When determining a molar mass from freczingpoint depression, it is possible to make each of the following errors (among others). In each case, predict whether the error would cause the reported molar mass to be greater or less than the actual molar mass. (a) There was dust on the balance, causing the mass of solute to appear greater than it actually was. (b) The water was measured by volume, assuming a density of \(1.00 \mathrm{~g} \cdot \mathrm{cm}^{-3}\), but the water was warmer and less dense than assumed. (c) The thermometer was not calibrated accurately, so the temperature of the freering point was actually \(0.5^{\circ} \mathrm{C}\) higher than recorded. (d) The solution was not stirred sufficiently so that not all the solute dissolved.

Short Answer

Expert verified
The reported molar mass would be: (a) less than actual, (b) greater than actual, (c) greater than actual, (d) less than actual.

Step by step solution

01

Analyzing Error (a)

If there was dust on the balance, the mass of solute would appear greater than it actually was. Using this inflated mass in the calculation of molar mass, obtained by dividing the mass of solute by the moles of solute (which is independent of the error), would lead to a calculated molar mass that is less than the actual molar mass. This is because the inflated mass would suggest more moles than were actually present.
02

Analyzing Error (b)

If water was measured by volume with the assumption of water density being 1.00 g/cm^3, but was actually warmer and less dense than assumed, the error would lead to a lower mass of water being used in the calculations than should have been. A lower mass would lead to a larger change in freezing point for a given amount of solute, which would cause an underestimation in the calculation of the molar mass, resulting in a reported molar mass that is greater than the actual molar mass.
03

Analyzing Error (c)

If the thermometer was not calibrated accurately, and the freezing point was actually 0.5°C higher than recorded, the freezing point depression would seem greater than it actually was. Since the molar mass is calculated by the change in freezing point, a greater change would imply fewer moles of solute for the given mass, leading to the calculation of a greater molar mass than the actual molar mass.
04

Analyzing Error (d)

If the solution was not stirred sufficiently and not all the solute dissolved, the effective concentration of the solute would be less than assumed. Since a smaller change in the freezing point would be observed, it would seem like the there are more moles of solute for the given mass, resulting in the calculation of a molar mass that is less than the actual molar mass.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Freezing Point Depression
Freezing point depression is a colligative property, which means it depends on the number of solute particles in a solvent, not on the type of particles. When a solute is dissolved in a solvent, the solution's freezing point is lowered compared to the pure solvent. This is due to the solute particles disrupting the formation of the solvent's crystal structure. The extent to which the freezing point is lowered is proportional to the molal concentration of the solution, which is the number of moles of solute per kilogram of solvent.

The equation that describes the relationship is given by \( \Delta T_f = i \cdot K_f \cdot m \), where \( \Delta T_f \) is the freezing point depression, \( i \) is the van't Hoff factor (which accounts for the number of particles the solute breaks into), \( K_f \) is the freezing point depression constant of the solvent, and \( m \) is the molality of the solution. Thus, the accurate determination of a solute's molar mass from freezing point depression requires precise measurements of the change in freezing point and the solution's concentration.
Analytical Error Analysis
Analytical error analysis involves identifying and understanding the various sources of errors that can occur during an experimental procedure and evaluating their potential impact on the results. In the context of molar mass determination by freezing point depression, several errors can influence the outcome. As examined in the textbook exercise, errors such as contamination of the balance (leading to an incorrect mass of solute), incorrect assumptions about solvent density (affecting solvent mass), and miscalibration of the thermometer (altering the observed freezing point) all result in the molar mass calculation being skewed.

Understanding the direction and magnitude of these errors is essential for accurate analytical measurements. Proper calibration of instruments, careful procedural execution, and taking into account variability in conditions such as temperature can minimize these errors. When conducting an experiment, scientists must identify potential errors, estimate their effects, and apply appropriate correction factors or uncertainty margins in their calculations and reported results.
Solution Concentration Calculations
Solution concentration calculations are critical for quantifying the amount of solute in a given volume or mass of solvent. For molar mass determination by freezing point depression, the key concentration measure is molality, defined as moles of solite per kilogram of solvent. Correctly calculating molality necessitates accurate mass measurements for both solute and solvent.

Error (b), as described in the exercise, illustrates how assuming water has a density of \(1.00 \text{ g/cm}^3\) without accounting for temperature changes could lead to an incorrect solvent mass, thus altering the solution's true concentration. This sort of analytical error would affect the molar mass calculation, as different concentrations lead to different extents of freezing point depression for a given solute. Hence, it is vital to measure all relevant properties accurately, including solute mass, solvent mass, and volume, to ensure precise and reliable solution concentration calculations.

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Most popular questions from this chapter

Hexane, \(\left(\mathrm{C}_{6} \mathrm{H}_{14}\right)\) and cyclohexane \(\left(\mathrm{C}_{6} \mathrm{H}_{12}\right)\) form an ideal solution. The vapor pressure of hexane is 151 Torr and that of cyclohexane is 98 Torr at \(25.0^{\circ} \mathrm{C}\). Calculate the vapor pressure of each of the following solutions and the mole fraction of each substance in the vapor phase above those solutions: (a) \(0.2 \mathrm{~S} \mathrm{~mol} \mathrm{C}_{6} \mathrm{H}_{14}\) mixed with \(0.65 \mathrm{~mol} \mathrm{C}_{6} \mathrm{H}_{12}\) (b) \(10.0 \mathrm{~g}\) of hexane mixed with \(10.0 \mathrm{~g}\) of cyclohexane.

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