How many grams of barium chloride `(BaCl_(2))` are needed to prepare `100cm^(3)` of 0.250 mL of a 0.50 M `BaCl_(2)` solution?

Explanation

To determine how many grams of barium chloride (BaCl₂) are needed to prepare 100 cm³ of a 0.50 M BaCl₂ solution, we can follow these steps:

### Step 1: Convert the volume from cm³ to L
Since molarity (M) is defined as moles of solute per liter of solution, we need to convert the volume from cm³ to liters.

\[ 100 \, \text{cm}^3 = 100 \, \text{mL} = 0.100 \, \text{L} \]

### Step 2: Calculate the number of moles of BaCl₂ required
Using the formula for molarity:

\[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in L}} \]

Rearranging the formula to find the number of moles:

\[ \text{moles of BaCl₂} = \text{Molarity} \times \text{Volume in L} \]

Substituting the values:

\[ \text{moles of BaCl₂} = 0.50 \, \text{M} \times 0.100 \, \text{L} = 0.050 \, \text{moles} \]

### Step 3: Calculate the mass of BaCl₂ needed
To find the mass, we use the formula:

\[ \text{mass} = \text{moles} \times \text{molar mass} \]

The molar mass of BaCl₂ can be calculated as follows:
- Barium (Ba) = 137.33 g/mol
- Chlorine (Cl) = 35.45 g/mol (and there are 2 Cl atoms)

\[ \text{Molar mass of BaCl₂} = 137.33 + (2 \times 35.45) = 137.33 + 70.90 = 208.23 \, \text{g/mol} \]

Now, substituting the values into the mass formula:

\[ \text{mass of BaCl₂} = 0.050 \, \text{moles} \times 208.23 \, \text{g/mol} = 10.4115 \, \text{g} \]

Rounding to two decimal places, we find:

\[ \text{mass of BaCl₂} \approx 10.41 \, \text{g} \]

### Final Answer:
To prepare 100 cm³ of a 0.50 M BaCl₂ solution, approximately **10.41 grams** of barium chloride are needed.

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