A 0.010 g sample of`Cr(NH_(3))_(4)(SO_(4))Cl` is dissolved in 25.0 nL of water and the osmotic pressure of the solution is 59.1 torr at `25^(@)C`. How many moles of ions are produced per mole of compound?

Options

A. 1
B. 4
C. 2
D. 3

Answer

To solve the problem step-by-step, we will follow the method outlined in the video transcript.

### Step 1: Understand the given data
We have:
- Mass of the compound \( = 0.010 \, \text{g} \)
- Volume of water \( = 25.0 \, \text{mL} \)
- Osmotic pressure \( \pi = 59.1 \, \text{torr} \)
- Temperature \( T = 25^\circ C = 298 \, \text{K} \)

### Step 2: Convert osmotic pressure from torr to atm
To convert torr to atm, we use the conversion factor:
\[ 1 \, \text{torr} = 0.00132 \, \text{atm} \]
Thus,
\[ \pi = 59.1 \, \text{torr} \times 0.00132 \, \text{atm/torr} = 0.078 \, \text{atm} \]

### Step 3: Calculate the molecular weight of the compound
The molecular formula is \( \text{Cr(NH}_3\text{)}_4\text{(SO}_4)\text{Cl} \). The molecular weight can be calculated as follows:
- Chromium (Cr) = 52.0 g/mol
- Nitrogen (N) = 14.0 g/mol
- Hydrogen (H) = 1.0 g/mol
- Sulfur (S) = 32.1 g/mol
- Oxygen (O) = 16.0 g/mol
- Chlorine (Cl) = 35.5 g/mol

Calculating the total:
\[ \text{Molecular weight} = 52.0 + (4 \times 14.0) + (4 \times 16.0) + 32.1 + 35.5 = 251.6 \, \text{g/mol} \]

### Step 4: Calculate the concentration of the solution
Using the formula for concentration \( C \):
\[ C = \frac{\text{mass}}{\text{molecular weight} \times \text{volume in L}} \]
Convert volume from mL to L:
\[ 25.0 \, \text{mL} = 0.025 \, \text{L} \]
Now calculate the concentration:
\[ C = \frac{0.010 \, \text{g}}{251.6 \, \text{g/mol} \times 0.025 \, \text{L}} = \frac{0.010}{6.29} \approx 0.00159 \, \text{mol/L} \]

### Step 5: Use the osmotic pressure formula
The formula for osmotic pressure is:
\[ \pi = CRTI \]
Where:
- \( R = 0.0821 \, \text{L atm/(mol K)} \)
- \( T = 298 \, \text{K} \)

Rearranging for \( I \):
\[ I = \frac{\pi}{CRT} \]
Substituting the values:
\[ I = \frac{0.078}{0.00159 \times 0.0821 \times 298} \]

### Step 6: Calculate \( I \)
Calculating the denominator:
\[ 0.00159 \times 0.0821 \times 298 \approx 0.0387 \]
Now calculate \( I \):
\[ I = \frac{0.078}{0.0387} \approx 2.02 \]

### Step 7: Conclusion
The Van't Hoff factor \( I \) indicates that the compound dissociates into approximately 2 moles of ions per mole of the compound.

### Final Answer
The number of moles of ions produced per mole of the compound is approximately **2**.

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