`A" " 5.0 g` sample of felspar containing `Na_(2)O,K_(2)O` and some inert impurity is dissolved in dilute `HCl` solution and `NaCl` and `KCl` formed are separated by fractional crystallization. During crystallization some less soluble impurities also come out. Mass of `NaCl, KCl` and impurity accompanying these salts was found to be `6.47 g`. Solid crystal was then re-dissolved and required `300mL` of `0.3M AgNO_(3)` for complete precipitation of chlorides. The precipitate this, obtained ws found to contain `4.23%` insoluble impurity. Determine mass percentage of `Na_(2)O` and `K_(2)O` in the original sample:

Explanation

To determine the mass percentage of \( Na_2O \) and \( K_2O \) in the original sample of felspar, we can follow these steps:

### Step 1: Calculate the moles of \( AgNO_3 \)
We know that the volume of \( AgNO_3 \) solution used is \( 300 \, mL \) and its molarity is \( 0.3 \, M \).

\[ \text{Moles of } AgNO_3 = \text{Molarity} \times \text{Volume (in L)} = 0.3 \, \text{mol/L} \times 0.300 \, \text{L} = 0.09 \, \text{mol} \]

### Step 2: Calculate the mass of \( AgCl \) precipitate
The reaction between \( AgNO_3 \) and \( NaCl \) or \( KCl \) produces \( AgCl \). The molar mass of \( AgCl \) is approximately \( 143.5 \, g/mol \).

\[ \text{Mass of } AgCl = \text{Moles} \times \text{Molar Mass} = 0.09 \, \text{mol} \times 143.5 \, g/mol = 12.915 \, g \]

### Step 3: Calculate the total mass of the precipitate
Given that the precipitate contains \( 4.23\% \) insoluble impurity, we can find the total mass of the precipitate using the mass of \( AgCl \).

Let \( x \) be the total mass of the precipitate. The mass of insoluble impurity is \( 0.0423x \).

Thus, we can write:

\[ x - 0.0423x = 12.915 \implies 0.9577x = 12.915 \implies x = \frac{12.915}{0.9577} \approx 13.485 \, g \]

### Step 4: Calculate the mass of the insoluble impurity
The mass of the insoluble impurity is:

\[ \text{Mass of impurity} = 0.0423 \times 13.485 \approx 0.57 \, g \]

### Step 5: Calculate the mass of \( NaCl \) and \( KCl \)
The total mass of \( NaCl + KCl + \text{impurity} = 6.47 \, g \).

Thus, the mass of \( NaCl + KCl \) is:

\[ \text{Mass of } NaCl + KCl = 6.47 \, g - 0.57 \, g = 5.9 \, g \]

### Step 6: Calculate the mass of \( NaCl \) and \( KCl \)
Assuming the masses of \( NaCl \) and \( KCl \) are \( m_{NaCl} \) and \( m_{KCl} \) respectively, we can set up the following equations:

Let \( m_{NaCl} + m_{KCl} = 5.9 \, g \).

Using the molar masses:
- Molar mass of \( NaCl = 58.5 \, g/mol \)
- Molar mass of \( KCl = 74.5 \, g/mol \)

We can express the moles of \( NaCl \) and \( KCl \):

\[ \text{Moles of } NaCl = \frac{m_{NaCl}}{58.5}, \quad \text{Moles of } KCl = \frac{m_{KCl}}{74.5} \]

### Step 7: Calculate the mass percentages of \( Na_2O \) and \( K_2O \)
Using the stoichiometry of the salts:

\[ \text{Mass of } Na_2O = \text{Moles of } NaCl \times 61.98 \, g/mol \quad (\text{molar mass of } Na_2O) \]
\[ \text{Mass of } K_2O = \text{Moles of } KCl \times 94.2 \, g/mol \quad (\text{molar mass of } K_2O) \]

Finally, calculate the mass percentages in the original sample:

\[ \text{Mass percentage of } Na_2O = \left( \frac{\text{Mass of } Na_2O}{5.0 \, g} \right) \times 100 \]
\[ \text{Mass percentage of } K_2O = \left( \frac{\text{Mass of } K_2O}{5.0 \, g} \right) \times 100 \]