Question

( triangle T_{f}=i cdot k f cdot r n )
( i rightarrow operatorname{van} t ) Hodt bactor
( 0 cdot 558=i cdot 1 cdot 86 cdot(0 cdot 1) )
( i i=3 int i+operatorname{men} p quad d i s sin alpha t i 0 n )
is occursng here (for i ( >1 ) always dissociation ocurs
( 3=frac{k}{1}left[begin{array}{c}because k=n u m b e r d t text { particle } g a text { tter dissociation }]end{array}right. )
No of miles of particles formed abter dissociation
[
2
]
From this case it can beseen theet
compound ig coclg: SNH ( _{3} ) since it gives one mote of ( left.[cos (n+1)]_{s} c lright] ) cl ( _{2} )
and 2 mole of cl on diss ocia rion
Hence the compound is ( operatorname{coc} l_{3} ) s s Nh, and ( [x=5] )
opion ( (b) ) is correct dear.

# point of the solution Sallple of Co NH CH CHÁNH) point of the solution is -0.87C. How ma of water is 1.86°C/molal (a) 2 (a) LL12) 664 (d) [Pt(H20) 2014].4H20 (C) [Pt(H20)3C13]C1-3H20 23. A complex is represented as CoCxNH. Its 0.1 molal solution in water showS AT = 0.558 K Kf for H, is 1.86 K molality -1. Assuming 100% ionisation of complex and co-ordination number of Co is six, calculate formula of complex: (b) [Co(NH2)5Cl]Cl2 (a) [CO(NH3)]C13 (d) none of these (c) [Co(NH3)4C12]CI inne will be highest for: (6) 3 84. An aqueous solution contains 3% and 1 freezing point of solution? (K = 1.86 (a) -1.172°C (b) -2.27°C

Solution