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

There are 0.018 grams of water present in the 10.0-gram sample of \(t\)-butanol.

Step-by-step solution

Identify given values and formula

We are given: - Freezing point of pure \(t\)-butanol: \(T_{f_{pure}} = 25.50^{\circ} \mathrm{C}\) - Freezing point depression constant: \(K_f = 9.1^{\circ} \mathrm{C} \cdot \mathrm{kg}/\) mol - Freezing point of the mixture: \(T_{f_{mixture}} = 24.59^{\circ} \mathrm{C}\) - Sample mass of \(t\)-butanol: \(m_{t-butanol} = 10.0\, \mathrm{g}\) The freezing point depression formula is: \(\Delta T_f = K_f \cdot m_w\), where - \(\Delta T_f = T_{f_{pure}} - T_{f_{mixture}}\) is the freezing point depression - \(m_w\) is the molality of water in the mixture (mol/kg)

Calculate the freezing point depression

Using the freezing point depression formula, we first find the depression of the freezing point: \(\Delta T_f = T_{f_{pure}} - T_{f_{mixture}} = 25.50^{\circ} \mathrm{C} - 24.59^{\circ} \mathrm{C} = 0.91^{\circ} \mathrm{C}\)

Calculate the molality of water in the mixture

Using the freezing point depression formula and the value of \(K_f\): \(m_w = \frac{\Delta T_f}{K_f} = \frac{0.91^{\circ} \mathrm{C}}{9.1^{\circ} \mathrm{C} \cdot \mathrm{kg}/ \mathrm{mol}} = 0.1 \, \mathrm{mol/kg}\)

Calculate the mass of water in the mixture

We know that the molality of water in the mixture is \(0.1 \, \mathrm{mol/kg}\). Since molality is defined as moles of solute (water) per kilogram of solvent (\(t\)-butanol), we can write: \(m_w = \frac{n_{\mathrm{water}}}{m_{t-butanol} / 1000}\), where \(n_{\mathrm{water}}\) is moles of water and we divided by 1000 to convert grams to kilograms. Now we can solve for the moles of water: \(n_{\mathrm{water}} = m_w \cdot \frac{m_{t-butanol}}{1000} = 0.1 \, \mathrm{mol/kg} \cdot \frac{10.0 \, \mathrm{g}}{1000} = 0.001 \, \mathrm{mol}\) Since the molar mass of water is 18.015 g/mol, we can now find the mass of water present in the sample: \(m_{\mathrm{water}} = n_{\mathrm{water}} \cdot M_{\mathrm{water}} = 0.001 \, \mathrm{mol} \cdot 18.015 \, \mathrm{g}/\mathrm{mol} = 0.018 \, \mathrm{g}\)

Conclusion

There are 0.018 grams of water present in the 10.0-gram sample of \(t\)-butanol.

Key concepts

t-butanol

T-butanol, or tertiary butanol, is a type of alcohol with a chemical formula of \( C_4H_{10}O \). It is a colorless liquid that usually changes to a white crystalline solid when it reaches a temperature below its freezing point of 25.50°C. One interesting property of t-butanol is that it's highly hygroscopic, meaning that it can absorb moisture from the air very efficiently. This characteristic can lead to changes in its freezing point, making it an excellent compound to study freezing point depression, a colligative property of solutions.

molality

Molality is a measure of the concentration of a solute in a solvent. It is defined as the number of moles of solute divided by the mass of the solvent in kilograms. Unlike molarity, which uses the volume of the solution, molality uses the mass of the solvent:
\[ m = \frac{n_{\text{solute}}}{m_{\text{solvent}}} \]
where \( n_{\text{solute}} \) is the number of moles of solute and \( m_{\text{solvent}} \) is the mass of the solvent in kilograms. This concept is especially useful in calculations involving temperature changes, such as freezing point depression, because it is not affected by volume changes due to temperature fluctuations.

solvent-solute interactions

Solvent-solute interactions refer to the forces and bonds that occur between the particles of solute and the particles of solvent. In the case of freezing point depression, these interactions help to determine how much the freezing point is lowered when a solute is added to a solvent.

• The primary interactions include hydrogen bonding, dipole-dipole interactions, and van der Waals forces.
• In t-butanol, the presence of water (the solute) affects how t-butanol molecules (the solvent) arrange themselves, leading to changes in its freezing point.

The strength and nature of these interactions will impact how significantly the freezing point is depressed.

cryoscopic constant

The cryoscopic constant, denoted as \( K_f \), is a property specific to each solvent. It predicts the degree of freezing point depression per molal concentration of solute. This constant is a measure of how sensitive a solvent is to the addition of a solute, with units in °C kg/mol.

• The equation for freezing point depression is \( \Delta T_f = K_f \cdot m \), where \( \Delta T_f \) represents the change in freezing point, \( K_f \) is the cryoscopic constant, and \( m \) is the molality.
• For t-butanol, \( K_f = 9.1^{\circ} \text{C} \cdot \text{kg/mol} \), showing it's quite responsive to solute addition.

Understanding the cryoscopic constant is crucial when calculating how much a solute will influence the freezing point.