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

explanation:

let the different atoms be x and y. each with four protons and same mass will be represented as follows.

mass number is number of protons + number of neutrons.

            and  

since, it is given that one has a negative charge while the other is neutral. therefore, there will be increase in number of protons and the atom will be represented as follows.

according to periodic table, mass number of beryllium is 9 and it has atomic number 4 so, the atoms will be and .  

the protons and neutrons will be located inside the nucleus whereas the electrons will be revolving around the nucleus.