0.3 g `[ML_(6)]Cl_(3)` of molar mass 267.46 g/mol is reacted with 0.125 M `AgNO_3` (aq) solution, calculate volume of `AgNO_3` required in ml.

Options

A. (a) 34.40 ml
B. (b) 26.92 ml
C. (c) 20.20 ml
D. (d) 30.10 ml

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Calculate the number of moles of the complex \([ML_6]Cl_3\)

Given:
- Mass of \([ML_6]Cl_3\) = 0.3 g
- Molar mass of \([ML_6]Cl_3\) = 267.46 g/mol

Using the formula for moles:
\[ \text{Moles} = \frac{\text{Given mass}}{\text{Molar mass}} \]
\[ \text{Moles} = \frac{0.3 \, \text{g}}{267.46 \, \text{g/mol}} = 0.001122 \, \text{mol} \]

### Step 2: Determine the moles of \(AgNO_3\) required

From the reaction, we know that:
- 1 mole of \([ML_6]Cl_3\) reacts with 3 moles of \(AgNO_3\).

Thus, the moles of \(AgNO_3\) needed can be calculated as:
\[ \text{Moles of } AgNO_3 = 3 \times \text{Moles of } [ML_6]Cl_3 \]
\[ \text{Moles of } AgNO_3 = 3 \times 0.001122 \, \text{mol} = 0.003366 \, \text{mol} \]

### Step 3: Use molarity to find the volume of \(AgNO_3\) solution needed

Given:
- Molarity of \(AgNO_3\) = 0.125 M

Using the formula for molarity:
\[ \text{Molarity} = \frac{\text{Number of moles}}{\text{Volume in liters}} \]
Rearranging the formula to find volume:
\[ \text{Volume in liters} = \frac{\text{Number of moles}}{\text{Molarity}} \]
Substituting the values:
\[ \text{Volume in liters} = \frac{0.003366 \, \text{mol}}{0.125 \, \text{mol/L}} = 0.026928 \, \text{L} \]

### Step 4: Convert volume from liters to milliliters

To convert liters to milliliters:
\[ \text{Volume in mL} = \text{Volume in liters} \times 1000 \]
\[ \text{Volume in mL} = 0.026928 \, \text{L} \times 1000 = 26.928 \, \text{mL} \]

### Final Answer
The volume of \(AgNO_3\) required is approximately **26.93 mL**.

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