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Chapter 9: Problem 103

Which one of the following aqueous solutions will exhibit highest boiling point? a. \(0.05 \mathrm{M}\) glucose b. \(0.01 \mathrm{M} \mathrm{KNO}_{3}\) c. \(0.015 \mathrm{M}\) urea d. \(0.01 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\)

Short Answer

Expert verified
The 0.05 M glucose solution will exhibit the highest boiling point.

Step by step solution

01

Understand Boiling Point Elevation

The boiling point of a solution is elevated based on the concentration of solute particles it contains. According to the colligative property, boiling point elevation can be found using the formula \( \Delta T_b = i \cdot K_b \cdot m \), where \( \Delta T_b \) is the boiling point elevation, \( i \) is the van't Hoff factor (number of particles the solute splits into), \( K_b \) is the ebullioscopic constant, and \( m \) is the molality of the solution.
02

Calculate for Each Option

First, determine the van't Hoff factor \( i \) for each solute: - Glucose: \( i = 1 \) (as it does not dissociate)- \( \mathrm{KNO}_{3} \): \( i = 2 \) (dissociates into \( \mathrm{K}^+ \) and \( \mathrm{NO}_3^- \))- Urea: \( i = 1 \) (as it does not dissociate)- \( \mathrm{Na}_{2} \mathrm{SO}_{4} \): \( i = 3 \) (dissociates into 2 \( \mathrm{Na}^+ \) and 1 \( \mathrm{SO}_4^{2-} \))Next, multiply the molarity with the van't Hoff factor to get the effective molarity for each:- Glucose: \( 0.05 \times 1 = 0.05 \)- \( \mathrm{KNO}_{3} \): \( 0.01 \times 2 = 0.02 \)- Urea: \( 0.015 \times 1 = 0.015 \)- \( \mathrm{Na}_{2} \mathrm{SO}_{4} \): \( 0.01 \times 3 = 0.03 \)
03

Compare Effective Molarities

The solution with the highest effective molarity will exhibit the highest boiling point because \( \Delta T_b \) is directly proportional to the effective molarity. - Glucose: 0.05- \( \mathrm{KNO}_{3} \): 0.02- Urea: 0.015- \( \mathrm{Na}_{2} \mathrm{SO}_{4} \): 0.03Among these, glucose has the highest effective molarity, so it will have the highest boiling point.

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Key Concepts

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

Colligative Properties
Colligative properties are fascinating because they depend solely on the number of solute particles in a solution rather than the type of particles. This means that different substances dissolved in water might affect properties like boiling points and freezing points, in similar ways, if they are present in the same concentration. Four main colligative properties are boiling point elevation, freezing point depression, vapor pressure lowering, and osmotic pressure. When a non-volatile solute is added to a solvent, it disrupts the ability of solvent molecules to leave the liquid state, thus increasing the boiling point of the solution.
  • This property is solely based on solute concentration and not on chemical identity.
  • It explains why saltwater boils at a higher temperature than pure water.
Understanding these properties helps in applications from cooking to designing chemical processes.
Van't Hoff Factor
The Van't Hoff Factor, denoted as 'i', is a crucial concept when dealing with colligative properties. It represents the number of particles into which a solute dissociates in solution. For instance, when table salt (NaCl) dissolves in water, it splits into two ions: Na鈦� and Cl鈦�, so its van't Hoff factor is 2. In contrast, a substance like glucose remains intact, so its van't Hoff factor is 1.
  • In terms of boiling point elevation, the higher the van鈥檛 Hoff factor, the more significant the elevation observed for a solution.
  • This is because more particles are present to interfere with the solvent molecules.
Calculating this factor becomes crucial when predicting the change in boiling or freezing points of solutions.
Ebullioscopic Constant
The ebullioscopic constant, represented as Kb, is specific to each solvent and indicates how much the boiling point will increase per molal concentration unit of a non-volatile solute. It is used in the equation for boiling point elevation \[ \Delta T_b = i \cdot K_b \cdot m \] where \( \Delta T_b \) is the increase in boiling point, \( i \) is the van鈥檛 Hoff factor, and \( m \) is the molality.
  • Each solvent has its unique ebullioscopic constant. For example, water has a Kb value of approximately 0.512掳C路kg/mol.
  • This constant ensures that the temperature change in the boiling point is proportional to the concentration of solute particles in the solvent.
Understanding Kb and incorporating it into calculations allows chemists to accurately predict how solutions will behave under different conditions.

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Most popular questions from this chapter

At \(80^{\circ} \mathrm{C}\), the vapour pressure of pure liquid 'A' is \(520 \mathrm{~mm} \mathrm{Hg}\) and that of pure liquid 'B' is \(1000 \mathrm{~mm}\) Hg. If a mixture solution of 'A' and 'B' boils at \(80^{\circ} \mathrm{C}\) and 1 atm pressure, the amount of 'A' in the mixture is \((1 \mathrm{~atm}=760 \mathrm{~mm} \mathrm{Hg})\). (a) 52 mol per cent (b) 34 mol per cent (c) 48 mol per cent (d) 50 mol per cent

A sugar syrup of weight \(214.2 \mathrm{~g}\) contains \(34.2 \mathrm{~g}\) of water. The molal concentration is a. \(0.55\) b. \(5.5\) c. 55 d. \(0.1\)

An aqueous solution of \(6.3 \mathrm{~g}\) oxalic acid dihydrate is made up to \(250 \mathrm{ml}\). The volume of \(0.1 \mathrm{~N} \mathrm{NaOH}\) required to completely neutralize \(10 \mathrm{ml}\) of this solution is a. \(40 \mathrm{ml}\) b. \(20 \mathrm{ml}\) c. \(10 \mathrm{ml}\) d. \(4 \mathrm{ml}\) [IIT 2001]

\(0.5\) molal aqueous solution of a weak acid \((H X)\) is \(20 \%\) ionized. If \(\mathrm{K}_{\mathrm{f}}\) for water is \(1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\), the lowering in freezing point of the solution is (a) \(-1.12 \mathrm{~K}\) (b) \(0.56 \mathrm{~K}\) (c) \(1.12 \mathrm{~K}\) (d) \(-0.56 \mathrm{~K}\)

'V' litre decinormal solution of \(\mathrm{KCl}\) is prepared. Half of the solution is converted into centinormal and added it into left decinormal solution. Then a. Number of milli equivalents of \(\mathrm{KCl}\) do not change b. Number of milli moles of \(\mathrm{KCl}\) are reduced by \(1 / 5\). c. Normality of the final solution becomes \(0.01\) \(\mathrm{N}\). d. Molarity of the final solution becomes \(0.018\) M.
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