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At the same temperature, each of the following solution has the same osmotic pressure except (a) \(0.140 \mathrm{M}\) -sucrose (b) \(0.07 \mathrm{M}-\mathrm{KCl}\) (c) \(0.070 \mathrm{M}-\mathrm{Ca}\left(\mathrm{NO}_{2}\right)_{2}\) (d) \(0.140 \mathrm{M}\) -urea

Short Answer

Expert verified
The solution with different osmotic pressure is (c) \(0.070 \mathrm{M}-\mathrm{Ca}\left(\mathrm{NO}_2\right)_2\).

Step by step solution

01

Understand the Concept of Osmotic Pressure

Osmotic pressure is a property of solutions that is dependent on the solute concentration and the number of particles in solution (ionization factor), not on the type of particles. It is given by the Van't Hoff equation for dilute solutions: \( \text{Osmotic Pressure} = i \times M \times R \times T \), where \(i\) is the van't Hoff factor (number of particles the solute dissociates into), \(M\) is the molarity, \(R\) is the gas constant, and \(T\) is the temperature.
02

Calculate the Van't Hoff Factor for Each Solute

For sucrose and urea, the van't Hoff factor \(i\) is 1 since they do not dissociate. For \(KCl\), it dissociates into \(K^+\) and \(Cl^-\), therefore \(i = 2\). For \(Ca(NO_2)_2\), it dissociates into \(Ca^{2+}\) and 2 \(NO_2^-\), so \(i = 3\).
03

Identify Which Solution Has a Different Osmotic Pressure

Calculate the products of the van't Hoff factor \(i\) and the molarity \(M\) for each solution. For sucrose and urea, the product is \(1 \times 0.140 = 0.140\). For \(KCl\), the product is \(2 \times 0.070 = 0.140\). However, for \(Ca(NO_2)_2\), the product is \(3 \times 0.070 = 0.210\), which is different from the rest. Hence, \(Ca(NO_2)_2\) has a different osmotic pressure.

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Key Concepts

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

Van't Hoff factor
The Van't Hoff factor, represented by the symbol 'i', is pivotal in understanding how solutes affect the properties of solutions. It indicates the number of particles a solute can produce when it dissolves in a solution. For substances that don't dissociate in solution, like sucrose and urea, the Van't Hoff factor is 1 because each molecule stays intact as a single particle. In contrast, salts like KCl and Ca(NO2)2 break apart into ions. KCl dissociates into two ions, K+ and Cl-, resulting in a Van't Hoff factor of 2. Similarly, Ca(NO2)2 separates into three ions (one Ca2+ and two NO2-), giving it a Van't Hoff factor of 3.

Understanding the van't Hoff factor is crucial for accurately determining a solution's osmotic pressure, boiling point elevation, and freezing point depression. Osmotic pressure, for instance, relies not just on the concentration of a solute but also on how many particles it turns into upon dissolution. Hence, a 1 M solution of a compound that dissociates into three ions (i=3) will exert higher osmotic pressure than a 1 M solution of a non-dissociating compound (i=1).
Molarity Calculation
Molarity calculation is a fundamental concept in chemistry, representing the concentration of a solution. Defined as moles of solute per liter of solution, it's expressed in units of moles per liter (M). To calculate molarity, you divide the number of moles of solute by the volume of the solution in liters.

For example, if we dissolve one mole of sucrose into one liter of water, the molarity of the sucrose solution is 1 M. Adjusting the volume affects the concentration proportionally; this concept is applied when diluting concentrated solutions or concentrating dilute solutions.

Being adept at molarity calculations is essential for determining how a solution will behave under different conditions and is especially required when using the Van't Hoff equation to calculate osmotic pressure.
Solution Properties
Solution properties like osmotic pressure, boiling point elevation, and freezing point depression are colligative, meaning they depend on the number of solute particles in a solvent, not the solute's identity. These properties provide insights into the behavior of solutions in various conditions.

