Ammonia in `0.224 g` of a compound `Zn(NH_(3))_(x)Cl_(2)` is neutralised by `30.7 mL` of `0.20 M HCl`. The value of `x` in the formula is

Options

A. 4
B. 5
C. 6
D. 8

Explanation

To find the value of \( x \) in the compound \( Zn(NH_3)_xCl_2 \), we can follow these steps:

### Step 1: Calculate the moles of HCl used
Given:
- Volume of HCl = \( 30.7 \, \text{mL} = 0.0307 \, \text{L} \)
- Molarity of HCl = \( 0.20 \, \text{M} \)

Using the formula for moles:
\[ \text{Moles of HCl} = \text{Molarity} \times \text{Volume (L)} \]
\[ \text{Moles of HCl} = 0.20 \, \text{mol/L} \times 0.0307 \, \text{L} = 0.00614 \, \text{mol} \]

### Step 2: Relate moles of HCl to moles of NH3
Since each mole of HCl neutralizes one mole of NH3, the moles of NH3 will also be \( 0.00614 \, \text{mol} \).

### Step 3: Calculate the molecular weight of the compound
The molecular weight of \( Zn(NH_3)_xCl_2 \) can be expressed as:
\[ \text{Molar mass} = \text{Molar mass of Zn} + x \times \text{Molar mass of NH}_3 + 2 \times \text{Molar mass of Cl} \]
Where:
- Molar mass of Zn = \( 65.30 \, \text{g/mol} \)
- Molar mass of NH3 = \( 14 + 3 \times 1 = 17 \, \text{g/mol} \)
- Molar mass of Cl = \( 35.5 \, \text{g/mol} \)

Thus,
\[ \text{Molar mass} = 65.30 + 17x + 2 \times 35.5 \]
\[ = 65.30 + 17x + 71 = 136.30 + 17x \]

### Step 4: Set up the equation using the mass of the compound
Given the mass of the compound is \( 0.224 \, \text{g} \), we can express the number of moles of the compound as:
\[ \text{Moles of compound} = \frac{0.224 \, \text{g}}{136.30 + 17x} \]

### Step 5: Relate moles of NH3 to moles of the compound
Since there are \( x \) moles of NH3 in one mole of the compound, we can write:
\[ x \times \text{Moles of compound} = 0.00614 \]
Substituting the expression for moles of the compound:
\[ x \times \frac{0.224}{136.30 + 17x} = 0.00614 \]

### Step 6: Solve for \( x \)
Rearranging gives:
\[ 0.224x = 0.00614(136.30 + 17x) \]
Expanding:
\[ 0.224x = 0.836282 + 0.10438x \]
Bringing all terms involving \( x \) to one side:
\[ 0.224x - 0.10438x = 0.836282 \]
\[ 0.11962x = 0.836282 \]
\[ x = \frac{0.836282}{0.11962} \approx 6.99 \]

### Step 7: Round to the nearest whole number
Since \( x \) must be a whole number, we round \( 6.99 \) to \( 7 \).

### Final Answer
The value of \( x \) in the formula \( Zn(NH_3)_xCl_2 \) is approximately \( 7 \).

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