`6.022 xx 10^(22)` molecules of `N_(2)` at NTP will occupy a volume of

Options

A. 22.4 litres
B. 2.24 litres
C. 6.02 litres
D. 6.02 mL

Answer

To solve the problem of how much volume `6.022 x 10^(22)` molecules of `N2` will occupy at NTP (Normal Temperature and Pressure), we can follow these steps:

### Step 1: Calculate the number of moles
To find the number of moles from the number of molecules, we use Avogadro's number, which is `6.022 x 10^(23)` molecules per mole.

\[ \text{Number of moles} = \frac{\text{Number of molecules}}{N_A} = \frac{6.022 \times 10^{22}}{6.022 \times 10^{23}} \]

### Step 2: Simplify the calculation
When we divide `6.022 x 10^(22)` by `6.022 x 10^(23)`, we can simplify it:

\[ \text{Number of moles} = \frac{6.022 \times 10^{22}}{6.022 \times 10^{23}} = \frac{1}{10} = 0.1 \text{ moles} \]

### Step 3: Calculate the volume occupied by the moles at NTP
At NTP, 1 mole of any gas occupies a volume of `22.4 liters`. Therefore, to find the volume occupied by `0.1` moles of `N2`, we use the formula:

\[ \text{Volume} = \text{Number of moles} \times 22.4 \text{ L} \]

Substituting the number of moles we calculated:

\[ \text{Volume} = 0.1 \text{ moles} \times 22.4 \text{ L/mole} \]

### Step 4: Perform the multiplication
Now we can calculate the volume:

\[ \text{Volume} = 2.24 \text{ L} \]

### Final Answer
Thus, `6.022 x 10^(22)` molecules of `N2` at NTP will occupy a volume of `2.24 liters`.

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