The osmotic pressure of a dilute solution of an ionic compound XY in water is four times that of a s

Q: Why do we use the van't Hoff factor here?

Because ionic compounds dissociate in water, increasing the total number of solute particles. Colligative properties like osmotic pressure depend strictly on the number of particles, not their identity. The van't Hoff factor ($i$) accounts for this multiplication.

Q: How did we get $i=3$ for $\text{BaCl}_2$?

When $\text{BaCl}_2$ dissolves completely, one molecule breaks apart into one $\text{Ba}^{2+}$ ion and two $\text{Cl}^-$ ions. $1 + 2 = 3$ total particles. Therefore, $i = 3$.

Q: Why did $R$ and $T$ cancel out in the equation?

Since both solutions are aqueous and no temperature difference is mentioned, we assume they are at the same room temperature $T$. $R$ is the universal gas constant. Since they appear on both sides of the equation as multipliers, they cancel out.

Q: What if the compound $XY$ was a weak electrolyte?

If it were a weak electrolyte, it wouldn't dissociate 100%. We would need its degree of dissociation ($\alpha$) to calculate the exact van't Hoff factor using the formula $i = 1 + (n-1)\alpha$.

Q: Can osmotic pressure be used to find molar mass?

Yes! In fact, measuring osmotic pressure is one of the most accurate methods for determining the molar mass of large macromolecules like proteins and polymers, because even a small concentration produces a measurable pressure.

Question

The osmotic pressure of a dilute solution of an ionic compound $XY$ in water is four times that of a solution of $0.01	ext{ M } 	ext{BaCl}_2$ in water. Assuming complete dissociation of the given ionic compounds in water, the concentration of $XY$ (in $	ext{mol L}^{-1}$) in solution is

Options:

(A) $4 	imes 10^{-2}$

(B) $16 	imes 10^{-4}$

(C) $4 	imes 10^{-4}$

(D) $6 	imes 10^{-2}$

Sigma Percentile · Accepted Answer

### The Invisible Force: Osmotic Pressure

Imagine two solutions separated by a semipermeable membrane. Nature always strives for balance, and solvent molecules will rush from the less concentrated side to the more concentrated side to equalize things. The pressure required to stop this flow is called **osmotic pressure** ($\pi$).

Because osmotic pressure is a **colligative property**, it doesn't care *what* the solute particles are; it only cares *how many* there are. The governing equation is beautifully simple:

$$\pi = iCRT$$

Here, $C$ is the molarity, $R$ is the universal gas constant, $T$ is the temperature, and $i$ is the all-important **van't Hoff factor**.

### The van't Hoff Factor: Counting the Pieces

When ionic compounds dissolve in water, they don't just sit there; they break apart. This means one mole of a compound can produce multiple moles of actual particles in the solution.

For our unknown compound $XY$, assuming complete dissociation, it splits into two ions:

$$XY \longrightarrow X^+ + Y^-$$

This gives us a van't Hoff factor of $i_{XY} = 2$.

For Barium Chloride ($	ext{BaCl}_2$), the split is even more dramatic. One molecule yields one Barium ion and two Chloride ions:

$$	ext{BaCl}_2 \longrightarrow 	ext{Ba}^{2+} + 2	ext{Cl}^-$$

This gives us a van't Hoff factor of $i_{	ext{BaCl}_2} = 3$.

### Setting Up the Equation

The problem hands us a golden key: the osmotic pressure of the $XY$ solution is exactly four times that of the $	ext{BaCl}_2$ solution. Let's translate this English sentence into mathematics:

$$\pi_{XY} = 4 \pi_{	ext{BaCl}_2}$$

Now, we substitute our osmotic pressure formula into both sides:

$$i_{XY} 	imes C_{XY} 	imes RT = 4 	imes (i_{	ext{BaCl}_2} 	imes C_{	ext{BaCl}_2} 	imes RT)$$

Since both solutions are in the same environment, the temperature $T$ is constant, and $R$ is always constant. They gracefully cancel out from both sides, leaving us with a pure relationship between particles and concentrations:

$$i_{XY} 	imes C_{XY} = 4 	imes i_{	ext{BaCl}_2} 	imes C_{	ext{BaCl}_2}$$

### The Final Calculation

Now, it's just a matter of plugging in the numbers we've gathered. We know $i_{XY} = 2$, $i_{	ext{BaCl}_2} = 3$, and the concentration of Barium Chloride is $0.01	ext{ M}$.

$$2 	imes C_{XY} = 4 	imes 3 	imes 0.01$$

$$2 	imes C_{XY} = 12 	imes 0.01$$

$$2 	imes C_{XY} = 0.12$$

Dividing both sides by 2, we find the hidden concentration:

$$C_{XY} = 0.06 	ext{ mol L}^{-1}$$

To match the options provided in the question, we convert this to scientific notation:

**$$C_{XY} = 6 	imes 10^{-2} 	ext{ mol L}^{-1}$$**

And there we have it! By understanding how ionic compounds multiply their presence in water, we easily cracked the code.

Select Answer:

Visualized Solution

$Visualizing the Solutions$

$Osmotic Pressure Equation$

$Relating the Pressures$

$Calculating van’t Hoff Factors (i)$

$Substituting Values$

$Final Concentration$

$What if dissociation wasn’t 100%?$

The Sigma Insight: Abnormal Molecular Mass and Distribution Law

The Invisible Force

The van't Hoff Factor

Setting Up the Equation

The Final Calculation

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Solutions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Osmotic Pressure Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Relating the Pressures

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating van’t Hoff Factors (i)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Concentration

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } What if dissociation wasn’t 100%?