Kb = 0.8 × 10–5
Kf = 1.6 × 10–4
Keq = Kf / Kb = 1.6X10-4
/ 0.8X10-5 = 20
molality = number of moles of solute / weight of solvent (in
kg)
= (1.8/180) / 0.25 = 0.01/0.25 = 0.04M
Compound : No. of
valance electron on the central atom
XeF2 : 10
AlCl3 : 6
SF6 : 12
SCl2 : 8
In the hydrocarbon
state of hybridisation of carbon 1,2,3,4 and 7 are in the
following sequence.
The above reaction is an example of which of the following
E = hν = hc/λ
=6.626 × 1034 J s × 3 ×108
m s1 / 45 ×109 m
= 4 .42 ×1018 J