The freezing point depression constants of the solvents cyclohexane andnaphthalene are \(20.1^{\circ} \mathrm{C} / \mathrm{m}\) and \(6.94^{\circ}\mathrm{C} / \mathrm{m},\) respectively. Which solvent will give a moreaccurate result if you are using freezing point depression to determine themolar mass of a substance that is soluble in either one? Why?
Short Answer
Expert verified
Cyclohexane will give a more accurate result because it has a higher freezing point depression constant.
Freezing point depression refers to the lowering of the freezing point of a solvent when a solute is dissolved in it. The extent of the freezing point depression is given by the formula: \[\triangle T_f = K_f \times m \] where \(\triangle T_f\) is the freezing point depression, \(K_f\) is the freezing point depression constant, and \(m\) is the molality of the solution.
02
- Compare Freezing Point Depression Constants
Given \(K_f\) for cyclohexane = 20.1°C/m, and \(K_f\) for naphthalene = 6.94°C/m. Larger \(K_f\) means a larger change in temperature per unit molality for a given amount of solute, making the changes easier to measure accurately.
03
- Evaluate Measurement Sensitivity
Since cyclohexane has a larger \(K_f\), the same molality will result in a larger freezing point depression. This larger change is easier to measure accurately, reducing experimental error.
04
- Conclusion
Cyclohexane, with a higher freezing point depression constant, will give a more accurate result in determining the molar mass of a dissolved substance because the resulting temperature change is more pronounced and easier to measure accurately.
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Freezing Point Depression Constant
Freezing point depression is when the freezing point of a solvent lowers after a solute is added. The extent of this depression depends on several factors. One of the most critical ones is the freezing point depression constant, denoted as \( K_f \). \( K_f \) indicates how much the freezing point of the solvent will drop per molal concentration of solute. A higher \( K_f \) means a more significant change in temperature per molal unit, making measurements more noticeable. In our example, cyclohexane has a \( K_f \) of 20.1°C/m, and naphthalene has \( K_f \) of 6.94°C/m. This means cyclohexane will show a greater temperature drop for the same amount of solute dissolved compared to naphthalene.
Molality
Molality is different from molarity. It measures the concentration of a solute in a solution based on the mass of the solvent, not the volume of the solution. Molality is defined as the number of moles of solute per kilogram of solvent and is denoted as 'm'. When using the formula for freezing point depression, \[ \triangle T_f = K_f \times m \], 'm' represents molality. This value helps in determining how much the freezing point has dropped. Since molality is mass-based, it is not affected by temperature changes, making it a reliable concentration measure, especially for experiments involving temperature variations like freezing point depression. \[ m = \frac{n_{\text{solute}}}{kg_{\text{solvent}}} \] where \( n_{\text{solute}} \) is the moles of solute and \( kg_{\text{solvent}} \) is the mass of the solvent in kilograms.
Experimental Accuracy
Accuracy in experiments is all about how close the measured value is to the true value. When measuring freezing point depression, higher accuracy is often achieved with a solvent having a larger freezing point depression constant. This leads to a larger temperature change for the same solute amount, making it easier to detect and measure. For example, with cyclohexane having a \( K_f \) of 20.1°C/m, the temperature drop will be more significant compared to naphthalene's \( K_f \) of 6.94°C/m. This larger change reduces possible errors in measurement, ensuring the results are more accurate. It makes cyclohexane a better solvent for determining molar mass when using freezing point depression.
