Question

An aqueous solution contains the amino acid glycine \(\left(\mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{COOH}\right)\). Assuming that the acid does not ionize in water, calculate the molality of the solution if it freezes at \(-1.1^{\circ} \mathrm{C}\)

Short answer

The molality of the solution is approximately 0.5914 mol/kg.

Step-by-step solution

Understanding the Problem

We need to find the molality of a glycine solution which has a freezing point depression. Given that its freezing point is -1.1°C. We will use the freezing point depression formula to find the molality.

Applying Freezing Point Depression

The freezing point depression \( \Delta T_f \) can be expressed as \( \Delta T_f = i \cdot K_f \cdot m \). Here, \( i =1 \) (since glycine does not ionize), \( K_f = 1.86^{\circ} \mathrm{C} \cdot \mathrm{kg/mol} \) (for water), and \( \Delta T_f = 1.1^{\circ} \mathrm{C} \) (since it freezes at -1.1°C, a decrease from 0°C, which is the freezing point of pure water).

Calculating Molality

Rearranging the equation, we get \( m = \frac{\Delta T_f}{i \cdot K_f} = \frac{1.1}{1 \cdot 1.86} \approx 0.5914 \). This gives us the molality of the solution.

Finalizing

The calculated molality suggests that there are approximately 0.5914 moles of glycine per kilogram of water, assuming glycine does not ionize.

Key concepts

Molality

Molality is a measure of the concentration of a solute in a solvent. It is defined as the number of moles of solute per kilogram of solvent. This unit is particularly useful in scenarios where temperature changes can affect the volume of the solution, such as in freezing point depression. Unlike molarity, which is influenced by temperature because it depends on the volume of the solution, molality remains constant since it is mass-based.

• This means, in any situation involving temperature change like with freezing or boiling points, molality is a more reliable measure to use.
• Molality is symbolized as "m" and is calculated using the formula: \[ m = \frac{\text{moles of solute}}{\text{kilograms of solvent}} \]

For example, in the exercise with glycine, the molality of the solution was solved using the freezing point depression equation. Given the freezing point depression of \(1.1^{\circ} \mathrm{C}\) and a cryoscopic constant of \(1.86^{\circ} \mathrm{C} \cdot \mathrm{kg/mol}\), the molality was calculated as \(0.5914\,\text{mol/kg}\.\)

Glycine

Glycine is a simple amino acid with the chemical formula \( \mathrm{NH}_{2}\mathrm{CH}_{2}\mathrm{COOH} \). In chemistry, it's considered a zwitterion, meaning it contains both a positive and a negative charge. However, it's important to note that, in certain solutions and exercises like the one given, it can be treated as if it does not ionize in water.

• Being the smallest and only non-chiral amino acid, glycine is interesting for studies on molecular behavior.
• In the given exercise, the importance of glycine not ionizing in its solution meant the van 't Hoff factor, \( i \), was simply "1".

This simplifies calculations related to colligative properties, as it simplifies the equation for freezing point depression used to determine molality.

Colligative Properties

Colligative properties are the physical changes that result from adding solutes to a solvent. They depend on the number of solute particles in a solution, not on the nature or type of particles. Common colligative properties include boiling point elevation, vapor pressure lowering, osmotic pressure, and freezing point depression.

• In freezing point depression, such as in our exercise, the addition of a solute lowers the freezing point of the solvent.
• The formula for freezing point depression is \[ \Delta T_f = i \cdot K_f \cdot m \]

Here, \( \Delta T_f \) is the change in freezing point, \( i \) the van 't Hoff factor (used to account for ionization; for non-ionizing substances like glycine, \( i = 1 \)), \( K_f \) the cryoscopic constant, and \( m \) the molality. In essence, colligative properties like freezing point depression help chemists predict how solute-solvent mixtures will behave under different conditions.