`Al(SO)_(4))_(3)` solution of 1 molal concentration is present in 1 litre solution of density 2.684 g/cc. How many moles `BaSO_(4)` would be precipated on adding excess `BaCl_(2)` in it?

Options

A. 2 moles
B. 3 moles
C. 6 moles
D. 12 moles

Answer

To solve the problem, we need to determine how many moles of barium sulfate (BaSO₄) will precipitate when excess barium chloride (BaCl₂) is added to a solution of aluminum sulfate (Al₂(SO₄)₃) with a given molality. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Reaction
The reaction between aluminum sulfate and barium chloride can be written as:
\[ \text{Al}_2(\text{SO}_4)_3 + 3 \text{BaCl}_2 \rightarrow 3 \text{BaSO}_4 \downarrow + 2 \text{AlCl}_3 \]

From this balanced equation, we can see that 1 mole of aluminum sulfate produces 3 moles of barium sulfate.

### Step 2: Calculate the Mass of the Solution
Given the density of the solution is 2.684 g/cm³ and the volume is 1 liter (1000 cm³):
\[ \text{Mass of solution} = \text{Density} \times \text{Volume} \]
\[ \text{Mass of solution} = 2.684 \, \text{g/cm}^3 \times 1000 \, \text{cm}^3 = 2684 \, \text{g} \]

### Step 3: Calculate the Moles of Aluminum Sulfate
The molality (m) of the solution is given as 1 molal, which means:
\[ m = \frac{\text{number of moles of solute}}{\text{mass of solvent in kg}} \]

Let \( x \) be the number of moles of aluminum sulfate. The mass of the solvent can be calculated as:
\[ \text{Mass of solvent} = \text{Mass of solution} - \text{Mass of solute} \]

The molar mass of aluminum sulfate (Al₂(SO₄)₃) is:
\[ \text{Molar mass of Al}_2(\text{SO}_4)_3 = 2(27) + 3(32 + 4(16)) = 342 \, \text{g/mol} \]

Thus, the mass of the solute (aluminum sulfate) is:
\[ \text{Mass of solute} = 342x \, \text{g} \]

Now, the mass of solvent in grams is:
\[ \text{Mass of solvent} = 2684 - 342x \]

To convert this to kilograms:
\[ \text{Mass of solvent in kg} = \frac{2684 - 342x}{1000} \]

Now we can set up the equation for molality:
\[ 1 = \frac{x}{\frac{2684 - 342x}{1000}} \]

### Step 4: Solve for \( x \)
Cross-multiplying gives:
\[ x \cdot 1000 = 2684 - 342x \]
\[ 1000x + 342x = 2684 \]
\[ 1342x = 2684 \]
\[ x = \frac{2684}{1342} = 2 \]

So, there are 2 moles of aluminum sulfate in the solution.

### Step 5: Calculate Moles of Barium Sulfate Precipitated
From the balanced equation, we know that 1 mole of aluminum sulfate produces 3 moles of barium sulfate. Therefore, for 2 moles of aluminum sulfate:
\[ \text{Moles of BaSO}_4 = 2 \times 3 = 6 \]

### Conclusion
The number of moles of barium sulfate that would be precipitated is **6 moles**.

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