A mole of any substance contains `6.023xx10^(23)` particles. The particles may be atom, molecule ions, electron, proton or neutron. One mole of atom is equal to 1 gm -atom which is equal to atomic weight of atom. 1 gm molecule of any gas is 1 mole of gas whose volume is 22.4 litre at N.T.P.
The volume of `3.011xx10^(23)` atoms of hydrogen gas at N.T.P. is

Options

A. 1.12 litres
B. 2.24 litres
C. 2.8 litres
D. 5.6 litres

Answer

To find the volume of `3.011 x 10^(23)` atoms of hydrogen gas at N.T.P. (Normal Temperature and Pressure), we will follow these steps:

### Step 1: Determine the number of molecules from the number of atoms.
Hydrogen gas (H₂) is diatomic, meaning each molecule of hydrogen gas consists of 2 atoms. Therefore, to find the number of molecules from the number of atoms, we divide the number of atoms by 2.

\[ \text{Number of molecules} = \frac{3.011 \times 10^{23} \text{ atoms}}{2} = 1.5055 \times 10^{23} \text{ molecules} \]

### Step 2: Calculate the number of moles of hydrogen gas.
Using Avogadro's number, which is approximately \(6.023 \times 10^{23}\) molecules per mole, we can find the number of moles of hydrogen gas.

\[ \text{Number of moles} = \frac{\text{Number of molecules}}{\text{Avogadro's number}} = \frac{1.5055 \times 10^{23}}{6.023 \times 10^{23}} \approx 0.25 \text{ moles} \]

### Step 3: Calculate the volume of hydrogen gas at N.T.P.
At N.T.P., 1 mole of any gas occupies a volume of 22.4 liters. Therefore, the volume occupied by 0.25 moles of hydrogen gas can be calculated as follows:

\[ \text{Volume} = \text{Number of moles} \times 22.4 \text{ L/mole} = 0.25 \text{ moles} \times 22.4 \text{ L/mole} = 5.6 \text{ liters} \]

### Final Answer:
The volume of `3.011 x 10^(23)` atoms of hydrogen gas at N.T.P. is **5.6 liters**.

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