The number of N atoms in 645 g of `C_(7)H_(5)N_(3)O_(6) ` is `x xx 10^(21)` . The value of x is ________. `(N_(A) = 6.02 xx 10^(23)mol^(-1))` (Nearest Integer )

Explanation

To find the number of nitrogen atoms in 645 g of \( C_7H_5N_3O_6 \), we can follow these steps:

### Step 1: Calculate the molar mass of \( C_7H_5N_3O_6 \)

The molar mass can be calculated using the atomic masses of carbon (C), hydrogen (H), nitrogen (N), and oxygen (O):

- Carbon (C): 12.01 g/mol
- Hydrogen (H): 1.008 g/mol
- Nitrogen (N): 14.01 g/mol
- Oxygen (O): 16.00 g/mol

The molar mass of \( C_7H_5N_3O_6 \) is calculated as follows:

\[ \text{Molar mass} = (7 \times 12.01) + (5 \times 1.008) + (3 \times 14.01) + (6 \times 16.00) \]

Calculating each part:

\[ = 84.07 + 5.04 + 42.03 + 96.00 = 227.14 \text{ g/mol} \]

### Step 2: Calculate the number of moles of \( C_7H_5N_3O_6 \) in 645 g

Using the formula:

\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]

Substituting the values:

\[ \text{Number of moles} = \frac{645 \text{ g}}{227.14 \text{ g/mol}} \approx 2.84 \text{ moles} \]

### Step 3: Calculate the number of nitrogen atoms

Each molecule of \( C_7H_5N_3O_6 \) contains 3 nitrogen atoms. Therefore, the total number of nitrogen atoms can be calculated as follows:

\[ \text{Total nitrogen atoms} = \text{Number of moles} \times \text{Avogadro's number} \times \text{Number of nitrogen atoms per molecule} \]

Substituting the values:

\[ \text{Total nitrogen atoms} = 2.84 \text{ moles} \times 6.022 \times 10^{23} \text{ atoms/mole} \times 3 \]

Calculating this:

\[ = 2.84 \times 6.022 \times 10^{23} \times 3 \approx 5.13 \times 10^{24} \text{ nitrogen atoms} \]

### Step 4: Express the number of nitrogen atoms in the form \( x \times 10^{21} \)

To express \( 5.13 \times 10^{24} \) in the form \( x \times 10^{21} \):

\[ 5.13 \times 10^{24} = 51.3 \times 10^{21} \]

Thus, the value of \( x \) is approximately 51.3. Rounding to the nearest integer gives:

\[ x \approx 51 \]

### Final Answer

The value of \( x \) is **51**.

---