The molecular weight of sodium chloride determined by measuring the osmotic pressure of its aqueous solution is
1) double the theoretical value
2) same as the theoretical value
3) half the theoretical value
4) three times the theoretical value

Options

A. double the theoretical value
B. same as the theoretical value
C. half the theoretical value
D. three times the theoretical value

Answer

To determine the molecular weight of sodium chloride (NaCl) using osmotic pressure, we need to consider how NaCl behaves in solution. Here’s a step-by-step solution to the question:

### Step 1: Understanding Osmotic Pressure
Osmotic pressure (π) is a colligative property that depends on the number of solute particles in a solution. The formula for osmotic pressure is given by:
\[ \pi = CRT \]
where:
- \( \pi \) = osmotic pressure
- \( C \) = concentration of the solution (in moles per liter)
- \( R \) = universal gas constant (0.0821 L·atm/(K·mol))
- \( T \) = temperature (in Kelvin)

### Step 2: Behavior of NaCl in Solution
Sodium chloride (NaCl) is an electrolyte, which means it dissociates into ions when dissolved in water:
\[ \text{NaCl} \rightarrow \text{Na}^+ + \text{Cl}^- \]
This dissociation means that one mole of NaCl produces two moles of particles (Na⁺ and Cl⁻).

### Step 3: Adjusting the Osmotic Pressure Equation
For a solution of an electrolyte like NaCl, the osmotic pressure can be modified to account for the number of particles produced upon dissociation. The modified equation becomes:
\[ \pi = iCRT \]
where \( i \) is the van 't Hoff factor, which represents the number of particles the solute breaks into. For NaCl, \( i = 2 \) because it dissociates into two ions.

### Step 4: Relating Osmotic Pressure to Molecular Weight
The theoretical molecular weight (M) of a solute can be calculated using the formula:
\[ M = \frac{RT}{\pi} \]
However, since NaCl dissociates into two particles, the effective osmotic pressure will be:
\[ \pi = 2CRT \]
Thus, the molecular weight determined from osmotic pressure will be:
\[ M = \frac{RT}{\pi/2} = \frac{2RT}{\pi} \]

### Step 5: Conclusion
From the above analysis, we can conclude that the molecular weight of sodium chloride determined by measuring the osmotic pressure of its aqueous solution is:
- **Double the theoretical value**.

### Final Answer
The correct option is:
**1) double the theoretical value**.

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