Problem 172 - A Complete Resource Book in Chemistry for JEE Main

Question

Consider separate solution of \(0.500 \mathrm{M} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{aq})\), \(0.100 \mathrm{M} \mathrm{Mg}_{3}\left(\mathrm{PO}_{4}\right)_{2}(\mathrm{aq}), 0.250 \mathrm{M} \mathrm{KBr}(\mathrm{aq})\) and \(0.125\) M \(\mathrm{Na}_{3} \mathrm{PO}_{4}\) (aq) at \(25^{\circ} \mathrm{C}\). Which statement is true about these solutions, assuming all salts to be strong electrolytes? [2014] (a) \(0.125 \mathrm{M} \mathrm{Na}_{3} \mathrm{PO}_{4}\) (aq) has the highest osmotic pressure. (b) \(0.500 \mathrm{M} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) (aq) has the highest osmotic pressure. (c) They all have the same osmotic pressure. (d) \(0.100 \mathrm{M} \mathrm{Mg}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) (aq) has the highest osmotic pressure.

Step-by-Step Solution

Answer

(c) They all have the same osmotic pressure.

Step 1: Understanding Osmotic Pressure

Osmotic pressure () is calculated using the formula \( 2 = iMRT \), where \( 2 \) is the osmotic pressure, \( i \) is the van't Hoff factor, \( M \) is the molarity, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin. For strong electrolytes, dissociation into ions increases \( i \), thus affecting osmotic pressure.

Step 2: Determine Van't Hoff Factors

For each solution, determine the van't Hoff factor \( i \): - \( 4 C_{2}H_{5}OH\) is a non-electrolyte, so \( i = 1 \). - \( 5 KBr \) dissociates into \( 2 \) ions: \( K^+ \) and \( Br^- \), so \( i = 2 \). - \( 6 Na_{3}PO_{4} \) dissociates into \( 4 \) ions: \( 3Na^+ \) and \( PO_4^{3-} \), so \( i = 4 \). - \( 7 Mg_{3}(PO_{4})_{2} \) dissociates into \( 5 \) ions: \( 3Mg^{2+} \) and \( 2PO_4^{3-} \), so \( i = 5 \).

Step 3: Calculate Osmotic Pressure for Each Solution

Use the formula \( 8 = iMRT \) to calculate osmotic pressure for each: - For \( C_{2}H_{5}OH\), \( 9_{1} = 1 \times 0.500M \times RT \) - For \( KBr \), \( 9_{2} = 2 \times 0.250M \times RT \) - For \( Na_{3}PO_{4} \), \( 9_{3} = 4 \times 0.125M \times RT \) - For \( Mg_{3}(PO_{4})_{2} \), \( 9_{4} = 5 \times 0.100M \times RT \)

Step 4: Compare Osmotic Pressures

Simplify and compare osmotic pressures: - \( 9_{1} = 0.500RT \) - \( 9_{2} = 0.500RT \) - \( 9_{3} = 0.500RT \) - \( 9_{4} = 0.500RT \) All solutions yield the same numerical value for osmotic pressure, \( 0.500RT \), meaning they all have the same osmotic pressure.

Step 5: Choose the Correct Statement

Based on the calculations, option (c) is true: they all have the same osmotic pressure. None of the individual solutions has a higher or lower osmotic pressure than the others.

Key Concepts

Van't Hoff Factor

The Van’t Hoff factor, denoted as \( i \), plays a crucial role in the determination of colligative properties, such as osmotic pressure, boiling point elevation, and freezing point depression. This factor represents the number of particles into which a solute dissociates in solution.
In our context, the Van’t Hoff factor impacts how the osmotic pressure of a solution is calculated using the formula \( \pi = iMRT \). **Non-electrolytes** such as \( C_2H_5OH \) do not dissociate in solution, leading to \( i = 1 \). **Electrolytes**, on the other hand, dissociate into multiple ions, influencing \( i \). For instance,

• Potassium bromide (\( KBr \)) dissociates into two ions, \( K^+ \) and \( Br^- \), with \( i = 2 \).
• Sodium phosphate (\( Na_3PO_4 \)) breaks down into four ions, resulting in \( i = 4 \).
• Magnesium phosphate (\( Mg_3(PO_4)_2 \)) dissociates into five ions, leading to \( i = 5 \).

The more ions a compound dissociates into, the higher the osmotic pressure exerted by the solution for a given molarity. This means understanding the nature of the solute is essential when calculating and comparing osmotic pressures.

Electrolytes and Non-electrolytes

Electrolytes and non-electrolytes are classifications of substances that either dissociate into ions when dissolved in water or do not, respectively. This classification is fundamental in understanding how substances behave in solution.
Electrolytes are compounds that dissociate into ions, allowing them to conduct electricity. Strong electrolytes, like \( KBr \), \( Na_3PO_4 \), and \( Mg_3(PO_4)_2 \), dissociate completely into their respective ions. This complete dissociation significantly impacts properties like osmotic pressure, as it directly affects the Van’t Hoff factor \( i \).
**Non-electrolytes**, however, do not dissociate into ions. An example is \( C_2H_5OH \) (ethanol), which remains intact in solution. It lacks the ability to conduct electricity and retains a Van’t Hoff factor \( i = 1 \).
The distinction between these two types of compounds determines their behavior in solutions, influencing various properties dependent on solute particle concentration such as osmotic pressure, boiling point, and freezing point changes.

Molarity in Solution Chemistry

Molarity is a measure of the concentration of a solute in a solution, expressed as the number of moles of solute per liter of solution. It is a crucial parameter in solution chemistry because it allows for the calculation of various properties, including osmotic pressure.
In our exercise, we are given the molarities of different solutions:

• \( C_2H_5OH \): 0.500 M
• \( KBr \): 0.250 M
• \( Na_3PO_4 \): 0.125 M
• \( Mg_3(PO_4)_2 \): 0.100 M

Using these molarity values, we can plug them into the osmotic pressure formula \( \pi = iMRT \). Here, \( M \) represents the molarity of each solution, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin. Molarity helps us determine the number of moles of solute in a given volume of solution, which is necessary for calculating properties like osmotic pressure.
Understanding molarity, alongside the dissociation characteristics represented by the Van’t Hoff factor, helps us accurately determine the strength and behavior of solutions for various chemical analyses.