The boiling point and freezing point has to be calculated. Concept Introduction: The depression in freezing point, the elevation of boiling point and osmotic pressure are together known as colligative properties. The elevation in boiling point can be given by the equation, ΔT=K b m solute Where, ΔT = change in boiling point elevation K b = molal boiling point elevation constant m solute = molality of solute The depression in freezing point can be given by the equation, ΔT=K f m solute Where, ΔT =change in freezing point depression K f = molal freezing point depression constant m solute = molality of solute
The boiling point and freezing point has to be calculated. Concept Introduction: The depression in freezing point, the elevation of boiling point and osmotic pressure are together known as colligative properties. The elevation in boiling point can be given by the equation, ΔT=K b m solute Where, ΔT = change in boiling point elevation K b = molal boiling point elevation constant m solute = molality of solute The depression in freezing point can be given by the equation, ΔT=K f m solute Where, ΔT =change in freezing point depression K f = molal freezing point depression constant m solute = molality of solute
Solution Summary: The author explains that the boiling point and the freezing point are together known as colligative properties. The molarity of ionized Formic acid solution is calculated by the equation.
Interpretation: The boiling point and freezing point has to be calculated.
Concept Introduction: The depression in freezing point, the elevation of boiling point and osmotic pressure are together known as colligative properties.
The elevation in boiling point can be given by the equation,
ΔT=Kbmsolute
Where,
ΔT= change in boiling point elevation
Kb = molal boiling point elevation constant
msolute = molality of solute
The depression in freezing point can be given by the equation,
Part II. Given two isomers: 2-methylpentane (A) and 2,2-dimethyl butane (B) answer the following:
(a) match structures of isomers given their mass spectra below (spectra A and spectra B)
(b) Draw the fragments given the following prominent peaks from
each spectrum:
Spectra A m/2 =43 and 1/2-57
spectra B m/2 = 43
(c) why is 1/2=57 peak in spectrum A more intense compared
to the same peak in spectrum B.
Relative abundance
Relative abundance
100
A
50
29
29
0
10
-0
-0
100
B
50
720
30
41
43
57
71
4-0
40
50
60 70
m/z
43
57
8-0
m/z = 86
M
90 100
71
m/z = 86
M
-O
0
10 20 30
40 50
60
70
80
-88
m/z
90
100
Part IV. C6H5 CH2CH2OH is an aromatic compound which was subjected to Electron Ionization - mass
spectrometry (El-MS) analysis. Prominent m/2 values: m/2 = 104 and m/2 = 9) was obtained.
Draw the structures of these fragments.
For each reaction shown below follow the curved arrows to complete each equationby showing the structure of the products. Identify the acid, the base, the conjugated acid andconjugated base. Consutl the pKa table and choose the direciton theequilibrium goes. However show the curved arrows. Please explain if possible.