The freezing point of an aqueous solution is -2.79°C. What is the boiling point of this solution?

1 Answer
Jun 16, 2016

Answer:

# 100.765 °C #

Explanation:

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 #