OK Goiitians here are the solns to the chem q bank

in case u want the question bank only then plz check my messages in my profile

u can find the questions there

this thread  consists of questions with their solns...........

 

 

 

100 ml of a liquid contained in an insulated container at a presssure of 1 bar. The pressure is steeply increased to 100 bar.
The volume of the liquid is decreased by  1ml at this constant pressure . Find the value of H and U.......

JEE 2004

U=q+W

for adiabatic process q=0; hence U=W

W= -p(v)

U= -100(99-100)=100

U=100 bar ml

H=U+(PV)

so H =100+(100x99-1x100)=9900 bar ml

A first order reaction A-B, requires activation energy of 70 Kj/mol
. When a 20 % solution of A was kept at 25 degree celsius for 20 minutes, 25% decomposition took place. What will be the percentage decomposition in the same time in a 30 % solution maintained at 40 degree celsius?
Assume that activation energy remains constant in this range of temperature
....
JEE 1993

According to Arrhenius eqn

k=Ae^-Ea/RT

by solving this eqn w r t the range of T1 and T2 absolute temperature , we get the foll eqn:

 

Ea=[2.303R T1 T2xlog₁₀(k2/k1)]/T2-T1

on solving this we get k2/k1=3.872

ratio of rate constant at two different temp

for first order reacn

k1= 2.303/t x log₁₀a/(a-x)

substituting the values we get k1= 0.014386 min^-1

now k2/k1 is known from above eqn(3.872)

so we can get k2

k2=0.05571 min^-1

now 0.05571 =[2.303 log (100/100-x)]/20

so we get x as 67.17%

so 67.17 % decomposition takes place at 40 ^oc.

OK now an easy one
The molecule having linear structure is
A. CO2
B. NO2
C. SO2
D. SiO2

ans is A.

CO2 has sp hybridisation

On the basis of ground state electronic configuration arrange the following molecules in increasing O-O bond length order

KO2, O2, O2[AsF6]

 

Ans is see the bond order is inversely proportional to bond length

bond order for O2+, O2,O2- can be found out from molecular orbital thoery

 

accordingly bond length will be in order of O2+<O2<O2-

Two students use same stock solution of ZnSO4 and CuSO4
. The emf of one cell is 0.03 V higher than the other. The conc of CuSO4 in the cell with higher emf value is 0.5 M. Find the cocn of CuSO4 in the other cell.
2.303 RT/F= 0.06

JEE 2003

 

Daniel cell is

Zn(s)/Zn+2(aq)||Cu+2(aq)/Cu(s)

let there be two Daniel cells with their E2-E1=0.03

and C2=C1

the cell reaction is

Zn(s)+Cu+2(aq)------Zn+2(aq)+Cu(s)

So,

E cell = E^o_cell -(2.303 RT/nF)log 10[Zn+2]/Cu+2

subs the given values

we get the eqn as

E1= E^ocell -0.06/2 log C1/C

E2= E⁰ cell -0.06/2 log C2/0.5

E2-E1=0.06/2(log C2/Cx 0.5/C1) (as C1=C2)

or 0.03 = 0.06/2 log 0.5/C

so C= 0.05 M

Ok from inorganic
Element A burns in nitrogen to give an ionic compound B.
Compound B reacts with water to give C and D.
A solution of C becomes milky on bubbling CO2. Identify A, B , C and D
JEE 1997

3M+N2------------M3N2

M3N2 +6H20

M3N2 + 6H2O----------- 3M(OH2)+2NH3

M(OH2)+CO2-------------MCO3+H20

M can be either Ca or Ba but not Mg as Mg(OH2) has very low sol in water.......