Carboxylic Acid and Acid DerivativesEasy
Question
A 0.020 m solution of each of the following compounds is prepared. Which solution would you expect to freeze at -0.149oC ? Kf (water) = 1.86 K kg. mol-1 :-
where, py = pyridine (unidentate),
en = ethylenediamine (bidentate),
and EDTA = ethylenediaminetetraacetic acid
(hexadentate)
where, py = pyridine (unidentate),
en = ethylenediamine (bidentate),
and EDTA = ethylenediaminetetraacetic acid
(hexadentate)
Options
A.[Co(en)2Cl2]Cl
B.Na[Co(EDTA)]
C.[Cr(py)5Cl]Cl2
D.[Cr(NH3)6]Cl3
Solution
As we know,
ᐃTf = i.Kf.m
So, Tof - Tf = Kf.m
0.149 = i × (1.86) × (0.020)
Hence, i = 4
∴ the compound having the value of i = 4 is
i.e., [Cr(NH3)]Cl3.
Which will have a freezing point of -0.149oC.
ᐃTf = i.Kf.m
So, Tof - Tf = Kf.m
0.149 = i × (1.86) × (0.020)
Hence, i = 4
∴ the compound having the value of i = 4 is
i.e., [Cr(NH3)]Cl3.
Which will have a freezing point of -0.149oC.
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