Cuprous sulphide and silver sulphide are isomorphous. The atomic mass of Cu is 63.57 g/mole. If the percentage composition of Sulphur in each of these is 20.14% and 12.94% respectively then the atomic mass of silver will be:

Options

A. 126.4 g/mole
B. 107.74 g/mole
C. 47 g/mole
D. 210.23 g/mole

Answer

To find the atomic mass of silver (Ag) in silver sulfide (Ag2S), we will use the percentage composition of sulfur in both cuprous sulfide (Cu2S) and silver sulfide, along with the known atomic mass of copper (Cu).

### Step-by-Step Solution:

1. **Identify the Given Data**:
- Atomic mass of Copper (Cu) = 63.57 g/mole
- Percentage of Sulfur (S) in Cu2S = 20.14%
- Percentage of Sulfur (S) in Ag2S = 12.94%

2. **Calculate the Molar Mass of Cu2S**:
- The formula for cuprous sulfide is Cu2S, which contains 2 moles of copper and 1 mole of sulfur.
- Molar mass of Cu2S = (2 × Atomic mass of Cu) + (1 × Atomic mass of S)
- We need to find the atomic mass of sulfur (S) first.

3. **Set Up the Equation for Cu2S**:
- Using the mass percentage formula:
\[ \text{Percentage of S in Cu2S} = \frac{\text{mass of S}}{\text{mass of Cu2S}} \times 100 \]
- Let the atomic mass of sulfur be \( M_S \). The mass of sulfur in Cu2S is \( M_S \) and the mass of copper is \( 2 \times 63.57 \).
- The total mass of Cu2S = \( 2 \times 63.57 + M_S \).

4. **Substituting Values for Cu2S**:
- From the percentage of sulfur:
\[ 20.14 = \frac{M_S}{(2 \times 63.57) + M_S} \times 100 \]
- Rearranging gives:
\[ 20.14 = \frac{M_S}{127.14 + M_S} \times 100 \]
- This simplifies to:
\[ 20.14 (127.14 + M_S) = 100 M_S \]
- Expanding and rearranging:
\[ 2564.98 + 20.14 M_S = 100 M_S \]
\[ 2564.98 = 100 M_S - 20.14 M_S \]
\[ 2564.98 = 79.86 M_S \]
\[ M_S = \frac{2564.98}{79.86} \approx 32.12 \text{ g/mole} \]

5. **Calculate the Molar Mass of Ag2S**:
- The formula for silver sulfide is Ag2S, which contains 2 moles of silver and 1 mole of sulfur.
- Molar mass of Ag2S = (2 × Atomic mass of Ag) + (1 × Atomic mass of S).

6. **Set Up the Equation for Ag2S**:
- Using the mass percentage formula again:
\[ \text{Percentage of S in Ag2S} = \frac{M_S}{(2 \times M_{Ag}) + M_S} \times 100 \]
- Substituting the known values:
\[ 12.94 = \frac{32.12}{(2 \times M_{Ag}) + 32.12} \times 100 \]
- Rearranging gives:
\[ 12.94 (2 M_{Ag} + 32.12) = 3200 \]
- Expanding and rearranging:
\[ 12.94 \times 2 M_{Ag} + 12.94 \times 32.12 = 3200 \]
\[ 25.88 M_{Ag} + 414.57 = 3200 \]
\[ 25.88 M_{Ag} = 3200 - 414.57 \]
\[ 25.88 M_{Ag} = 2785.43 \]
\[ M_{Ag} = \frac{2785.43}{25.88} \approx 107.74 \text{ g/mole} \]

### Final Answer:
The atomic mass of silver (Ag) is approximately **107.74 g/mole**.