The impure 6g of NaCl is dissolved in water and then treated with excess of silver nitrate solution. The mass of precipitate of silver chloride is found to be 14g. The % purity of NaCl solution would be:

Options

A. 0.95
B. 0.85
C. 0.75
D. 0.65

Answer

To solve the problem of determining the percent purity of the NaCl solution, we will follow these steps:

### Step 1: Write the balanced chemical equation
When sodium chloride (NaCl) reacts with silver nitrate (AgNO₃), silver chloride (AgCl) precipitate and sodium nitrate (NaNO₃) are formed. The balanced equation is:

\[ \text{NaCl} + \text{AgNO}_3 \rightarrow \text{AgCl} + \text{NaNO}_3 \]

### Step 2: Determine the molar masses
- Molar mass of NaCl = 58.5 g/mol
- Molar mass of AgCl = 143.5 g/mol

### Step 3: Find the amount of NaCl required to produce 14 g of AgCl
Using stoichiometry, we know that 1 mole of NaCl produces 1 mole of AgCl. Therefore, we can set up a proportion to find out how much NaCl is needed to produce 14 g of AgCl.

First, calculate the number of moles of AgCl produced:

\[ \text{Moles of AgCl} = \frac{\text{mass}}{\text{molar mass}} = \frac{14 \text{ g}}{143.5 \text{ g/mol}} \approx 0.0974 \text{ moles} \]

Since the stoichiometry is 1:1, the moles of NaCl required will also be approximately 0.0974 moles.

Now, calculate the mass of NaCl required:

\[ \text{Mass of NaCl} = \text{moles} \times \text{molar mass} = 0.0974 \text{ moles} \times 58.5 \text{ g/mol} \approx 5.7 \text{ g} \]

### Step 4: Calculate the percent purity of NaCl
The percent purity of the NaCl solution can be calculated using the formula:

\[ \text{Percent Purity} = \left( \frac{\text{mass of pure NaCl}}{\text{total mass of impure NaCl}} \right) \times 100 \]

Substituting the values we have:

\[ \text{Percent Purity} = \left( \frac{5.7 \text{ g}}{6 \text{ g}} \right) \times 100 \approx 95\% \]

### Final Answer
The percent purity of the NaCl solution is approximately **95%**.

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