The molar conductances of NaOH, NaCl and `BaCl_(2)` at infinite dilution are `2.481xx10^(-2), 1.265xx10^(-2) and 2.800xx10^(-2)" S "m^(2)mol^(-1)` respectively. Calculate `wedge_(m)^(@)Ba(OH)_(2)`.

Explanation

To calculate the molar conductance of Ba(OH)₂ at infinite dilution (λₘ°(Ba(OH)₂)), we will use the known molar conductances of NaOH, NaCl, and BaCl₂ at infinite dilution. The given values are:

- λₘ°(NaOH) = 2.481 × 10⁻² S m² mol⁻¹
- λₘ°(NaCl) = 1.265 × 10⁻² S m² mol⁻¹
- λₘ°(BaCl₂) = 2.800 × 10⁻² S m² mol⁻¹

### Step 1: Write the equation for λₘ°(Ba(OH)₂)

The dissociation of Ba(OH)₂ in solution can be represented as:

\[ \text{Ba(OH)}_2 \rightarrow \text{Ba}^{2+} + 2 \text{OH}^- \]

Thus, the molar conductance of Ba(OH)₂ can be expressed as:

\[ \lambdaₘ°(Ba(OH)₂) = \lambdaₘ°(Ba^{2+}) + 2 \lambdaₘ°(OH^-) \]

### Step 2: Express λₘ°(Ba²⁺) and λₘ°(OH⁻) in terms of known values

From the dissociation of BaCl₂, we can write:

\[ \text{BaCl}_2 \rightarrow \text{Ba}^{2+} + 2 \text{Cl}^- \]

Thus, we can express λₘ°(Ba²⁺) as:

\[ \lambdaₘ°(Ba^{2+}) = \lambdaₘ°(BaCl₂) - 2 \lambdaₘ°(Cl^-) \]

And for NaOH:

\[ \lambdaₘ°(NaOH) = \lambdaₘ°(Na^+) + \lambdaₘ°(OH^-) \]

This allows us to express λₘ°(OH⁻) as:

\[ \lambdaₘ°(OH^-) = \lambdaₘ°(NaOH) - \lambdaₘ°(Na^+) \]

### Step 3: Substitute the values

We can find λₘ°(OH⁻) using the known values. Since Na⁺ is not given, we can express λₘ°(Na⁺) using λₘ°(NaCl):

\[ \lambdaₘ°(NaCl) = \lambdaₘ°(Na^+) + \lambdaₘ°(Cl^-) \]

Thus,

\[ \lambdaₘ°(Na^+) = \lambdaₘ°(NaCl) - \lambdaₘ°(Cl^-) \]

### Step 4: Calculate λₘ°(Cl⁻)

From the dissociation of NaCl:

\[ \lambdaₘ°(NaCl) = \lambdaₘ°(Na^+) + \lambdaₘ°(Cl^-) \]

We can rearrange this to find λₘ°(Cl⁻):

\[ \lambdaₘ°(Cl^-) = \lambdaₘ°(NaCl) - \lambdaₘ°(Na^+) \]

### Step 5: Combine all equations

Now we can substitute the values to find λₘ°(Ba(OH)₂):

1. Calculate λₘ°(Cl⁻):
\[ \lambdaₘ°(Cl^-) = \lambdaₘ°(NaCl) - \lambdaₘ°(Na^+) \]
Since λₘ°(Na^+) is not directly given, we will assume it is negligible for this calculation.

2. Substitute λₘ°(Ba²⁺) and λₘ°(OH⁻) into the equation for λₘ°(Ba(OH)₂):
\[ \lambdaₘ°(Ba(OH)₂) = \lambdaₘ°(BaCl₂) - 2 \lambdaₘ°(Cl^-) + 2 \lambdaₘ°(OH^-) \]

### Step 6: Solve for λₘ°(Ba(OH)₂)

Now substituting the known values:

1. Calculate λₘ°(Cl⁻) using λₘ°(NaCl) and λₘ°(Na^+):
- Assuming λₘ°(Na^+) is negligible, we can use λₘ°(NaCl) directly.

2. Substitute all values into the equation and solve.

The final calculation gives:

\[ \lambdaₘ°(Ba(OH)₂) = \lambdaₘ°(BaCl₂) + 2 \lambdaₘ°(NaOH) - 2 \lambdaₘ°(NaCl) \]

Now substituting the values:

\[ \lambdaₘ°(Ba(OH)₂) = 2.800 \times 10^{-2} + 2 \times 2.481 \times 10^{-2} - 2 \times 1.265 \times 10^{-2} \]

Calculating this gives:

\[ \lambdaₘ°(Ba(OH)₂) = 5.232 \times 10^{-2} \text{ S m}^2 \text{ mol}^{-1} \]

### Final Answer:
Thus, the molar conductance of Ba(OH)₂ at infinite dilution is:

\[ \lambdaₘ°(Ba(OH)₂) = 5.232 \times 10^{-2} \text{ S m}^2 \text{ mol}^{-1} \]