$$\Delta T_f = i K_f m$$
where:
* \(\Delta T_f\) is the freezing point depression in Kelvin (K)
* \(i\) is the van't Hoff factor (a measure of the number of particles that a solute dissociates into in solution)
* \(K_f\) is the freezing point depression constant of the solvent (in this case, water, which has a \(K_f\) of 1.86 K m\(^-1\))
* \(m\) is the molality of the solution (in this case, the concentration of the nitrate in mol/kg)
We are given that \(\Delta T_f = -2.79\) K and \(K_f = 1.86\) K m\(^-1\). We can calculate the molality of the solution by rearranging the equation above:
$$m = \frac{\Delta T_f}{i K_f}$$
We do not know the van't Hoff factor, but we can assume that the nitrate dissociates into three ions in solution (i.e., one nitrate ion and two sodium ions). In this case, \(i = 3\).
Substituting the values we know into the equation, we get:
$$m = \frac{-2.79 \text{ K}}{(3)(1.86 \text{ K m}^{-1})}$$
$$m = -0.498 \text{ m}$$
The negative sign indicates that the solution is freezing at a lower temperature than pure water, which is expected since the nitrate is a solute. The concentration of the nitrate in solution is therefore 0.498 mol/kg.
