First, you have to find the molality of particles in the solution using the following formula:
Delta T= K_fxxmxx i
where:
m -> " is the molality of the solute expressed in " (mol)/(Kg)
Delta T ->" is the difference between the freezing point of the "
"solution"\ (T_f)" and that of the pure solvent "(T_i) .
\Delta T = T_f- T_i
i -> " is the Van'to Hoff factor. It is related to the number of"
"particles ( or ions) of the solute dissolved in the solution."
K_f->"is the freezing point depression constant of the solvent."
"in aqueous solutions the solvent is water and " K_f = 1.86\°C*Kg*mol^-1
Delta T= - K_f\ xxunderbrace (mxxi)_("molality of particles")
Delta T= - K_f\ xxunderbrace (mxxi)_(m')
Delta T= - K_fxx\ m'
m'=- (Delta T)/ K_f
m'=- ((T_f-T_i))/ K_f
m'= -((-2.79 °C-0°C))/(1.86°C.Kg.mol^-1)
m'= (2.79 cancel (°C))/(1.86cancel (°C.)Kg.mol^-1)
m'= 1.5 \ (mol)/(Kg)
Once the molality of the solute is determined, the boiling point of the solution could be determined.
The change in the boiling point is determined using the following formula:
Delta T= K_b\ xx mxx i
Where K_b is the boiling point elevation constant for water.
K_b = 0.510 \°C* Kg*mol^-1
Delta T= K_b\ xxunderbrace (mxx i)_(m')
Delta T= K_b\ xxm'
Delta T= 0.510 \°C* Kg*mol^-1 xx1.5 \ (mol)/(Kg)
Delta T= 0.765 °C
Delta T = T_f- T_i
Note that T_i is the normal boiling point of water and T_f is the temperature at which the solution boils.
0.765 °C= T_f- 100.000 °C
T_f=100.765 °C
