula to solve for \( i \): \( i = \frac{\Delta T_f}{K_f \cdot m} \). Substitute the known values of \( \Delta T_f \), \( K_f \), and \( m \) to find \( i \).. The van't Hoff factor \( i \) is related to the degree of dissociation \( \alpha \) by the equation: \( i = 1 + \alpha \cdot (n - 1) \), where \( n \) is the number of particles the solute dissociates into. For HNO2, \( n = 2 \) (HNO2 dissociates into H+ and NO2-). Solve for \( \alpha \) using the calculated \( i \).. The fraction of HNO2 that has dissociated is equal to the degree of dissociation \( \alpha \). Express \( \alpha \) as a percentage to find the fraction of HNO2 that has dissociated.>

When HNO 2 is dissolved in water, it partially dissociates according to the equation HNO 2 ⇌ H + + NO 2 - . A solution is prepared that contains 7.050 g of HNO 2 in 1.000 kg of water. Its freezing point is -0.2929 °C. Calculate the fraction of HNO 2 that has dissociated.

Welcome back everyone. When nitrous acid is dissolved in water, it partially dissociates according to the equation ho two in equilibrium with H plus and no two negative A solution is prepared that contains 7.050 g of ho two in 1 kg of water. Its freezing point is negative 0.2929 °C. Calculate the fraction of HO two that has associated. There are four answer choices. A 0.050 B 0.50 C 0.25 ND 0.025. We're looking at the process of freezing. So let's simply recall the freezing point depression law which states that the change in the freezing point delta TF is equal to the van Hog factor high multiplied by the freezing point constant KF multiplied by morality. M. Now, essentially what we're trying to identify in this problem is I minus one. But let's first understand what I represents. Well, I represents the degree of the association. Let's suppose that our substance dissolves into two ions. And basically this means that ideally we would have an I of two. So let's first of all solve for I and then we will understand it better. I is essentially delta TF divided by the product between KF and M. Now, what is the change in the freezing point? Well, since we're looking at an aqueous solution, because HNO two is an acid, the change in the freezing point will be 0.2929 because the original freezing point of water is zero °C. So we're using this valley and then we are dividing by KF which is 1.86 for water degrees Celsius per molo. And then we need to multiply by the morality. So let's identify what it is. Let's remember that morality is equal to the number of moles of solute. In this case, H two, we're taking this quantity of most and dividing by the kilograms of solvent, which is water. In this case, to identify the number of moles of H and 02, we need to take its mass 7.050 g and divided by the molar mass. So we're taking most on top grams on the bottom one mole of H and 02 is equivalent to 47.02 g. Now, on the bottom, we're going to include kilograms of water. That's 1 kg is given in the problem. And what we understand is that the morality of the solution is zero 0.1499 well done. So we can now use this number for the calculation of the van hog factor. So we are going to put 0.1499. And now we can calculate I, so we get the high value of 1.050. Now, what does this tell us? Let's remember that we have an equation H two in equilibrium with the ions specifically H plus and no two negative. Ideally, if this association was complete, we would have a van half hectare of two. How do we know that? Well, essentially one mole of starting material dissociates into two ions completely. But here we have a partial dissociation. Now, the van ho factor is, in this case is 1.050 right? Since we can consider this as a weak electrolyte, we know that non electrolytes have even half factor of one. And here we have even half factor which is greater than one. So logically, the fraction associated is simply I minus one because that access corresponds to the association. And if we take 1.050 and subtract one, we get 0.050. So this axis above one represents the degree of the association. And specifically, if we look at the answer choices, we can determine the correct answer. Now, which one of them is the correct choice. Well, it's a 0.050. Thank you for watching.

Ch.13 - Solutions

Problem 114

Chapter 13, Problem 114

When HNO₂ is dissolved in water, it partially dissociates according to the equation HNO₂ ⇌ H⁺ + NO₂^-. A solution is prepared that contains 7.050 g of HNO₂ in 1.000 kg of water. Its freezing point is -0.2929 °C. Calculate the fraction of HNO₂ that has dissociated.

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<Step 1: Determine the molality of the solution.> Calculate the molality (m) of the solution using the formula: \( m = \frac{\text{moles of solute}}{\text{kilograms of solvent}} \). First, find the moles of HNO2 by dividing the mass of HNO2 (7.050 g) by its molar mass (47.013 g/mol). Then, divide the moles of HNO2 by the mass of water in kilograms (1.000 kg).

<Step 2: Use the freezing point depression formula.> The freezing point depression (\( \Delta T_f \)) is given by the formula: \( \Delta T_f = i \cdot K_f \cdot m \), where \( i \) is the van't Hoff factor, \( K_f \) is the freezing point depression constant for water (1.86 °C kg/mol), and \( m \) is the molality. Since the freezing point is -0.2929 °C, \( \Delta T_f = 0.2929 \) °C.

<Step 3: Solve for the van't Hoff factor (i).> Rearrange the freezing point depression formula to solve for \( i \): \( i = \frac{\Delta T_f}{K_f \cdot m} \). Substitute the known values of \( \Delta T_f \), \( K_f \), and \( m \) to find \( i \).

<Step 4: Determine the degree of dissociation (\( \alpha \)).> The van't Hoff factor \( i \) is related to the degree of dissociation \( \alpha \) by the equation: \( i = 1 + \alpha \cdot (n - 1) \), where \( n \) is the number of particles the solute dissociates into. For HNO2, \( n = 2 \) (HNO2 dissociates into H+ and NO2-). Solve for \( \alpha \) using the calculated \( i \).

<Step 5: Calculate the fraction of HNO2 dissociated.> The fraction of HNO2 that has dissociated is equal to the degree of dissociation \( \alpha \). Express \( \alpha \) as a percentage to find the fraction of HNO2 that has dissociated.>

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Dissociation of Acids

Dissociation refers to the process by which an acid separates into its constituent ions in solution. For weak acids like HNO2, this process is not complete, meaning only a fraction of the acid molecules dissociate into hydrogen ions (H+) and nitrite ions (NO2-). Understanding the degree of dissociation is crucial for calculating properties such as freezing point depression.

Freezing Point Depression

Freezing point depression is a colligative property that describes how the freezing point of a solvent decreases when a solute is added. The extent of this depression depends on the number of solute particles in solution, which can be calculated using the formula ΔTf = i * Kf * m, where 'i' is the van 't Hoff factor, 'Kf' is the freezing point depression constant, and 'm' is the molality of the solution.

Molality

Molality is a measure of concentration defined as the number of moles of solute per kilogram of solvent. It is expressed in moles per kilogram (mol/kg) and is particularly useful in colligative property calculations because it remains unaffected by temperature changes. In this problem, calculating the molality of the HNO2 solution is essential for determining the extent of freezing point depression and, subsequently, the fraction of dissociated HNO2.

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Molality

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Open Question

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When HNO2 is dissolved in water, it partially dissociates according to the equation HNO2 ⇌ H+ + NO2-. A solution is prepared that contains 7.050 g of HNO2 in 1.000 kg of water. Its freezing point is -0.2929 °C. Calculate the fraction of HNO2 that has dissociated.

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Key Concepts

Dissociation of Acids

Freezing Point Depression

Molality

When HNO₂ is dissolved in water, it partially dissociates according to the equation HNO₂ ⇌ H⁺ + NO₂^-. A solution is prepared that contains 7.050 g of HNO₂ in 1.000 kg of water. Its freezing point is -0.2929 °C. Calculate the fraction of HNO₂ that has dissociated.