0.98 g of the metal sulphate was dissolved in water and excess of barium chloride was added. The precipitated barium sulphate weighted 0.95 g. Calculate the equivalent weight of the metal.

Answer

To calculate the equivalent weight of the metal in the given problem, we will follow these steps:

### Step 1: Understand the Reaction
When the metal sulfate is dissolved in water and barium chloride is added, barium sulfate (BaSO4) precipitates out. The reaction can be summarized as:
\[ \text{Metal Sulfate} + \text{Barium Chloride} \rightarrow \text{Barium Sulfate} + \text{Metal Chloride} \]

### Step 2: Identify Given Data
- Mass of metal sulfate (M) = 0.98 g
- Mass of barium sulfate (BaSO4) precipitate = 0.95 g

### Step 3: Calculate Molar Mass of Barium Sulfate
The molar mass of barium sulfate (BaSO4) can be calculated as follows:
- Molar mass of Ba = 137.33 g/mol
- Molar mass of S = 32.07 g/mol
- Molar mass of O = 16.00 g/mol (4 O atoms)

Thus,
\[ \text{Molar mass of BaSO4} = 137.33 + 32.07 + (4 \times 16.00) = 137.33 + 32.07 + 64.00 = 233.40 \text{ g/mol} \]

### Step 4: Calculate Moles of Barium Sulfate
Using the mass of barium sulfate precipitate:
\[ \text{Moles of BaSO4} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{0.95 \text{ g}}{233.40 \text{ g/mol}} \approx 0.00407 \text{ mol} \]

### Step 5: Determine Moles of Sulfate Ion
Since one mole of BaSO4 corresponds to one mole of sulfate ion (SO4^2-), the moles of sulfate ion will also be 0.00407 mol.

### Step 6: Calculate Equivalent Weight of Sulfate Ion
The equivalent weight of sulfate ion can be calculated as:
\[ \text{Equivalent weight of SO4}^{2-} = \frac{\text{Molar mass of SO4}^{2-}}{\text{Valency}} = \frac{96 \text{ g/mol}}{2} = 48 \text{ g/equiv} \]

### Step 7: Calculate the Total Equivalent of Barium and Chloride Ions
- Molar mass of Ba^2+ = 137.33 g/mol, so:
\[ \text{Equivalent weight of Ba}^{2+} = \frac{137.33 \text{ g/mol}}{2} = 68.66 \text{ g/equiv} \]
- Molar mass of Cl^- = 35.5 g/mol, so:
\[ \text{Equivalent weight of Cl^{-}} = \frac{35.5 \text{ g/mol}}{1} = 35.5 \text{ g/equiv} \]

### Step 8: Set Up the Equation
Using the law of equivalent proportions:
\[ \frac{\text{Mass of metal sulfate}}{\text{Mass of BaSO4}} = \frac{\text{Equivalent of metal} + \text{Equivalent of SO4}^{2-}}{\text{Equivalent of Ba}^{2+} + \text{Equivalent of Cl^{-}}} \]
Substituting the known values:
\[ \frac{0.98}{0.95} = \frac{E_m + 48}{68.66 + 35.5} \]

### Step 9: Solve for Equivalent Weight of Metal
Calculating the sum of equivalents:
\[ 68.66 + 35.5 = 104.16 \]
Now substituting back:
\[ \frac{0.98}{0.95} = \frac{E_m + 48}{104.16} \]
Cross-multiplying gives:
\[ 0.98 \times 104.16 = 0.95 \times (E_m + 48) \]
Calculating:
\[ 102.05 = 0.95E_m + 45.6 \]
Rearranging gives:
\[ 0.95E_m = 102.05 - 45.6 = 56.45 \]
Now, solving for \(E_m\):
\[ E_m = \frac{56.45}{0.95} \approx 59.42 \text{ g/equiv} \]

### Final Answer
The equivalent weight of the metal is approximately **59.42 g/equiv**.