Chemistry NCERT Exemplar Solutions Class 12th Chapter Five

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A
alok kumar singh

Contributor-Level 10

Moles of Fe3O4 =   4 . 6 4 * 1 0 3 2 3 2 = 2 0

              Moles of CO =   2 . 5 2 * 1 0 3 2 8 = 9 0

              So limiting reagent = Fe3O4

              So moles of Fe formed = 60

              Weight of Fe = 60 * 56 = 3360

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Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

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V
Vishal Baghel

Contributor-Level 10

No. of compounds containing asymmetric carbon are

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V
Vishal Baghel

Contributor-Level 10

Millie q. of H2SO4 used by NH3 = 12.5 * 1 * 2 = 25

So millimoles of N = 25

Moles of N = 25 * 10-3

Wt of N = 14 * 25 * 10-3

% of N =    1 4 * 2 5 * 1 0 3 0 . 5 5 * 1 0 0 = 6 3 . 6 6 6 4

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Vishal Baghel

Contributor-Level 10

 3MnO42+4H+2MnO4+MnO2+2H2O

No. of unpaired electron in Mn7+ = 0

μ=0B.M

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Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

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V
Vishal Baghel

Contributor-Level 10

 ln (K2K1)=EaR (1T11T2)

ln (K2K1)=5326118.3* (10310*300)

Where, K2 is at 310 K and K1 at 300K

ln (K2K1)=6.9=3*ln10

ln (K2K1)=ln103

K2 = K1 * 103

K1 = K2 * 10-3

 x = 1

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Vishal Baghel

Contributor-Level 10

At anode (oxidation)

2H2O O2 (g) + 4H+ + 4e-

At cathode (Reduction)

2H+ + 2e- H2 (g)

No. of gm- equivalents = i*t96500=0.1*2*60*6096500=0.00746

VO2=0.007464*22.7=0.0423L

VH2=0.007462*22.7=0.0846L Vtotal127mL

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Vishal Baghel

Contributor-Level 10

Mass = d * v = 1.02 * 1.2 = 1.224gm

Moles of acetic acid = 0.0204 moles in 2L

So molality = 0.0102 mol/kg

Δ T f = i * K f * m

i = 1 + a for acetic acid

0.0198  = (1 + a) * 1.85 * 0.0102

α = 0.04928 = 5%

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Vishal Baghel

Contributor-Level 10

85gm NH3 = 5 moles of NH3

Enthalpy change for 1 mol = 23.4 kJ

Then enthalpy change for 5 mol = 23.4 * 5 = 117 kJ

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