5 g of `SrCl_(2).6H_(2)O` gave 3.442 gm of dry `SrSO_(4)`, then eq. wt. of Sr will be

Options

A. `43.8`
B. 50
C. `87.4`
D. 100

Answer

To find the equivalent weight of Strontium (Sr) based on the given data, we can follow these steps:

### Step 1: Determine the moles of SrSO4 produced
We know that 5 g of SrCl2·6H2O produces 3.442 g of SrSO4. First, we need to calculate the molar mass of SrSO4.

- The molar mass of Sr (Strontium) = 87.62 g/mol
- The molar mass of S (Sulfur) = 32.07 g/mol
- The molar mass of O (Oxygen) = 16.00 g/mol (4 O atoms in SO4)

Calculating the molar mass of SrSO4:
\[ \text{Molar mass of SrSO4} = 87.62 + 32.07 + (4 \times 16.00) \]
\[ = 87.62 + 32.07 + 64.00 \]
\[ = 183.69 \text{ g/mol} \]

### Step 2: Calculate moles of SrSO4 produced
Now, we can find the number of moles of SrSO4 produced using its mass and molar mass:
\[ \text{Moles of SrSO4} = \frac{\text{mass}}{\text{molar mass}} = \frac{3.442 \text{ g}}{183.69 \text{ g/mol}} \]
\[ \approx 0.0188 \text{ moles} \]

### Step 3: Determine moles of Sr used
Since each mole of SrCl2 produces one mole of SrSO4, the moles of Sr used will also be approximately 0.0188 moles.

### Step 4: Calculate the equivalent weight of Sr
The equivalent weight of an element can be calculated using the formula:
\[ \text{Equivalent weight} = \frac{\text{Atomic mass}}{\text{n factor}} \]

For Sr in SrSO4, the oxidation state is +2, which means the n factor is 2.

Using the atomic mass of Sr (87.62 g/mol):
\[ \text{Equivalent weight of Sr} = \frac{87.62 \text{ g/mol}}{2} \]
\[ = 43.81 \text{ g/equiv} \]

### Conclusion
The equivalent weight of Strontium (Sr) is approximately 43.81 g/equiv.

### Final Answer
The equivalent weight of Sr is approximately **43.8 g/equiv**.

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