A 1.0g sample of a pure organic compound cotaining chlorine is fused with `Na_(2)O_(2)` to convert chlorine to NaCl. The sample is then dissolved in water, and the chloride precipitated with `AgNO_(3)`, giving 1.96 g of AgCl. If the molecular mass of organic compound is 147, how many chlorine does each molecule contain ?

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem step by step, we need to determine how many chlorine atoms are present in each molecule of the organic compound.

### Step 1: Calculate the moles of AgCl produced.
We know that the mass of AgCl produced is 1.96 g. The molar mass of AgCl can be calculated as follows:
- Molar mass of Ag = 108 g/mol
- Molar mass of Cl = 35.5 g/mol
- Therefore, molar mass of AgCl = 108 g/mol + 35.5 g/mol = 143.5 g/mol.

Now, we can calculate the moles of AgCl:
\[ \text{Moles of AgCl} = \frac{\text{mass of AgCl}}{\text{molar mass of AgCl}} = \frac{1.96 \text{ g}}{143.5 \text{ g/mol}} \approx 0.01365 \text{ mol}. \]

### Step 2: Relate moles of AgCl to moles of chlorine in the organic compound.
From the reaction, we know that each mole of AgCl corresponds to one mole of Cl. Therefore, the moles of Cl in the organic compound is also 0.01365 mol.

### Step 3: Calculate the moles of the organic compound.
The mass of the organic compound is given as 1.0 g, and its molecular mass is 147 g/mol. We can calculate the moles of the organic compound:
\[ \text{Moles of organic compound} = \frac{\text{mass of organic compound}}{\text{molar mass of organic compound}} = \frac{1.0 \text{ g}}{147 \text{ g/mol}} \approx 0.0068 \text{ mol}. \]

### Step 4: Determine the number of chlorine atoms per molecule.
Let \( n \) be the number of chlorine atoms in each molecule of the organic compound. According to the principle of conservation of mass, the moles of Cl produced from the organic compound should equal the moles of AgCl formed:
\[ n \times \text{moles of organic compound} = \text{moles of Cl} \implies n \times 0.0068 \text{ mol} = 0.01365 \text{ mol}. \]
Now, solving for \( n \):
\[ n = \frac{0.01365 \text{ mol}}{0.0068 \text{ mol}} \approx 2.01. \]

Since \( n \) must be a whole number, we round it to the nearest whole number, which is 2.

### Conclusion
Each molecule of the organic compound contains **2 chlorine atoms**.

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