Chemistry NCERT Exemplar Solutions Class 12th Chapter One

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P
Payal Gupta

Contributor-Level 10

 Molality=molesofsolute*1000wtofsolvent (gm)

Molality=35*1000*1.4636.5*100 = 14.0 M

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V
Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

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V
Vishal Baghel

Contributor-Level 10

  [Co (H2O)6]Cl2Co2+3d7t2g5eg2

Number of unpaired electrons = 3

[Cr (H2O)6]Cl3Cr3+3d3t2g3eg0

Number of unpaired electrons = 3

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Vishal Baghel

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 t1/2α [Reactant]t=0 (1n),  where n = order of reaction.

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Vishal Baghel

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Fe3O4 + 8H+ + 8e -> 3Fe + 4H2O

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Vishal Baghel

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K e q = ( 1 2 * 1 0 1 5 ) 1 / 2 = 2 . 2 4 * 1 0 ? 8

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Vishal Baghel

Contributor-Level 10

Remaining volume of solution = 400 ml

Mass of HNO3 = 25.2 – 11.5 = 13.7

Molarity =  (13.7/63) (400/1000) = 0.54 M = 54 * 10-2 M

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Vishal Baghel

Contributor-Level 10

 C (s)+O2 (g)CO2 (g)Δng=0

qC=20*2=40kJ

Number of moles of C = 2.412=0.2moles

ΔHC=40kJ0.2mole=200kJ/mole

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V
Vishal Baghel

Contributor-Level 10

Xe = Xn

MeVe = MnVn

VeVn=MnMe1758.24

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