To determine which of the given options contains the same number of molecules as $1.0 \mathrm{~g}$ of $\mathrm{H}_2$, we need to use the concept of moles and Avogadro's number. First, let's calculate the number of moles in $1.0 \mathrm{~g}$ of $\mathrm{H}_2$.

The molar mass of $\mathrm{H}_2$ is $2 \mathrm{~g/mol}$. Therefore, the number of moles of $\mathrm{H}_2$ in $1.0 \mathrm{~g}$ can be calculated as:

$$ \text{Number of moles of } \mathrm{H}_2 = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{1.0 \mathrm{~g}}{2 \mathrm{~g/mol}} = 0.5 \mathrm{~mol} $$

Next, we need to find out which of the given options corresponds to the same number of moles. Let's calculate the number of moles for each option:

Option A: $$14 \mathrm{~g} \text{ of } \mathrm{N}_2$$.

The molar mass of $\mathrm{N}_2$ is $28 \mathrm{~g/mol}$. The number of moles of $\mathrm{N}_2$ in $14 \mathrm{~g}$ is:

$$ \text{Number of moles of } \mathrm{N}_2 = \frac{14 \mathrm{~g}}{28 \mathrm{~g/mol}} = 0.5 \mathrm{~mol} $$

Option B: $$18 \mathrm{~g} \text{ of } \mathrm{H}_2 \mathrm{O}$$.

The molar mass of $$\mathrm{H}_2 \mathrm{O}$$ is $18 \mathrm{~g/mol}$. The number of moles of $$\mathrm{H}_2 \mathrm{O}$$ in $18 \mathrm{~g}$ is:

$$ \text{Number of moles of } \mathrm{H}_2 \mathrm{O} = \frac{18 \mathrm{~g}}{18 \mathrm{~g/mol}} = 1 \mathrm{~mol} $$

Option C: $$16 \mathrm{~g} \text{ of } \mathrm{CO}$$.

The molar mass of $\mathrm{CO}$ is $28 \mathrm{~g/mol}$. The number of moles of $\mathrm{CO}$ in $16 \mathrm{~g}$ is:

$$ \text{Number of moles of } \mathrm{CO} = \frac{16 \mathrm{~g}}{28 \mathrm{~g/mol}} = 0.571 \mathrm{~mol} $$

Option D: $$28 \mathrm{~g} \text{ of } \mathrm{N}_2$$.

The molar mass of $\mathrm{N}_2$ is $28 \mathrm{~g/mol}$. The number of moles of $\mathrm{N}_2$ in $28 \mathrm{~g}$ is:

$$ \text{Number of moles of } \mathrm{N}_2 = \frac{28 \mathrm{~g}}{28 \mathrm{~g/mol}} = 1 \mathrm{~mol} $$

Comparing these calculated values, we can see that $0.5 \mathrm{~mol}$ of $\mathrm{N}_2$ (Option A) is equal to $0.5 \mathrm{~mol}$ of $\mathrm{H}_2$. Therefore, the correct answer is:

Option A: $14 \mathrm{~g}$ of $\mathrm{N}_2$.