Arrange the following in decreasing order of number of molecules contained in:

(A) \(16\text{ g of }O_2\)

(B) \(16\text{ g of }CO_2\)

(C) \(16\text{ g of }CO\)

(D) \(16\text{ g of }H_2\)

Choose the correct order from the options given below:

D, C, A, B

The correct answer is option 2. D, C, A, B.

To compare the number of molecules contained in the given masses of different gases, we can use the concept of molar mass and Avogadro's number.

The molar mass of \(O_2\) (oxygen gas) is \(32\) g/mol (since the molar mass of oxygen is \(16\) g/mol, and \(O_2\) is \(2\) times that).

The molar mass of \(CO_2\) (carbon dioxide) is \(44\) g/mol (carbon \(12\) g/mol, oxygen \(16 \times 2\) g/mol).

The molar mass of \(CO\) (carbon monoxide) is \(28\) g/mol (carbon \(12\) g/mol, oxygen \(16\) g/mol).

The molar mass of \(H_2\) (hydrogen gas) is \(2\) g/mol.

Now, using Avogadro's number (\(6.022 \times 10^{23}\) molecules/mol), we can calculate the number of molecules in each case:

(A) \(16\) g of \(O_2\) contains \(16 \, \text{g} \times \frac{1 \, \text{mol}}{32 \, \text{g}} \times (6.022 \times 10^{23}) = 3.011\times 10^{23}\) molecules.

(B) \(16\) g of \(CO_2\) contains \(16 \, \text{g} \times \frac{1 \, \text{mol}}{44 \, \text{g}} \times (6.022 \times 10^{23}) = 2.189\times 10^{23}\) molecules.

(C) \(16\) g of \(CO\) contains \(16 \, \text{g} \times \frac{1 \, \text{mol}}{28 \, \text{g}} \times (6.022 \times 10^{23}) = 3.44\times 10^{23}\) molecules.

(D) \(16\) g of \(H_2\) contains \(16 \, \text{g} \times \frac{1 \, \text{mol}}{2 \, \text{g}} \times (6.022 \times 10^{23}) = 48.176\times 10^{23}\) molecules.

Now, comparing the values, we find that the decreasing order of the number of molecules contained is \(D > C > A > B\)

So, the correct order is option 2: \(D, C, A, B\).