The density of 3 M solution of `Na_2S_2O_3` is 1.25g/mL . What is % by weight of `Na_2S_2O_3` ?

Options

A. 36.24
B. 37.92
C. 40.24
D. 38.24

Answer

To solve the problem of finding the percentage by weight of `Na2S2O3` in a 3 M solution with a density of 1.25 g/mL, we can follow these steps:

### Step 1: Calculate the mass of `Na2S2O3` in the solution
- **Molarity (M)** is defined as moles of solute per liter of solution. Here, the molarity of `Na2S2O3` is given as 3 M.
- Therefore, in 1 liter (1000 mL) of solution, the moles of `Na2S2O3` = 3 moles.

### Step 2: Calculate the mass of `Na2S2O3`
- The molecular weight of `Na2S2O3` (sodium thiosulfate) is 158 g/mol.
- To find the mass of `Na2S2O3` in 3 moles:
\[ \text{Mass of } Na2S2O3 = \text{Moles} \times \text{Molecular Weight} = 3 \, \text{moles} \times 158 \, \text{g/mol} = 474 \, \text{g} \]

### Step 3: Calculate the mass of the solution
- The density of the solution is given as 1.25 g/mL.
- Therefore, the mass of 1 liter (1000 mL) of the solution can be calculated as:
\[ \text{Mass of solution} = \text{Density} \times \text{Volume} = 1.25 \, \text{g/mL} \times 1000 \, \text{mL} = 1250 \, \text{g} \]

### Step 4: Calculate the mass of the solvent
- The mass of the solvent (water) can be calculated by subtracting the mass of `Na2S2O3` from the total mass of the solution:
\[ \text{Mass of solvent} = \text{Mass of solution} - \text{Mass of } Na2S2O3 = 1250 \, \text{g} - 474 \, \text{g} = 776 \, \text{g} \]

### Step 5: Calculate the percentage by weight of `Na2S2O3`
- The percentage by weight of `Na2S2O3` in the solution is given by the formula:
\[ \text{Percentage by weight} = \left( \frac{\text{Mass of } Na2S2O3}{\text{Mass of solution}} \right) \times 100 \]
- Substituting the values:
\[ \text{Percentage by weight} = \left( \frac{474 \, \text{g}}{1250 \, \text{g}} \right) \times 100 \approx 37.92\% \]

### Final Answer:
The percentage by weight of `Na2S2O3` in the solution is approximately **37.92%**.

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