Osmotic pressure, for instance, is the pressure required to stop the osmotic flow of water into a solution through a semipermeable membrane. It is directly proportional to the solute concentration, determined by both the molarity and the number of particles formed when the solute dissolves, thus bringing the Van't Hoff factor into play again.

Using these properties, chemists can discern many aspects of solutions, such as determining molecular weights, purity levels, and the solute's impact on the solution's behavior under various temperature conditions. An awareness of solution properties is invaluable for anyone studying chemistry, as it links fundamental concepts of molarity and the Van’t Hoff factor to real-world applications.

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Most popular questions from this chapter

The boiling point and freezing point of a solvent 'A' are \(90.0^{\circ} \mathrm{C}\) and \(3.5^{\circ} \mathrm{C}\), respectively. \(K_{\mathrm{f}}\) and \(K_{\mathrm{b}}\) values of the solvent are \(17.5\) and \(5.0 \mathrm{~K}-\mathrm{kg} / \mathrm{mol}\), respectively. What is the boiling point of a solution of 'B' (non-volatile, nonelectrolyte solute) in 'A', if the solution freezes at \(2.8^{\circ} \mathrm{C} ?\) (a) \(90.0^{\circ} \mathrm{C}\) (b) \(89.8^{\circ} \mathrm{C}\) (c) \(90.2^{\circ} \mathrm{C}\) (d) \(90.7^{\circ} \mathrm{C}\)

A liquid mixture of ' \(\mathrm{A}\) ' and 'B' (assume ideal solution) is placed in a cylinder and piston arrangement. The piston is slowly pulled out isothermally so that the volume of liquid decreases and that of the vapour increases. At the instant when the quantity of the liquid still remaining is negligibly small, the mole fraction of 'A' in the vapour is \(0.4\). If \(P_{\mathrm{A}}^{\circ}=0.4 \mathrm{~atm}\), \(P_{\mathrm{B}}^{\circ}=1.2 \mathrm{~atm}\) at this temperature, the total pressure at which the liquid has almost evaporated, is (a) \(0.667\) atm (b) \(1.5 \mathrm{~atm}\) (c) \(0.8 \mathrm{~atm}\) (d) \(0.545 \mathrm{~atm}\)

\(\mathrm{PtCl}_{4} 6 \mathrm{H}_{2} \mathrm{O}\) can exist as a hydrated complex. \(1.0\) molal aqueous solution has depression in freezing point of \(3.72^{\circ} \mathrm{C}\). Assume \(100 \%\) ionization and \(K_{\ell}\) of water \(=1.86^{\circ} \mathrm{C} \mathrm{mol}^{-1} \mathrm{~kg}\). The complex is (a) \(\left[\mathrm{Pt}\left(\mathrm{H}_{2} \mathrm{O}\right)\right] \mathrm{Cl}_{4}\) (b) \(\left[\mathrm{Pt}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{Cl}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O}\) (c) \(\left[\mathrm{Pt}\left(\mathrm{H}_{2} \mathrm{O}\right)_{3} \mathrm{Cl}_{3}\right] \mathrm{Cl} \cdot 3 \mathrm{H}_{2} \mathrm{O}\) (d) \(\left[\mathrm{Pt}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2} \mathrm{Cl}_{4}\right] \cdot 4 \mathrm{H}_{2} \mathrm{O}\)

The degree of dissociation \((\alpha)\) of a weak electrolyte, \(A_{x} B_{y}\), is related to Van't Hoff factor (i) by the expression (a) \(\alpha=\frac{i-1}{x+y-1}\) (b) \(\alpha=\frac{i-1}{x+y+1}\) (c) \(\alpha=\frac{x+y-1}{i-1}\) (d) \(\alpha=\frac{x+y+1}{i-1}\)

An ideal solution was obtained by mixing methanol and ethanol. If the partial vapour pressures of methanol and ethanol are \(2.8\) and \(4.2 \mathrm{kPa}\) respectively, the mole fraction of methanol in the vapour at equilibrium is (a) \(0.67\) (b) \(0.4\) (c) \(0.6\) (d) \(0.33\)

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