7 g of a sample of sodium chloride on treatment with excess of silver nitrate gave 14.35 g of AgCl. The percentage of NaCl in the sample is

Options

A. 80
B. 50
C. 65.8
D. 83.5

Answer

To determine the percentage of sodium chloride (NaCl) in the sample, we can follow these steps:

### Step 1: Write the chemical reaction
The reaction between sodium chloride and silver nitrate can be represented as follows:
\[ \text{NaCl} + \text{AgNO}_3 \rightarrow \text{AgCl} + \text{NaNO}_3 \]

### Step 2: Calculate the molar masses
- Molar mass of NaCl:
- Sodium (Na) = 23 g/mol
- Chlorine (Cl) = 35.5 g/mol
- Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

- Molar mass of AgCl:
- Silver (Ag) = 108 g/mol
- Chlorine (Cl) = 35.5 g/mol
- Molar mass of AgCl = 108 + 35.5 = 143.5 g/mol

### Step 3: Relate the masses of NaCl and AgCl
From the stoichiometry of the reaction:
- 58.5 g of NaCl produces 143.5 g of AgCl.
- We need to find out how much NaCl corresponds to the 14.35 g of AgCl produced.

### Step 4: Set up the proportion
Using the stoichiometric relationship:
\[ \frac{58.5 \text{ g NaCl}}{143.5 \text{ g AgCl}} = \frac{x \text{ g NaCl}}{14.35 \text{ g AgCl}} \]

### Step 5: Solve for x (mass of NaCl)
Cross-multiplying gives:
\[ x = \frac{58.5 \times 14.35}{143.5} \]
Calculating this:
\[ x = \frac{839.475}{143.5} \approx 5.85 \text{ g NaCl} \]

### Step 6: Calculate the percentage of NaCl in the sample
Now, we can find the percentage of NaCl in the original sample:
\[ \text{Percentage of NaCl} = \left( \frac{\text{mass of NaCl}}{\text{mass of sample}} \right) \times 100 \]
Substituting the values:
\[ \text{Percentage of NaCl} = \left( \frac{5.85 \text{ g}}{7 \text{ g}} \right) \times 100 \approx 83.57\% \]

### Final Answer
The percentage of NaCl in the sample is approximately **83.57%**.

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