To identify the correct option, we first need to understand the relationship between the freezing point depression ($$\Delta \mathrm{T}_{\mathrm{f}}$$) and the molal freezing point depression constant ($$\mathrm{K}_{\mathrm{f}}$$) of the solvent. This relationship is given by the colligative property equation for freezing point depression:

$$ \Delta \mathrm{T}_{\mathrm{f}} = \mathrm{K}_{\mathrm{f}} \times m $$

where:

$$\Delta \mathrm{T}_{\mathrm{f}}$$ is the freezing point depression,
$$\mathrm{K}_{\mathrm{f}}$$ is the cryoscopic constant (also known as the molal freezing point depression constant), and
$$m$$ is the molality of the solution.

The key to solving this problem is noting that the values for $$\Delta \mathrm{T}_{\mathrm{f}}$$ and $$\mathrm{K}_{\mathrm{f}}$$ need to be the same, which means:

$$ \Delta \mathrm{T}_{\mathrm{f}} = \mathrm{K}_{\mathrm{f}} \Rightarrow \mathrm{K}_{\mathrm{f}} = m $$

Thus, the molality of the solution needs to be 1 mol/kg since the only situation where the equality holds is if the molality $$m$$ is equal to 1.

Now, let's analyze the options:

Option A: $$1 \mathrm{~m}$$ - This represents a concentration of 1 molal, which means 1 mole of solute per 1 kilogram of solvent.
Option B: $$1 \mathrm{~M}$$ - This represents a concentration of 1 molar, which is 1 mole of solute per 1 liter of solution. Molarity and molality are not the same and the density of the solution would matter to convert between the two.
Option C: $$1 \mathrm{~N}$$ - This represents a 1 normal solution, which is related to the equivalent concept of moles of reactive species. Normality can vary based on the equivalent factor of the solute, and is not purely a measure of the number of moles of solute per liter of solution.
Option D: $$\frac{\mathrm{N}}{10}$$ - This represents one-tenth the normality of the solution, which would be equivalent to a 0.1 N solution.

From the analysis above, we can conclude that the correct option is A, as it provides a molality of 1 which is necessary for $$\Delta \mathrm{T}_{\mathrm{f}}$$ to equal $$\mathrm{K}_{\mathrm{f}}$$ when they are numerically the same:

$$\Delta \mathrm{T}_{\mathrm{f}} = \mathrm{K}_{\mathrm{f}} = 1 \mathrm{~m}$$

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