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Chapter 5: Problem 37

Which one of the following solution has least vapour pressure? (a) \(0.01 \mathrm{M} \mathrm{CaCl}_{2}\) (b) \(0.01 \mathrm{M}\) glucose (c) \(0.01 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\) (d) \(0.01 \mathrm{M} \mathrm{Na}_{3} \mathrm{PO}_{4}\)

Short Answer

Expert verified
Solution (d) 0.01 M Na鈧働O鈧� has the least vapour pressure.

Step by step solution

01

Understanding Vapour Pressure

Vapour pressure is the pressure exerted by vapour in equilibrium with its liquid or solid phase. In solutions, the presence of solute decreases the vapour pressure of the solvent, and this effect is governed by the number of solute particles in the solution.
02

Analyzing Colligative Properties

Colligative properties, such as vapour pressure lowering, depend on the number of solute particles. For electrolyte solutions, it is essential to consider the number of ions that the solute dissociates into. The more ions produced, the more the vapour pressure is lowered.
03

Calculating Ion Dissociation

Calculate the number of ions formed from each solute when dissolved: - Calcium chloride ( ext{CaCl}_{2}) dissociates into 3 ions: ext{Ca}^{2+} and 2 ext{Cl}^{-}. - Glucose does not dissociate into ions; it's a nonelectrolyte. - Sodium sulfate ( ext{Na}_{2} ext{SO}_{4}) gives 3 ions: 2 ext{Na}^{+} and ext{SO}_{4}^{2-}. - Sodium phosphate ( ext{Na}_{3} ext{PO}_{4}) dissociates into 4 ions: 3 ext{Na}^{+} and ext{PO}_{4}^{3-}.
04

Determining the Solution with Least Vapour Pressure

To find the solution with the least vapour pressure, look for the one with the most ions because it will lower the vapour pressure the most. From the calculations: - ext{CaCl}_{2} has 3 ions. - ext{Na}_{2} ext{SO}_{4} has 3 ions. - ext{Na}_{3} ext{PO}_{4} has 4 ions. - Glucose has 0 ions. Hence, the solution with ext{Na}_{3} ext{PO}_{4} has the least vapour pressure.

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Key Concepts

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

Vapour Pressure Lowering
Vapour pressure lowering is a colligative property, which means it depends on the number of solute particles in a solution and not on their identity. When a non-volatile solute is added to a solvent, the solvent's vapour pressure decreases. This happens because the solute particles occupy space at the surface of the liquid, preventing some solvent molecules from evaporating into the gas phase.
This effect can be understood using Raoult's Law, which states that the vapour pressure of a solution is directly proportional to the mole fraction of the solvent. So, the more solute particles present, the lower the mole fraction of the solvent and, consequently, the lower the vapour pressure.
  • The more particles present in the solution, the more the vapour pressure is lowered.
  • This property is independent of the type of particles, focusing only on the quantity.
Ion Dissociation
Ion dissociation refers to the process where a compound separates into its component ions when it dissolves in a solvent. Ionic compounds, such as electrolytes, tend to dissociate into multiple ions. The extent of dissociation varies among different compounds and is crucial in determining their effect on colligative properties like vapour pressure lowering.
For example, when calcium chloride ( CaCl_2 ) dissolves in water, it dissociates into three ions: one calcium ion ( Ca^{2+} ) and two chloride ions ( Cl^{-} ). Similarly, sodium sulfate ( Na_2SO_4 ) separates into three ions, whereas sodium phosphate ( Na_3PO_4 ) dissociates into four ions.
  • Ions increase the total number of particles in solution, impacting colligative properties like vapour pressure lowering.
  • The more ions produced, the more significant the effect on lowering the solvent's vapour pressure.
Electrolyte Solutions
Electrolyte solutions are those in which the solute dissociates into ions when dissolved in a solvent, typically water. These solutions are important in chemistry because they exhibit enhanced colligative properties compared to nonelectrolyte solutions.
Electrolytes can further be categorized into strong and weak electrolytes based on their dissociation extent. Strong electrolytes, like sodium chloride ( NaCl ) and calcium chloride ( CaCl_2 ), almost completely dissociate into their ions in a solution, leading to a higher impact on properties like boiling point elevation and vapour pressure lowering.
  • In strong electrolyte solutions, almost all the solute molecules dissociate into ions.
  • The presence of many ions causes significant changes in the solution's properties.
Understanding how electrolyte solutions work can help you predict physical changes in solutions and solve related chemistry problems effectively.

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Most popular questions from this chapter

A protein has been isolated as a salt with the formula \(\mathrm{Na}_{20} \mathrm{P}\) (this notation means that there are \(20 \mathrm{Na}^{+}\)ions associated with a negatively charged protein \(\mathrm{P}^{20-}\) ). The osmotic pressure of a \(10.0 \mathrm{ml}\) solution containing \(0.225 \mathrm{~g}\) of the protein is \(0.257 \mathrm{~atm}\) at \(25^{\circ} \mathrm{C}\). The actual molar mass of the protein from these data is approximately. (a) \(2.6 \times 10^{5} \mathrm{~g} \mathrm{~mol}^{-1}\) (b) \(4.5 \times 10^{4} \mathrm{~g} \mathrm{~mol}^{-1}\) (c) \(4.5 \times 10^{6} \mathrm{~g} \mathrm{~mol}^{-1}\) (d) \(2.3 \times 10^{3} \mathrm{~g} \mathrm{~mol}^{-1}\)

The solution which has the lowest freezing point is (a) \(0.1 \mathrm{M}\) potassium nitrate (b) \(0.1 \mathrm{M}\) aluminium sulphate (c) \(0.1 \mathrm{M}\) potassium chloride (d) \(0.1 \mathrm{M}\) potassium sulphate

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) \(154.2\)

By dissolving \(10 \mathrm{~g}\) of a non-volatile solute in \(100 \mathrm{~g}\) of benzene, the boiling point rises by \(1^{\circ} \mathrm{C}\). The molecular mass of solute is \(\left[\mathrm{K}_{\mathrm{b}}\right.\) for benzene \(\left.=2.53 \mathrm{~K} \mathrm{~m}^{-1}\right]\) (a) \(235 \mathrm{~g}\) (b) \(352 \mathrm{~g}\) (c) \(250 \mathrm{~g}\) (d) \(253 \mathrm{~g}\)

A \(0.2\) molal aqueous solution of a weak acid (HX) is \(20 \%\) ionized. The freezing point of this solution is (Given \(\mathrm{K}_{\mathrm{r}}=1.86^{\circ} \mathrm{C} \mathrm{kg} \mathrm{mol}^{-1}\) for water \()\) (a) \(-0.45^{\circ} \mathrm{C}\) (b) \(-0.90^{\circ} \mathrm{C}\) (c) \(-0.21^{\circ} \mathrm{C}\) (d) \(-0.43^{\circ} \mathrm{C}\)
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