Question

The following solutions are prepared by dissolving the requisite amount of solute in water to obtain the desired concentrations. Rank the solutions according to their respective osmotic pressures in decreasing order assuming the complete dissociation of ionic compounds. Rank from highest to lowest osmotic pressure. To rank items as equivalent, overlap them. 1M C6H12O6, 1M MgCl2, 2M CH3OH, 1M NaCl

Answer 1

Answer:

C₆H₁₂O₆ < NaCl ≈ CH₃OH < MgCl₂

Step-by-step explanation:

The formula for osmotic pressure (Π) is

Π = icRT

If T is constant,

Π = kic

C₆H₁₂O₆:

Π = k × 1 × 1 = k

MgCl₂:

Π = k × 3 × 1 = 3k

CH₃OH:

Π = k × 1 × 2 = 2k

NaCl:

Π = k  × 2 × 1 = 2k

The order of osmotic pressures is  

C₆H₁₂O₆ < NaCl ≈ CH₃OH < MgCl₂

Answer 2

MgCl₂ > NaCl = CH₃OH > C₆H₁₂O₆ or MgCl₂ > CH₃OH = NaCl > C₆H₁₂O₆.

Further explanation

Given:

• 1 M C₆H₁₂O₆
• 1 M MgCl₂
• 2 M CH₃OH
• 1 M NaCl

Question:

Rank the solutions according to their respective osmotic pressures in decreasing order assuming the complete dissociation of ionic compounds.

The Process:

The osmotic pressure of a nonelectrolyte solution is calculated as follows:

$\boxed{ \ \pi = MRT \ }$

The osmotic pressure of an electrolyte solution is calculated as follows:

$\boxed{ \ \pi = MRTi \ }$  

The van't Hoff factor is i = 1 + (n - 1)α, with  

• n = number of ions, and  
• α = degree of dissociation.

In our problem, assuming the complete dissociation of ionic compounds results in α = 100% and i = n.

From the information above, each type of solution can be prepared as follows:

• C₆H₁₂O₆ (glucose) and CH₃OH (methanol) are non-electrolyte solutions.
• MgCl₂ and NaCl are strong electrolyte solutions.
• $\boxed{ \ MgCl_2 \rightarrow Mg^{2+} + 2Cl^- \ } \rightarrow \boxed{ \ i = n = 3 \ ions \ }$
• $\boxed{ \ NaCl \rightarrow Na^{+} + Cl^- \ } \rightarrow \boxed{ \ i = n = 2 \ ions \ }$

Now we compare the amount of osmotic pressure from each solution.

• 1 M C₆H₁₂O₆ ⇒ $\boxed{ \ \pi = 1 \times RT \ } \rightarrow \boxed{ \ \pi = RT \ }$ in atm.
• 1 M MgCl₂ ⇒ $\boxed{ \ \pi = 1 \times RT \times 3 \ } \rightarrow \boxed{ \ \pi = 3RT \ }$ in atm.
• 2 M CH₃OH ⇒ $\boxed{ \ \pi = 2 \times RT \ } \rightarrow \boxed{ \ \pi = 2RT \ }$ in atm.
• 1 M NaCl ⇒ $\boxed{ \ \pi = 1 \times RT \times 2 \ } \rightarrow \boxed{ \ \pi = 2RT \ }$ in atm.

From the above results, it can be observed that 2 M CH₃OH and 1 M NaCl have the same osmotic pressure.

Thus, the rank of the solutions according to their respective osmotic pressures in decreasing orders is MgCl₂ > NaCl = CH₃OH > C₆H₁₂O₆ or MgCl₂ > CH₃OH = NaCl > C₆H₁₂O₆.

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Notes:

• Colligative properties are physical properties of a solution that depend on the amount of solute expressed as concentration. One of the colligative properties is osmotic pressure (π).
• Osmosis is a spontaneous process in which a solvent molecule passes through a semipermeable membrane from a dilute solution (lower solute concentration) to a more concentrated solution (higher solute concentration).
• The pressure that causes the osmosis process to stop is considered osmotic pressure. We can also observe osmotic pressure as the external pressure needed to prevent the osmosis process. The required external pressure is the same as the osmotic pressure of the solution.

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