The equivalent weight of compounds
(a)`Na_(2)SO_(4)" "(b) Na_(3)PO_(4).12H_(2)O" "(c ) Ca_(3)(PO_(4))_(2)`
resepctively are x,y and z. The correct set of molecular weight will be :-

Options

A. `x/2,y/3,z/6`
B. `2x,3y,6z`
C. `x,y/3,z/3`
D. `2x,3y,3z`

Answer

To find the molecular weights of the compounds given their equivalent weights, we will use the relationship between equivalent weight (E) and molecular weight (M), which is given by the formula:

\[ E = \frac{M}{n} \]

Where \( n \) is the number of equivalents (or the n-factor) for the compound. We will calculate the n-factor for each compound and then use it to find the molecular weight.

### Step-by-Step Solution:

#### (a) For \( \text{Na}_2\text{SO}_4 \):
1. **Determine the n-factor**:
- Sodium (\( \text{Na} \)) has a +1 charge, and there are 2 sodium ions, contributing +2.
- Sulfate (\( \text{SO}_4^{2-} \)) has a -2 charge.
- The total charge is +2 from sodium and -2 from sulfate, so the n-factor is 2.

2. **Use the formula to find molecular weight**:
\[ E = \frac{M}{n} \implies E = \frac{M}{2} \implies M = 2E \]
- Given \( E = X \), we have:
\[ M = 2X \]

#### (b) For \( \text{Na}_3\text{PO}_4 \cdot 12\text{H}_2\text{O} \):
1. **Determine the n-factor**:
- Sodium (\( \text{Na} \)) has a +1 charge, and there are 3 sodium ions, contributing +3.
- Phosphate (\( \text{PO}_4^{3-} \)) has a -3 charge.
- The total charge is +3 from sodium and -3 from phosphate, so the n-factor is 3.

2. **Use the formula to find molecular weight**:
\[ E = \frac{M}{n} \implies E = \frac{M}{3} \implies M = 3E \]
- Given \( E = Y \), we have:
\[ M = 3Y \]

#### (c) For \( \text{Ca}_3(\text{PO}_4)_2 \):
1. **Determine the n-factor**:
- Calcium (\( \text{Ca} \)) has a +2 charge, and there are 3 calcium ions, contributing +6.
- Phosphate (\( \text{PO}_4^{3-} \)) has a -3 charge, and there are 2 phosphate ions, contributing -6.
- The total charge is +6 from calcium and -6 from phosphate, so the n-factor is 6.

2. **Use the formula to find molecular weight**:
\[ E = \frac{M}{n} \implies E = \frac{M}{6} \implies M = 6E \]
- Given \( E = Z \), we have:
\[ M = 6Z \]

### Summary of Molecular Weights:
- For \( \text{Na}_2\text{SO}_4 \): \( M = 2X \)
- For \( \text{Na}_3\text{PO}_4 \cdot 12\text{H}_2\text{O} \): \( M = 3Y \)
- For \( \text{Ca}_3(\text{PO}_4)_2 \): \( M = 6Z \)

### Final Answer:
The correct set of molecular weights will be \( 2X, 3Y, 6Z \).