A reddish positive charged sol is obtained by adding small quantity of `FeCl_(3)` solution to freshly prepared and well washed `Fe(OH)_(3)` precipitate. How does it take place?
It is due to adsorptions of `Fe^(3+)` ions on the surface of `Fe(OH)_(3)` which gives colloidal sol.
`Fe(OH)_(3) ("ppt.")+Fe^(3+)("ions adsorbed") rarr [Fe(OH)_(3)]Fe^(3+)` ("colloidal sol.)
Correct Answer - 37.7 g
`DeltaT_(b)=(100.42^(@)C-100)^(@)C=0.42^(@)C=0.42" K",`
`M_(B)=92" g mol"^(-1),W_(A)=500 g =0.5" kg",`
`K_(b)=0.152" k kg mol"^(-1)`
`W_(B)=(M_(B)xxDeltaT_(b)xxW_(A))/(K_(b))=((92g mol^(-1))xx(0.42K)xx(0.5kg))/((0.512" k kg mol"^(-1)))=37.7 g`.
Correct Answer - A`to`p,r ; B`to`p,q ; C`to`p,s ; D`to`q,s,t
(A)`2Bi^(3+)+3H_2StoBi_2S_3darr(black)+6H^(+)`
`Bi^(3+)to3I^(-)toBiI_3darr` (black)
(B)`Cu^(2+)+H_2StoCuSdarr(black)+2H^(+)`
`Cu^(2+)+2SCN^(-)toCu(SCN)_2darr`(black)
which one standing slowly changes to white
`2Cu(SCN)_2to2Cu(SCN)darr(white)+(SCN)_2`
`Cu_2I_2` is white precipitate and `Cu_2[Fe(CN)_6]` brown precipitate...
Correct Answer - A
Marble `(CaCO_(3))` do not react, adsorb, absorb, or dissolve `Br_(2)`.As such there is no change is colour of `Br_(2)`. Remaining dissolves or absorb or adsorb bromine.
`N_(kmNo_(4))=(20)/(1000)xx(1)/(50)`
`n_(Fe^(+2))xxn_(KMnO_(4))`
`=5xx(20)/(1000)xx(1)/(50)=0.002`
`n_(N)_(2)H_(4))` is 10 ml `=(1)/(4) n_(Fe^(+2))=(0.002)/(4)`
`n_(N_(2)H_(4))` is 1L`=(0.002)/(4)xx100=(0.2)/(4)`
mass of `N_(2)H_(6)SO_(4)` in `1L=(0.2)/(4)xx130=6.5gm`
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