What is the boiling point of a solution of 1.20g of glycerol, C3H8O3, in 20.0 g of water? What is the freezing point?

TB solution = 100.3315 °C

T freezing solution = -1.209 °C

Given

1.2 g glycerol

20 g water

Required

the boiling point

the freezing point

Solution

ΔTB = KB . m

ΔTF = KF . m

mass water = 20 g = 0.02 kg

Kb water = 0.51 °C/m

KF water = 1.86 °C/m

mol glycerol :

= 1.2 g : 92,09382 g/mol

= 0.013

molal  : the number of moles of solute in 1 kg of solvent  

molal of solution :

= 0.013 : 0.02

= 0.65 m

ΔTB = KB . m

ΔTB = TB solution - TB solvent  

TB solution - 100 = 0.51 x 0.65

TB solution - 100 = 0.3315

TB solution = 100.3315 °C

ΔTF = KF . m

ΔT = KF . m

ΔTF = 1.86 x 0.65

ΔTF = 1.209

T_f  = -1.209 °C

hi my dude or dudette

a. energy is absorbed in the reaction.

if in a given reaction the energy of products is greater than the energy of reactants it means that reaction is endothermic and during the reaction, the energy is absorbed by reactants. in an endothermic reaction, the products have more stored chemical energy than the reactants. and in this reaction the reactants are more stable than products before the compounds with less energy are always more stable.