Which of the following relations is/are correct for solutions ?

Options

A. 3 N `Al_(2)(SO_(4))_(3)=0.5M Al_(2)(SO_(4))_(3)`
B. `3 M H_(2)SO_(4)=6 N H_(2)SO_(4)`
C. `1 M H_(3)PO_(4)=1//3 N H_(3)PO_(4)`
D. `1 M Al_(2)(SO_(4))_(3)=6 N Al(SO_(4))_(3)`

Explanation

To determine which of the given relations between normality and molarity for various solutions are correct, we will analyze each option step by step.

### Step-by-Step Solution:

1. **Understanding Normality and Molarity:**
- Normality (N) is defined as the number of equivalents of solute per liter of solution.
- Molarity (M) is defined as the number of moles of solute per liter of solution.
- The relationship between normality and molarity can be expressed as:
\[ \text{Normality} = \text{Molarity} \times n \]
where \( n \) is the number of equivalents (or the n-factor).

2. **Analyzing the First Relation:**
- **Relation:** \( 3 \, \text{N} \, \text{Al}_2(\text{SO}_4)_3 = 0.5 \, \text{M} \, \text{Al}_2(\text{SO}_4)_3 \)
- **Finding n-factor:**
- The formula \( \text{Al}_2(\text{SO}_4)_3 \) contains 2 Al³⁺ ions and 3 SO₄²⁻ ions.
- Total charge = \( 2 \times 3 + 3 \times 2 = 6 \).
- Thus, \( n = 6 \).
- **Calculating Molarity:**
\[ 3 \, \text{N} = \text{Molarity} \times 6 \implies \text{Molarity} = \frac{3}{6} = 0.5 \, \text{M} \]
- **Conclusion:** This relation is **correct**.

3. **Analyzing the Second Relation:**
- **Relation:** \( 3 \, \text{M} \, \text{H}_2\text{SO}_4 = 6 \, \text{N} \, \text{H}_2\text{SO}_4 \)
- **Finding n-factor:**
- For sulfuric acid \( \text{H}_2\text{SO}_4 \), it can donate 2 H⁺ ions.
- Thus, \( n = 2 \).
- **Calculating Normality:**
\[ \text{Normality} = 3 \, \text{M} \times 2 = 6 \, \text{N} \]
- **Conclusion:** This relation is **correct**.

4. **Analyzing the Third Relation:**
- **Relation:** \( 1 \, \text{M} \, \text{H}_3\text{PO}_4 = \frac{1}{3} \, \text{N} \, \text{H}_3\text{PO}_4 \)
- **Finding n-factor:**
- For phosphoric acid \( \text{H}_3\text{PO}_4 \), it can donate 3 H⁺ ions.
- Thus, \( n = 3 \).
- **Calculating Normality:**
\[ \text{Normality} = 1 \, \text{M} \times 3 = 3 \, \text{N} \]
- **Conclusion:** This relation is **incorrect** because it states \( 1 \, \text{M} = \frac{1}{3} \, \text{N} \).

5. **Analyzing the Fourth Relation:**
- **Relation:** \( 1 \, \text{M} \, \text{Al}_2(\text{SO}_4)_3 = 6 \, \text{N} \, \text{Al}_2(\text{SO}_4)_3 \)
- **Finding n-factor:**
- As calculated earlier, \( n = 6 \).
- **Calculating Normality:**
\[ \text{Normality} = 1 \, \text{M} \times 6 = 6 \, \text{N} \]
- **Conclusion:** This relation is **correct**.

### Final Conclusion:
- The correct relations are:
- \( 3 \, \text{N} \, \text{Al}_2(\text{SO}_4)_3 = 0.5 \, \text{M} \, \text{Al}_2(\text{SO}_4)_3 \) (Correct)
- \( 3 \, \text{M} \, \text{H}_2\text{SO}_4 = 6 \, \text{N} \, \text{H}_2\text{SO}_4 \) (Correct)
- \( 1 \, \text{M} \, \text{H}_3\text{PO}_4 = \frac{1}{3} \, \text{N} \, \text{H}_3\text{PO}_4 \) (Incorrect)
- \( 1 \, \text{M} \, \text{Al}_2(\text{SO}_4)_3 = 6 \, \text{N} \, \text{Al}_2(\text{SO}_4)_3 \) (Correct)