`8.0575 xx 10^(-12)` kg of Glauber's salt are dissolved in water to obtain `1 dm^3` of a solution of density `1077.2 kg m^(-3)`. Calculate the molarity.

Answer

To calculate the molarity of the solution, we will follow these steps:

### Step 1: Convert the mass of Glauber's salt from kg to grams.
Given mass of Glauber's salt = \( 8.0575 \times 10^{-12} \) kg

To convert this to grams:
\[ \text{Mass in grams} = 8.0575 \times 10^{-12} \, \text{kg} \times 1000 \, \text{g/kg} = 8.0575 \times 10^{-9} \, \text{g} \]

### Step 2: Determine the molar mass of Glauber's salt (Na2SO4·10H2O).
The formula for Glauber's salt is Na2SO4·10H2O. We need to calculate its molar mass:

- Molar mass of Na (Sodium) = 23 g/mol
- Molar mass of S (Sulfur) = 32 g/mol
- Molar mass of O (Oxygen) = 16 g/mol
- Molar mass of H2O (Water) = 18 g/mol

Calculating the molar mass:
\[ \text{Molar mass of Na2SO4} = (2 \times 23) + 32 + (4 \times 16) = 46 + 32 + 64 = 142 \, \text{g/mol} \]
\[ \text{Molar mass of 10H2O} = 10 \times 18 = 180 \, \text{g/mol} \]
\[ \text{Total molar mass of Glauber's salt} = 142 + 180 = 322 \, \text{g/mol} \]

### Step 3: Calculate the mass of sodium sulfate in the given mass of Glauber's salt.
Using the proportion of the mass of sodium sulfate in Glauber's salt:
\[ \text{Mass of Na2SO4} = \left( \frac{\text{Molar mass of Na2SO4}}{\text{Molar mass of Glauber's salt}} \right) \times \text{mass of Glauber's salt} \]
\[ \text{Mass of Na2SO4} = \left( \frac{142}{322} \right) \times 8.0575 \times 10^{-9} \, \text{g} \approx 3.39 \times 10^{-9} \, \text{g} \]

### Step 4: Convert the mass of sodium sulfate to moles.
Using the molar mass of sodium sulfate:
\[ \text{Moles of Na2SO4} = \frac{\text{Mass of Na2SO4}}{\text{Molar mass of Na2SO4}} = \frac{3.39 \times 10^{-9} \, \text{g}}{142 \, \text{g/mol}} \approx 2.39 \times 10^{-11} \, \text{mol} \]

### Step 5: Calculate the molarity of the solution.
Molarity (M) is defined as the number of moles of solute per liter of solution:
\[ \text{Molarity} = \frac{\text{Moles of Na2SO4}}{\text{Volume of solution in liters}} = \frac{2.39 \times 10^{-11} \, \text{mol}}{1 \, \text{L}} = 2.39 \times 10^{-11} \, \text{M} \]

### Final Answer:
The molarity of the solution is approximately \( 2.39 \times 10^{-11} \, \text{M} \).

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