A certain weight of sodium Iodine and sodium chloride mixture when treated with sulphuric acid was found to give the same weight (as that of mixture) of sodium sulphate. The percentage composition of NaCl is

Options

A. 18.86
B. 52.23
C. 35.57
D. 71.14

Answer

To solve the problem, we need to determine the percentage composition of sodium chloride (NaCl) in a mixture of sodium iodide (NaI) and sodium chloride that produces sodium sulfate (Na2SO4) when treated with sulfuric acid (H2SO4).

### Step-by-Step Solution:

1. **Define the Variables**:
- Let the total weight of the mixture be 1 gram.
- Let the weight of sodium chloride (NaCl) be \( x \) grams.
- Therefore, the weight of sodium iodide (NaI) will be \( 1 - x \) grams.

2. **Write the Reactions**:
- The reaction of sodium chloride with sulfuric acid:
\[ 2 \text{NaCl} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2 \text{HCl} \]
- The reaction of sodium iodide with sulfuric acid:
\[ 2 \text{NaI} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2 \text{HI} \]

3. **Calculate the Amount of Na2SO4 Produced**:
- From the reaction of NaCl:
- The molar mass of NaCl = 58.5 g/mol.
- From \( x \) grams of NaCl, the moles of NaCl = \( \frac{x}{58.5} \).
- The moles of Na2SO4 produced from NaCl = \( \frac{x}{2 \times 58.5} \).
- The mass of Na2SO4 produced = \( \frac{x \times 142}{117} \) grams (since the molar mass of Na2SO4 = 142 g/mol).

- From the reaction of NaI:
- The molar mass of NaI = 150 g/mol.
- From \( 1 - x \) grams of NaI, the moles of NaI = \( \frac{1 - x}{150} \).
- The moles of Na2SO4 produced from NaI = \( \frac{1 - x}{2 \times 150} \).
- The mass of Na2SO4 produced = \( \frac{(1 - x) \times 142}{300} \) grams.

4. **Set Up the Equation**:
- According to the problem, the total mass of Na2SO4 produced is equal to the mass of the mixture (1 gram):
\[ \frac{142x}{117} + \frac{142(1 - x)}{300} = 1 \]

5. **Cross Multiply and Simplify**:
- Multiply through by the least common multiple (LCM) of the denominators (117 and 300):
\[ 142x \cdot 300 + 142(1 - x) \cdot 117 = 117 \cdot 300 \]
- Simplifying this equation will give:
\[ 42600x + 16614 - 16614x = 35100 \]
- Combine like terms:
\[ -12314x + 16614 = 35100 \]
- Rearranging gives:
\[ -12314x = 35100 - 16614 \]
\[ -12314x = 18486 \]
- Solving for \( x \):
\[ x = \frac{18486}{12314} \approx 0.7114 \]

6. **Calculate the Percentage Composition of NaCl**:
- The percentage of NaCl in the mixture:
\[ \text{Percentage of NaCl} = x \times 100 = 0.7114 \times 100 \approx 71.14\% \]

### Final Answer:
The percentage composition of sodium chloride (NaCl) in the mixture is approximately **71.14%**.

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