Calculate the equivalent weight of following assuming molecular weight M
(i) `H_(3)BO_(3)`
(ii) `H_(3)PO_(2)`
(iii) `FeSO_(4).(NH_(4))_(2).6H_(2)O`
(iv) `K_(2)SO_(4)Cr_(2)(SO_(4))_(3).24H_(2)O`
(v) `Al_(2)O_(3)`
(vi) `HNO_(3)`
(vii) `MgCO_(3)`
(Viii) `CaSO_(4)`

Answer

To calculate the equivalent weight of the given compounds, we will use the formula:

\[ \text{Equivalent Weight} = \frac{\text{Molecular Weight (M)}}{\text{n factor}} \]

Where the n factor depends on the type of reaction the compound undergoes (e.g., the number of moles of H+ ions that can be donated, the number of electrons that can be gained or lost, etc.).

Now, let's calculate the equivalent weight for each compound step by step.

### (i) H₃BO₃ (Boric Acid)
1. **Identify the n factor**: Boric acid can donate 3 H⁺ ions.
2. **Calculate the equivalent weight**:
\[ \text{Equivalent Weight} = \frac{M}{3} \]

### (ii) H₃PO₂ (Phosphorous Acid)
1. **Identify the n factor**: Phosphorous acid can donate 1 H⁺ ion.
2. **Calculate the equivalent weight**:
\[ \text{Equivalent Weight} = \frac{M}{1} = M \]

### (iii) FeSO₄·(NH₄)₂·6H₂O (Mohr's Salt)
1. **Identify the n factor**: Mohr's salt can donate 1 Fe²⁺ ion.
2. **Calculate the equivalent weight**:
\[ \text{Equivalent Weight} = \frac{M}{1} = M \]

### (iv) K₂SO₄·Cr₂(SO₄)₃·24H₂O (Chrome Alum)
1. **Identify the n factor**: Chrome alum can donate 8 H⁺ ions.
2. **Calculate the equivalent weight**:
\[ \text{Equivalent Weight} = \frac{M}{8} \]

### (v) Al₂O₃ (Aluminum Oxide)
1. **Identify the n factor**: Aluminum oxide can donate 6 electrons.
2. **Calculate the equivalent weight**:
\[ \text{Equivalent Weight} = \frac{M}{6} \]

### (vi) HNO₃ (Nitric Acid)
1. **Identify the n factor**: Nitric acid can donate 1 H⁺ ion.
2. **Calculate the equivalent weight**:
\[ \text{Equivalent Weight} = \frac{M}{1} = M \]

### (vii) MgCO₃ (Magnesium Carbonate)
1. **Identify the n factor**: Magnesium carbonate can donate 2 H⁺ ions.
2. **Calculate the equivalent weight**:
\[ \text{Equivalent Weight} = \frac{M}{2} \]

### (viii) CaSO₄ (Calcium Sulfate)
1. **Identify the n factor**: Calcium sulfate can donate 2 H⁺ ions.
2. **Calculate the equivalent weight**:
\[ \text{Equivalent Weight} = \frac{M}{2} \]

### Summary of Equivalent Weights
- (i) H₃BO₃: \(\frac{M}{3}\)
- (ii) H₃PO₂: \(M\)
- (iii) FeSO₄·(NH₄)₂·6H₂O: \(M\)
- (iv) K₂SO₄·Cr₂(SO₄)₃·24H₂O: \(\frac{M}{8}\)
- (v) Al₂O₃: \(\frac{M}{6}\)
- (vi) HNO₃: \(M\)
- (vii) MgCO₃: \(\frac{M}{2}\)
- (viii) CaSO₄: \(\frac{M}{2}\)