A mixture of salts was treated to determine the barium constent. A 0.230 g sample was dissolved and treated with excess potassium chromate. The precipitate of `BaCrO_(4)` was dissolved in dil HCl to convert `BaCrO_(4)` to `Cr_(2)O_(7)^(2-)` anion. The solution treated with excess sodium iodide and trilodide produced was titrated with 21 mL of 0.095 M sodium thiosulphate. Calculate the `%` of `BaCl_(2). 2H_(2)O` in the sample. [Given Molar mass of `BaCl_(2.2H2O)=244]`

Options

A. `70.5%`
B. `30.25%`
C. `15.12%`
D. `110%`

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Calculate the milliequivalents of sodium thiosulfate used in the titration.
The volume of sodium thiosulfate used is 21 mL and its molarity is 0.095 M.

\[ \text{Milliequivalents of Na}_2\text{S}_2\text{O}_3 = \text{Volume (mL)} \times \text{Molarity (mol/L)} = 21 \, \text{mL} \times 0.095 \, \text{mol/L} = 1.995 \, \text{mmol} \]

### Step 2: Relate the milliequivalents of sodium thiosulfate to the milliequivalents of dichromate ion.
The reaction between dichromate ion \((\text{Cr}_2\text{O}_7^{2-})\) and sodium thiosulfate involves a 1:6 stoichiometry. Therefore, we can find the milliequivalents of \(\text{Cr}_2\text{O}_7^{2-}\):

\[ \text{Milliequivalents of } \text{Cr}_2\text{O}_7^{2-} = \frac{\text{Milliequivalents of Na}_2\text{S}_2\text{O}_3}{6} = \frac{1.995}{6} \approx 0.3325 \, \text{mmol} \]

### Step 3: Calculate the moles of barium chromate \((\text{BaCrO}_4)\) formed.
Since 1 mole of \(\text{BaCrO}_4\) produces 1 mole of \(\text{Cr}_2\text{O}_7^{2-}\), the moles of \(\text{BaCrO}_4\) formed is equal to the moles of \(\text{Cr}_2\text{O}_7^{2-}\):

\[ \text{Moles of } \text{BaCrO}_4 = 0.3325 \, \text{mmol} = 0.0003325 \, \text{mol} \]

### Step 4: Calculate the mass of barium in the sample.
The molar mass of \(\text{BaCrO}_4\) is approximately 197 + 52 + 16*4 = 236 g/mol. The mass of barium can be calculated using the moles of \(\text{BaCrO}_4\):

\[ \text{Mass of Ba} = \text{Moles of BaCrO}_4 \times \text{Molar Mass of Ba} = 0.0003325 \, \text{mol} \times 137 \, \text{g/mol} \approx 0.0455 \, \text{g} \]

### Step 5: Calculate the mass of \(\text{BaCl}_2 \cdot 2\text{H}_2\text{O}\) in the sample.
The molar mass of \(\text{BaCl}_2 \cdot 2\text{H}_2\text{O}\) is given as 244 g/mol. The mass of \(\text{BaCl}_2 \cdot 2\text{H}_2\text{O}\) can be calculated using the mass of barium:

\[ \text{Mass of BaCl}_2 \cdot 2\text{H}_2\text{O} = \frac{\text{Mass of Ba}}{\text{Molar Mass of Ba}} \times \text{Molar Mass of BaCl}_2 \cdot 2\text{H}_2\text{O} \]

\[ \text{Mass of BaCl}_2 \cdot 2\text{H}_2\text{O} = \frac{0.0455 \, \text{g}}{137 \, \text{g/mol}} \times 244 \, \text{g/mol} \approx 0.0804 \, \text{g} \]

### Step 6: Calculate the percentage of \(\text{BaCl}_2 \cdot 2\text{H}_2\text{O}\) in the sample.
To find the percentage of \(\text{BaCl}_2 \cdot 2\text{H}_2\text{O}\) in the original 0.230 g sample:

\[ \text{Percentage} = \left( \frac{\text{Mass of BaCl}_2 \cdot 2\text{H}_2\text{O}}{\text{Total mass of sample}} \right) \times 100 \]

\[ \text{Percentage} = \left( \frac{0.0804 \, \text{g}}{0.230 \, \text{g}} \right) \times 100 \approx 34.96\% \]

### Final Result
The percentage of \(\text{BaCl}_2 \cdot 2\text{H}_2\text{O}\) in the sample is approximately **34.96%**.

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