Which of the following relations is incorrect for solutions?

Options

A. (a)`3N Al_(2)(SO_(4))_(3)=0.5 Mal_(2)(SO_(4))_(3)`
B. (b)`3 M H_(2)SO_(4)=6 N H_(2)SO_(4)`
C. (c )`1 M H_(3)PO_(4)=1//3 N H_(3)PO_(4)`
D. (d)`1 M Al_(2)(SO_(4))_(3)=6 N Al_(2) (SO_(4))_(3)`

Explanation

To determine which of the given relations is incorrect for solutions, we need to analyze the relationships between normality, molarity, and the n-factor of the solute involved. The relationship can be expressed as:

\[ \text{Normality (N)} = \text{n-factor} \times \text{Molarity (M)} \]

Let's evaluate each option step by step:

### Step 1: Analyze Option A
- Given: Normality = 3, Molarity = 0.5
- For \( \text{Al}_2(\text{SO}_4)_3 \), the n-factor = 6 (because it can donate 6 moles of H⁺ ions).

Check the relation:
\[ 3 = 6 \times 0.5 \]
\[ 3 = 3 \] (True)

**Conclusion**: Option A is correct.

### Step 2: Analyze Option B
- Given: Normality = 6, Molarity = 3
- For \( \text{H}_2\text{SO}_4 \), the n-factor = 2 (because it can donate 2 moles of H⁺ ions).

Check the relation:
\[ 6 = 2 \times 3 \]
\[ 6 = 6 \] (True)

**Conclusion**: Option B is correct.

### Step 3: Analyze Option C
- Given: Normality = \( \frac{1}{3} \), Molarity = 1
- For \( \text{H}_3\text{PO}_4 \), the n-factor = 3 (because it can donate 3 moles of H⁺ ions).

Check the relation:
\[ \frac{1}{3} = 3 \times 1 \]
\[ \frac{1}{3} \neq 3 \] (False)

**Conclusion**: Option C is incorrect.

### Step 4: Analyze Option D
- Given: Normality = 6, Molarity = 1
- For \( \text{Al}_2(\text{SO}_4)_3 \), the n-factor = 6.

Check the relation:
\[ 6 = 6 \times 1 \]
\[ 6 = 6 \] (True)

**Conclusion**: Option D is correct.

### Final Answer
The incorrect relation among the options is **Option C**.

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