To find the value of 'n' in the compound \( Na_2SO_4 \cdot nH_2O \), we can follow these steps:
### Step 1: Determine the molar mass of \( Na_2SO_4 \)
The molar mass of sodium sulfate (\( Na_2SO_4 \)) can be calculated as follows:
- Sodium (Na): \( 23 \, \text{g/mol} \times 2 = 46 \, \text{g/mol} \)
- Sulfur (S): \( 32 \, \text{g/mol} \)
- Oxygen (O): \( 16 \, \text{g/mol} \times 4 = 64 \, \text{g/mol} \)
Adding these together:
\[
\text{Molar mass of } Na_2SO_4 = 46 + 32 + 64 = 142 \, \text{g/mol}
\]
### Step 2: Determine the molar mass of \( nH_2O \)
The molar mass of water (\( H_2O \)) is:
- Hydrogen (H): \( 1 \, \text{g/mol} \times 2 = 2 \, \text{g/mol} \)
- Oxygen (O): \( 16 \, \text{g/mol} \)
Thus, the molar mass of water is:
\[
\text{Molar mass of } H_2O = 2 + 16 = 18 \, \text{g/mol}
\]
Therefore, the molar mass of \( nH_2O \) is:
\[
\text{Molar mass of } nH_2O = 18n \, \text{g/mol}
\]
### Step 3: Write the equation for total mass
The total mass of the hydrate \( Na_2SO_4 \cdot nH_2O \) is given as:
\[
\text{Total mass} = \text{mass of } Na_2SO_4 + \text{mass of } nH_2O
\]
This can be expressed as:
\[
26.8 \, \text{g} = 142 \, \text{g/mol} + 18n \, \text{g/mol}
\]
### Step 4: Set up the equation for water mass
We know that the mass of water in the hydrate is given as 12.6 g. Therefore, we can write:
\[
\text{mass of water} = n \times \text{mass of } H_2O = n \times 18
\]
Thus,
\[
n \times 18 = 12.6
\]
### Step 5: Solve for 'n'
From the equation \( n \times 18 = 12.6 \):
\[
n = \frac{12.6}{18}
\]
Calculating this gives:
\[
n = 0.7
\]
### Step 6: Substitute back to find the total mass
Now, substituting \( n \) back into the total mass equation:
\[
26.8 = 142 + 18n
\]
Substituting \( n = 0.7 \):
\[
26.8 = 142 + 18 \times 0.7
\]
Calculating \( 18 \times 0.7 = 12.6 \):
\[
26.8 = 142 + 12.6
\]
This confirms our calculations.
### Final Result
Thus, the value of \( n \) is:
\[
\boxed{7}
\]
