Estimate the freezing and normal boiling points of 0.25 m aqueous solutions of nh4no3 nicl3 al2so43

The items are answered below and are numbered separately for each compound. 

The freezing point of impure solution is calculated through the equation,
     Tf = Tfw - (Kf)(m)

where Tf is the freezing point, Tfw is the freezing point of water, Kf is the freezing point constant and m is the molality. For water, Kf is equal to 1.86°C/m. In this regard, it is assumed that m as the unit of 0.25 is molarity.

1. NH4NO3
    Tf = 0°C - (1.86°C/m)(0.25 M)(2) = -0.93°C

2. NiCl3 
    Tf = 0°C - (1.86°C/m)(0.25 M)(4) = -1.86°C

3. Al2(SO4)3
      Tf = 0°C - (1.86 °C/m)(0.25 M)(5) = -2.325°C

For boiling points, 
    Tb = Tbw + (Kb)(m)
For water, Kb is equal to 0.51°C/m.

1. NH4NO3
     Tb = 100°C + (0.51°C/m)(0.25 M)(2) = 100.255°C

2. NiCl3
     Tb = 100°C + (0.51°C/m)(0.25 M)(4) = 100.51°C

3. Al2(SO4)3
    Tb = 100°C + (0.51°C/m)(0.25 M)(5) = 100.6375°C

the answer is true.

this statement is true.