MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: R

There are no items in this cart.
Continue Shopping
Menu
Get instant 20% OFF on Online Material.
coupon code: MOB20 | View Course list

  • Complete JEE Main/Advanced Course and Test Series
  • OFFERED PRICE: R 15,000
  • View Details
Get extra R 4,500 off
USE CODE: Venkat30

IIT JEE 2009 Chemistry Paper2 Code 1 Solutions

INTEGER ANSWER TYPE

12.    The oxidation number of Mn in the product of alkaline oxidative fusion of MnO2 is

Sol.   6.

         Below is the reaction

        2MnO2 + 4KOH + O2 --> 2K2MnO4 + 2H2O

     (Potassium manganate)

         O.S. of Mn = +6 in K2MnO4 

13.    The number of water molecule(s) directly bonded to the metal centre in CuSO4 . 5H2O is

Sol.   4.

         Since Cu shows its secondary valency as 4.Therefore , total no. of ligands will be 4 in the complex.

        CuSO4 . 5H2O --> [Cu(H2O)4]SO4 . H2O

        So, water molecules directly attached to Cu are 4.

14.    The coordination number of Al in the crystalline state of AlCl3 is

Sol.   6

        Coordination number of Al is 6. It exists in ccp lattice with 6 coordinate layer structure.

15.    In a constant volume calorimeter, 3.5 g of a gas with molecular weight 28 was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase form 298.0 K to 298.45 K due to the combustion process. Given that the heat capacity of the calorimeter is 2.5 kJ K-1, the numerical value for the enthalpy of combustion of the gas in kJ mol-1 is

Sol.   9

        Energy release at constant volume due to combustion of 3.5 gm of a gas = 2.5 × 0.45

        Hence energy released due to the combustion of 28 gm (i.e., 1 mole) of a gas = 2.5 × 0.45 × 28/3.5 = 9 kJ mol-1

16.    The dissociation constant of a substituted benzoic acid at 25oC is 
1.0 × 10-4. The pH of a 0.01 M solution of its sodium salt is

Sol.   8

        Ka (C6H5COOH) = 1 × 10-4

pH of 0.01 M C6H5COONa

C6H5COO- + H2O -------> C6H5COOH + OH-1

   0.01(1-h)                                 0.01 h            0.01 h

Kh = Kw/Ka =(0.01 h2)/(1-h) 

10-14/10-4 = 10-2 h2/(1-h) (1 – h ≈ 1)

[OH-] = 0.01 h = 0.01 × 10-4 = 10-6

[H+] = 10-8

pH = 8

<< Back || Next >>

  • Complete JEE Main/Advanced Course and Test Series
  • OFFERED PRICE: R 15,000
  • View Details
Get extra R 4,500 off
USE CODE: Venkat30