The freezing point depression and boiling point elevation of a solution are colligative properties, which depend on the number of particles in solution, not on their chemical nature. The freezing point depression and boiling point elevation of a solution are given by the following equations:

$\Delta T_f= K_f m$

$\Delta T_b=K_b m$

where $\Delta T_f$ is the freezing point depression, $\Delta T_b$ is the boiling point elevation, $K_f$ is the freezing point depression constant for the solvent ($1.86^\circ C/m$ for water), $K_b$ is the boiling point elevation constant for the solvent ($0.512^\circ C/m$ for water) and $m$ is the molality of the solution.

To calculate the freezing point depression and boiling point elevation of a 21.2 g NaCl in 135mL water solution, we first need to calculate the molality of the solution.

$m=\frac {moles \ of \ NaCl}{kg \ of \ water}$

First we need to convert grams of NaCl to mols:

$M NaCl = \frac{ 21.2 \ g}{58.44 g/mol} = 0.363 mol$

The mass in Kg of the solvent (water) is:

$$135 \ g \ H_2 O \times \frac{1 Kg}{1000 \ g} = 0.135 Kg$$

Therefore the molality is:

$$m= \frac{0.363 \ mol}{0.135 \ Kg}= 2.69 $$

Now we can calculate the freezing point depression and boiling point elevation:

$\Delta T_f= K_f m = (1.86 ^\circ C/m) (2.69 m) = 5.006^\circ C$

$\Delta T_b=K_b m = (0.512 ^\circ C/m) (2.69 m) = 1.38^\circ C$

Finally we calculate the new freezing and boiling points:

Freezing point: $0^\circ C - 5.006^\circ C$ \(=-5.006 ^oC \)

Boiling point: $100^\circ C + 1.38^\circ C$ \(=101.38 ^oC \)