10 g of `CaCO_(3)` contains :

Options

A. 10 moles of `CaCO_(3)`
B. 0.1 g atom of Ca
C. `6 xx 10^(23)` atoms of Ca
D. 0.1 equivalent of Ca

Explanation

To solve the question "10 g of CaCO₃ contains:", we will go through the calculations step by step.

### Step 1: Calculate the molar mass of CaCO₃
The molar mass of calcium carbonate (CaCO₃) can be calculated by adding the atomic masses of its constituent elements:
- Calcium (Ca): 40 g/mol
- Carbon (C): 12 g/mol
- Oxygen (O): 16 g/mol (and there are 3 oxygen atoms in CaCO₃)

So, the molar mass of CaCO₃ is:
\[ \text{Molar mass of CaCO₃} = 40 + 12 + (3 \times 16) = 40 + 12 + 48 = 100 \text{ g/mol} \]

### Step 2: Calculate the number of moles of CaCO₃ in 10 g
Using the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
We can find the number of moles of CaCO₃:
\[ \text{Number of moles of CaCO₃} = \frac{10 \text{ g}}{100 \text{ g/mol}} = 0.1 \text{ moles} \]

### Step 3: Calculate the number of gram atoms of calcium
In one mole of CaCO₃, there is one mole of calcium (Ca). Therefore, the number of moles of calcium in 0.1 moles of CaCO₃ is also 0.1 moles. Since 1 mole of an element is equivalent to 1 gram atom, we have:
\[ \text{Gram atoms of calcium} = 0.1 \text{ moles of Ca} = 0.1 \text{ gram atoms of Ca} \]

### Step 4: Calculate the number of atoms of calcium
Using Avogadro's number (approximately \(6.022 \times 10^{23}\) atoms/mole), we can calculate the number of atoms of calcium:
\[ \text{Number of atoms of calcium} = 0.1 \text{ moles} \times 6.022 \times 10^{23} \text{ atoms/mole} = 6.022 \times 10^{22} \text{ atoms} \]

### Step 5: Calculate the equivalents of calcium
To find the equivalents of calcium, we need to consider the valency of calcium in CaCO₃. Calcium has a valency of 2 (Ca²⁺). The formula for equivalents is:
\[ \text{Equivalents} = \text{Number of moles} \times \text{n-factor} \]
Where the n-factor for calcium is 2:
\[ \text{Equivalents of calcium} = 0.1 \text{ moles} \times 2 = 0.2 \text{ equivalents} \]

### Summary of Results
- Number of moles of CaCO₃: 0.1 moles
- Gram atoms of calcium: 0.1 gram atoms
- Number of atoms of calcium: \(6.022 \times 10^{22}\) atoms
- Equivalents of calcium: 0.2 equivalents

### Conclusion
From the options provided:
- 10 moles of CaCO₃: **Incorrect**
- 0.1 gram atom of calcium: **Correct**
- \(6 \times 10^{23}\) atoms of calcium: **Incorrect**
- 0.1 equivalent of calcium: **Incorrect**

Thus, the correct answer is **0.1 gram atom of calcium**.