Question

The freezing point of \(t\) -butanol is \(25.50^{\circ} \mathrm{C}\) and \(K_{\mathrm{f}}\) is \(9.1^{\circ} \mathrm{C} \cdot \mathrm{kg} /\) mol. Usually \(t\) -butanol absorbs water on exposure to air. If the freezing point of a \(10.0\) -g sample of \(t\) -butanol is \(24.59^{\circ} \mathrm{C}\), how many grams of water are present in the sample?

Short answer

The short answer to the question is: \(0.018\) g of water is present in the \(10.0\) g sample of t-butanol.

Step-by-step solution

Calculate the Freezing Point Depression

First, we'll calculate the change in freezing point (ΔTf) by subtracting the freezing point of the t-butanol sample with unknown water content from the pure t-butanol freezing point. ΔTf = Tf(pure) - Tf(sample) ΔTf = 25.50 - 24.59 ΔTf = 0.91 °C

Calculate the Molality

Next, we will use the freezing point depression equation to find the molality. Rearranging the equation to solve for molality (m) gives: molality = ΔTf / Kf Now substitute the given values and calculate the molality: molality = 0.91 °C / 9.1 °C⋅kg/mol molality = 0.1 mol/kg

Determine the Moles of Water Present in the Sample

As we have the molality, we know there are 0.1 moles of water per kg of t-butanol. Now, we calculate the moles of water present in the 10.0 g sample: moles of water = molality × (mass of t-butanol / 1000) moles of water = 0.1 mol/kg × (10.0 g / 1000) moles of water = 0.001 mol

Calculate the Grams of Water Present in the Sample

To calculate the grams of water present in the sample, we multiply the moles of water by the molar mass of water (18.015 g/mol): grams of water = moles of water × molar mass of water grams of water = 0.001 mol × 18.015 g/mol grams of water = 0.018 g So, there are 0.018 g of water present in the 10.0 g sample of t-butanol.

Key concepts

Molality

Molality is a way to express the concentration of a solution. It's particularly useful when dealing with colligative properties, like freezing point depression. Molality is defined as the number of moles of solute per kilogram of solvent. This is different from molarity, which uses the volume of the solvent in liters.
Why is this difference important? Well, using kilograms helps because mass doesn't change with temperature, unlike volume. This is crucial for experiments like freezing point depression which can involve temperature changes.
To calculate molality in a problem, we often rearrange the freezing point depression formula \[ \Delta T_f = K_f \times m \] where \( \Delta T_f \) is the change in freezing point, \( K_f \) is the cryoscopic constant, and \( m \) is the molality. Thus, molality can be found using \[ m = \frac{\Delta T_f}{K_f} \].
For example, if the change in freezing point \( \Delta T_f \) is 0.91 °C and the \( K_f \) is 9.1 °C/kg/mol, then molality becomes 0.1 mol/kg.

Kf (cryoscopic constant)

The cryoscopic constant, represented as \( K_f \), is a constant that relates to how a solute affects the freezing point of a solvent. Each solvent has its unique \( K_f \) value, depending on its properties and how much it changes temperature when a solute is dissolved.
In practical terms, \( K_f \) allows us to calculate the extent to which a solution's freezing point is lowered. This is particularly important in cases such as predicting how much an antifreeze compound will lower the freezing point of water.
In the problem of t-butanol, the text specifies that the \( K_f \) for t-butanol is 9.1°C⋅kg/mol. This means that for every mole of solute per kilogram of t-butanol, the freezing point is affected by 9.1 degrees Celsius. Thus, the magnitude of \( K_f \) expresses how sensitive the solvent is to changes caused by solute particles.

t-butanol

T-butanol, also known as tert-butanol, is an organic compound used as a solvent and an intermediate in chemical synthesis. It's distinct in that it can absorb moisture from the air, which is why it's important to account for water content in solutions of t-butanol.
It has a relatively high freezing point for an alcohol at 25.50°C, making it a useful substance for experiments involving freezing point depression. The presence of impurities, like absorbed water, can affect its freezing point.
In this particular problem, a 10.0 g sample of t-butanol is assessed for water content due to its exposure to ambient air. Understanding the properties of t-butanol helps us interpret the significance and scale of changes in freezing point caused by contaminants like water. Doing so helps solve issues related to purity and quality in both industrial and laboratory settings.