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By dissolving \(5 \mathrm{~g}\) substance in \(50 \mathrm{~g}\) of water, the decrease in freezing point is \(1.2^{\circ} \mathrm{C}\). The molal depression constant is \(1.85^{\circ} \mathrm{kg} \mathrm{mol}^{-1} .\) The molecular weight of substance is (a) \(105.4\) (b) \(118.2\) (c) \(137.2\) (d) \(154.2\)

Short Answer

Expert verified
We are given:1. Mass of the solute (substance) = \(5 \text{ g}\).2. Mass of solvent (water) = \(50 \text{ g} = 0.05 \text{ kg}\).3. Depression in freezing point, \(\Delta T_f = 1.2 ^\circ \text{C}\).4. Molal depression constant, \(K_f = 1.85 ^\circ \text{kg} \text{mol}^{-1}\).

Step by step solution

01

Understand the Given Values

We are given:1. Mass of the solute (substance) = \(5 \text{ g}\).2. Mass of solvent (water) = \(50 \text{ g} = 0.05 \text{ kg}\).3. Depression in freezing point, \(\Delta T_f = 1.2 ^\circ \text{C}\).4. Molal depression constant, \(K_f = 1.85 ^\circ \text{kg} \text{mol}^{-1}\).

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Key Concepts

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

Molal Depression Constant
The Molal Depression Constant, often represented as \(K_f\), is a crucial factor in calculating how the freezing point of a solvent is lowered when a solute is added. This constant is unique to each solvent. In simple terms, \(K_f\) helps determine how much the freezing point will drop per molal concentration of a solute present.

For instance, water has a \(K_f\) value of \(1.86\, ^\circ \text{C}\, \text{kg}\, \text{mol}^{-1}\), which means that for every mole of substance dissolved in 1 kg of water, the freezing point will decrease by \(1.86\, ^\circ \text{C}\). The value you are using depends on the nature of both the solute and the solvent. This constant helps us compare the effect of different solutes on the freezing point of a specific solvent.

Understanding \(K_f\) becomes particularly useful when we need to predict how a solution will behave under cold conditions or when trying to identify a solute based on its effect on the solvent's freezing point.
Depression in Freezing Point
The Depression in Freezing Point, or \(\Delta T_f\), is the change in the temperature at which a liquid freezes once a solute is added. When a solute dissolves in a solvent, it disrupts the solvent's natural ability to form a solid structure. As a result, the solvent's freezing point lowers.

This change is calculated using the formula: \[\Delta T_f = i \cdot K_f \cdot m\] where:
  • \(i\) is the van 't Hoff factor (equal to 1 for non-electrolytes).
  • \(K_f\) is the molal depression constant.
  • \(m\) (molality) is the moles of solute per kilogram of solvent.
This formula shows that the more solute is present, the greater the decrease in the freezing point. Like wearing layers in the cold lowers your body's exposure to the temperature, adding more solute molecules 'shields' the solvent molecules from reaching their freezing point.

Knowing \(\Delta T_f\) is crucial for various real-world applications, such as preventing roads from freezing or designing antifreeze for cars.
Molecular Weight Calculation
Calculating the molecular weight of a substance through the freezing point depression is based on relationships involving the amount of solute and how it affects the freezing point. The formula that ties everything together is:\[M = \frac{K_f \cdot w}{\Delta T_f \cdot W}\] where:
  • \(M\) is the molecular weight of the solute.
  • \(K_f\) is the molal depression constant.
  • \(w\) is the mass of solute (in grams).
  • \(\Delta T_f\) is the observed freezing point depression (in °C).
  • \(W\) is the mass of the solvent (in kilograms).
By plugging in the values, you compute \(M\), giving the molecular weight of the unknown substance. In practice, measuring \(\Delta T_f\) can help identify unknown solutes based on these principles.

This method is a handy tool in chemistry to find unknown substances by using their effects on the freezing point of a known solvent.

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Most popular questions from this chapter

The vapour pressure of a solution of \(5 \mathrm{~g}\) of non electrolyte in \(100 \mathrm{~g}\) of water at a particular temperature is \(2985 \mathrm{Nm}^{-2}\). The vapour pressure of pure water at that temperature is \(3000 \mathrm{Nm}^{-2}\). The molecular weight of the solute is (a) 180 (b) 90 (c) 270 (d) 360

Two liquids \(\mathrm{X}\) and \(\mathrm{Y}\) form an ideal solution. At \(300 \mathrm{~K}\), vapour pressure of the solutions containing 1 mol of \(X\) and 3 mol of \(Y\) is \(550 \mathrm{mmHg}\). At the same temperature, if 1 mol of \(Y\) is further added to this solu-tion, vapour pressure of the solution increases by 10 \(\mathrm{mmHg}\). Vapour pressure (in \(\mathrm{mmHg}\) ) of \(\mathrm{X}\) and \(\mathrm{Y}\) in their pure states will be, respectively: [2009] (a) 300 and 400 (b) 400 and 600 (c) 500 and 600 (d) 200 and 300

Which of the following statement is true about ideal solutions? (a) the volume of mixing is zero (b) the enthalpy of mixing is zero (c) both \(\mathrm{A}\) and \(\mathrm{B}\) (d) none of these

Benzene and toluene form nearly ideal solutions. At \(20^{\circ} \mathrm{C}\), the vapour pressure of benzene is 75 torr and that of toluene is 22 torr. The partial vapour pressure of benzene at \(20^{\circ} \mathrm{C}\) for a solution containing \(78 \mathrm{~g}\) of benzene and \(46 \mathrm{~g}\) of toluene in torr is (a) 25 (b) 50 (c) \(37.5\) (d) \(53.5\)

During osmosis, flow of water through a semipermeable membrane is (a) from both sides of semi-permeable membrane with unequal flow rates (b) from solution having lower concentration only (c) from solution having higher concentration only (d) from both sides of semi-permeable membrane with equal flow rates

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