Question

The osmotic coefficient of a nonelectrolyte is related to the freezing point depression by the expression, \(\phi=\Delta T_{f} /\) \(\left(m \cdot K_{e}\right) .\) The depression in freezing point of \(0.4\) molal aqueous solution of sucrose is \(0.93^{\circ} \mathrm{C}\). The osmotic coefficient is \(\left(K_{\mathrm{f}}\right.\) of water \(=1.86 \mathrm{~K}-\mathrm{kg} / \mathrm{mol}\) ) (a) \(0.8\) (b) \(1.0\) (c) \(1.25\) (d) \(0.125\)

Short answer

The osmotic coefficient is approximately 1.25.

Step-by-step solution

Understand the freezing point depression formula

The osmotic coefficient, \(\phi\), is related to the freezing point depression, \(\Delta T_f\), the molality of the solution, \(m\), and the cryoscopic constant, \(K_f\), by the formula \(\phi = \frac{\Delta T_f}{m \cdot K_f}\). To solve for \(\phi\), we need to plug in the known values for \(\Delta T_f\), \(m\), and \(K_f\).

Identify given values

We are given the freezing point depression \(\Delta T_f = 0.93^\circ C\), the molality of the solution \(m = 0.4\) molal, and the cryoscopic constant of water \(K_f = 1.86 \mathrm{K\cdot kg/mol}\).

Calculate the osmotic coefficient

Substitute the given values into the formula to get \(\phi = \frac{0.93}{0.4 \cdot 1.86}\). Now calculate \(\phi\) by performing the arithmetic.

Perform the arithmetic

Calculate the osmotic coefficient: \(\phi = \frac{0.93}{0.4 \cdot 1.86} = \frac{0.93}{0.744} \(approximately \).\) Divide 0.93 by 0.744 to find the value of \(\phi\).

Key concepts

Osmotic Coefficient

Understanding the osmotic coefficient is crucial for comprehending various phenomena in physical chemistry, particularly those involving solutions. The osmotic coefficient, denoted by the symbol \(\phi\), is a measure of the deviation of a real solution from ideal behavior. For nonelectrolyte solutions, it indicates how the presence of a solute affects the properties of the solvent, such as its freezing point.

The relationship between the osmotic coefficient and freezing point depression is encapsulated in the equation: \(\phi = \frac{\Delta T_f}{m \cdot K_f}\). In the context of the problem at hand, where a sucrose solution's freezing point is depressed, understanding this coefficient helps to quantify the effect of sucrose on the water's freezing point.

Notably, the osmotic coefficient is less than or equal to 1 for most solutions. A coefficient of 1 implies ideal behavior, where the solution behaves according to Raoult's law, without any interactions affecting the solute's impact. Values below 1 reflect the various interactions between solvent and solute molecules in real solutions.

Cryoscopic Constant

The cryoscopic constant, often symbolized as \(K_f\), plays a key role in understanding colligative properties, which are properties of solutions that depend on the number, not the identity, of solute particles. In simpler terms, \(K_f\) helps us determine how much the freezing point of a solvent will be lowered when a solute is dissolved in it.

\(K_f\) is unique to each solvent and is defined as the freezing point depression observed when one mole of solute is dissolved in one kilogram of solvent. Thus, for water, a \(K_f\) value of \(1.86\, \text{K} \cdot \text{kg/mol}\) means that the freezing point of water will decrease by \(1.86^\circ\text{C}\) for every mole of a non-volatile solute dissolved in one kilogram of water.

The cryoscopic constant is crucial in the solution provided because it allows for the calculation of the osmotic coefficient, \(\phi\), by applying the known freezing point depression \(\Delta T_f\) of the solution and the molality, \(m\), reflecting the number of moles of sucrose present in the solution.

Molality

Molality is a measure of the concentration of a solution, represented as the number of moles of solute per kilogram of solvent. Not to be confused with molarity, which is the number of moles per liter of solution, molality is especially useful in freezing point depression and boiling point elevation calculations because it, unlike molarity, does not change with temperature.

In the original exercise, molality is denoted as \(m\) and given as \(0.4\) molal, indicating that there are \(0.4\) moles of sucrose dissolved in \(1\) kilogram of the water solvent. Molality is used in the formula to find the osmotic coefficient because it directly relates to the number of solute particles present, which in turn affects the colligative property of freezing point depression.

It is important to note that when calculating molality, using the mass of the solvent (not the total solution) ensures that temperature fluctuations do not affect the concentration value, thus providing a more accurate measure for temperature-dependent studies.

Physical Chemistry

Physical chemistry explores the underlying physical principles governing the behavior of molecules and atoms, and how these principles give rise to chemical properties and reactions. This branch of chemistry is integral in understanding colligative properties like freezing point depression because it combines principles of thermodynamics, kinetics, and quantum mechanics to explain and predict the behavior of solutions.

The exploration of topics such as the osmotic coefficient and cryoscopic constant illustrates the real-world application of physical chemistry principles. By involving constants like the cryoscopic \(K_f\) and units of concentration like molality, physical chemistry provides the quantitative tools needed to analyze how molecular interactions in a solution can affect its physical properties.

This field of study is all about connections—how the microscopic properties of individual particles translate into the macroscopic properties observed in the laboratory. It's these connections that allow for the comprehension of phenomena like why the addition of a solute lowers the freezing point of a solvent, as demonstrated in the original exercise.