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Most popular questions from this chapter
A biochemical engineer isolates a bacterial gene fragment and dissolves a 10.0-mg sample in enough water to make \(30.0 \mathrm{~mL}\) of solution. Theosmotic pressure of the solution is 0.340 torr at \(25^{\circ} \mathrm{C}\). (a) What is the molar mass of the gene fragment? (b) If the solution density is \(0.997 \mathrm{~g} / \mathrm{mL}\), how large isthe freezing point depression for this solution \(\left(K_{\mathrm{f}}\right.\)of water \(\left.=1.86^{\circ} \mathrm{C} / \mathrm{m}\right) ?\)What is the strongest type of intermolecular force between solute and solventin each solution? (a) \(\mathrm{C}_{6} \mathrm{H}_{14}(l)\) in \(\mathrm{C}_{8} \mathrm{H}_{18}(l)\) (b) \(\mathrm{H}_{2} \mathrm{C}=\mathrm{O}(g)\) in \(\mathrm{CH}_{3}\mathrm{OH}(l)\) (c) \(\operatorname{Br}_{2}(l)\) in \(\mathrm{CCl}_{4}(l)\)At an air-water interface, fatty acids such as oleic acid lie in a one-molecule-thick layer (a monolayer), with the heads in the water and the tailsperpendicular in the air. When \(2.50 \mathrm{mg}\) of oleic acid is placed on awater surface, it forms a circular monolayer \(38.6 \mathrm{~cm}\) in diameter.Find the surface area (in \(\mathrm{cm}^{2}\) ) occupied by one molecule\((\mathscr{l l}\) of oleic acid \(=283 \mathrm{~g} / \mathrm{mol}\) ).Name three intermolecular forces that stabilize the shape of a soluble,globular protein, and explain how they act.A small protein has a molar mass of \(1.50 \times 10^{4} \mathrm{~g} /\mathrm{mol}\). What is the osmotic pressure exerted at \(24.0^{\circ}\mathrm{C}\) by \(25.0 \mathrm{~mL}\) of an aqueous solution that contains \(37.5\mathrm{mg}\) of the protein?
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The sources of error in a freezing point depression experiment can include impurities in the solvent or solute, inaccuracies in temperature measurement, improper stirring, and experimental technique.
Another example of freezing point depression of a solvent can be observed in vodka. It can be considered to be a solution of ethanol in water, and its freezing point is lower than water but higher than pure ethanol.
Freezing point depression is a colligative property observed in solutions that results from the introduction of solute molecules to a solvent. The freezing points of solutions are all lower than that of the pure solvent and is directly proportional to the molality of the solute.
The equation for freezing point depression is T=i*Kf*m, where T is the change in freezing point, i is the van't Hoff factor, Kf is the freezing point depression constant, and m is the molality of the solution.
The freezing point depression examples are mentioned below. The freezing point of seawater is below zero Celsius. Seawater remains liquid at temperatures lower than that of the freezing point of pure water.
Physical and chemical changes: both physical and chemical changes alter the freezing point of a substance. Pressure: Changing the pressure also affects the freezing point of a substance. If the answering pressure is lower than 1atm, the temperature at which the substance freezes also decreases.
The conclusion of the "Colligative Properties: Freezing-point Depression and Molar Mass" lab indicates that the freezing-point depression of a solution is directly proportional to the molality of the solute particles, providing a method to determine the molar mass of an unknown solute.
Freezing-point depression is what causes sea water (a mixture of salt and other compounds in water) to remain liquid at temperatures below 0 °C (32 °F), the freezing point of pure water.
In depression of freezing point experiment, vapour pressure of solution is less than that of pure solvent as well as only solvent molecules solidify at freezing point.
Freezing-point depression is the decrease of the freezing point of a solvent on addition of a non-volatile solute. Examples include salt in water, alcohol in water, or the mixing of two solids such as impurities into a finely powdered drug.
The main ratio that we use to find the angle of depression is tangent. The angle of depression may be found by using this formula: tan y = opposite/adjacent. The opposite side in this case is usually the height of the observer or height in terms of location, for example, the height of a plane in the air.
The freezing point depression due to the presence of a solute is also a colligative property. That is, the amount of change in the freezing point is related to the number of particles of solute in a solution and is not related to the chemical composition of the solute.
Depression in freezing point is a colligative property which depends on number of particles. Among given choices K2SO4 gives maximum number of ions, so it will have maximum depression in freezing point.
The freezing point of a solution can be calculated using colligative properties. In this case, we can use the freezing point depression equation ATf = Kf * m to calculate the change in freezing point. The freezing point constant, Kf, depends on the solvent used. For water, the freezing point constant is 1.86°C kg.
The change in the freezing point is proportional to the amount of solute added. This phenomenon is called freezing point depression. The change in the freezing point is defined as: ∆Tf = Tf,solution − Tf,solvent. ∆Tf is negative because the temperature of the solution is lower than that of the pure solvent.
Final answer: Calculate the expected freezing-point depression by multiplying the van't Hoff factor, the freezing-point depression constant, and the molality. For a 0.200 m KNO3 solution, it's expected to be 0.744°C, though real-world results may vary due to non-ideal behavior.
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