When any solution passess through a cation exchange resin that is in acidic form, H ion of the resin is replaced by cations of the solution. A solution containing 0.319 g of an isomer with molecular formula `CrCl_(3).6H_(2)O` is passed through a cation exchange resin in acidic form. The eluted solution requires 19 `cm^(3)` fo 0.125 N NaOH. The isomer is

Options

A. triaquatichloro chromium (III) chloride trihydrate
B. hexaaqua chromium (III) chloride
C. pentaaquamonochloro chromium (III) chloride monohydrate
D. tetraaquadichloro chromium (III) chloride dihydrate.

Answer

To solve the problem step by step, we will follow the reasoning laid out in the video transcript while providing a detailed explanation for each step.

### Step 1: Calculate the number of moles of the complex
Given:
- Mass of the isomer = 0.319 g
- Molecular formula = CrCl₃·6H₂O

First, we need to calculate the molar mass of CrCl₃·6H₂O.

1. **Calculate the molar mass:**
- Molar mass of Cr = 52.00 g/mol
- Molar mass of Cl = 35.45 g/mol (3 Cl atoms = 3 × 35.45 = 106.35 g/mol)
- Molar mass of H₂O = 18.02 g/mol (6 H₂O = 6 × 18.02 = 108.12 g/mol)

Total molar mass = 52.00 + 106.35 + 108.12 = 266.47 g/mol

2. **Calculate the number of moles:**
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{0.319 \text{ g}}{266.47 \text{ g/mol}} \approx 0.0012 \text{ moles} \]

### Step 2: Calculate the number of millimoles of NaOH used
Given:
- Volume of NaOH solution = 19 cm³ = 19 mL
- Normality of NaOH = 0.125 N

1. **Calculate the millimoles of NaOH:**
\[ \text{Millimoles of NaOH} = \text{Normality} \times \text{Volume (in L)} = 0.125 \text{ N} \times 0.019 \text{ L} = 0.002375 \text{ moles} = 2.375 \text{ millimoles} \]

### Step 3: Relate the moles of complex to the moles of NaOH
From the problem, we know that each mole of the complex will release 3 moles of H⁺ ions (due to the presence of 3 ionizable chloride ions).

1. **Set up the relationship:**
\[ \text{If 1 mole of complex gives 3 moles of H⁺, then:} \]
\[ \text{Moles of H⁺} = 3 \times \text{moles of complex} \]

2. **Calculate moles of H⁺ from NaOH:**
Since NaOH is a strong base, it will neutralize the H⁺ ions:
\[ \text{Moles of H⁺} = \text{Millimoles of NaOH} = 2.375 \text{ millimoles} \]

3. **Calculate moles of complex:**
\[ \text{Moles of complex} = \frac{\text{Moles of H⁺}}{3} = \frac{2.375 \text{ millimoles}}{3} \approx 0.7917 \text{ millimoles} \]

### Step 4: Determine the isomer
From the calculations, we have established that there are 3 ionizable chloride ions in the complex. Thus, the formula for the complex can be written as:
\[ \text{[Cr(H₂O)₆]}^{3+} \text{Cl}₃ \]
This indicates that the isomer is **Hexa-aqua chromium(III) chloride**.

### Final Answer
The isomer is **Hexa-aqua chromium(III) chloride (CrCl₃·6H₂O)**.

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