10. Prototipos y diseños
10.2 Diseños de interfaces (Hi-Fi)
3. A hydrated metallic salt A, light green in colour, gives a white anhydrous residue B after being heated gradually. B is soluble in water and its aqueous solution reacts with NO to give a dark brown compound C. B on strong heating gives a brown residue and a mixture of two gases E and F. The gaseous mixture, when passed through acidified permanganate, discharge the pink colour and when passed through acidified BaCl2 solution, gives a white precipitate. Identify A, B, C, D, E and F.
[IIT-1988]
Sol. The given observations are as follows.
(i)
Gaseous mixture (E) + (F)
acidified KMnO4
BaCl2 solution
Pink colour is discharged White precipitate (iv)
The observation (ii) shows that B must be ferrous sulphate since with NO, it gives dark brown compound according to the reaction
[Fe(H2O)6]2+ + NO → The observation (iii) is
2FeSO4 → observation (iv), namely,
colour
Hence, the various compounds are (A) FeSO4.7H2O (B) FeSO4
(C) [Fe(H2O)5NO]SO4 (D) Fe2O3
(E) and (F) SO2 and SO3
4. Compound (A) is a light green crystalline solid. It gives the following tests :
(i) It dissolves in dilute H2SO4 without evolving any gas.
(ii) A drop of KMnO4 is added to the above solution.
The pink colour disappears.
(iii) Compound (A) is heated strongly. Gases (B) and (C) with pungent smell came out. A brown residue (D) is left behind.
(iv) The gas mixture (B) and (C) is passed into dichromate solution. The solution turns green.
(v) The green solution from step (iv) gives a white ppt. (E) with a solution of Ba(NO3)2.
(vi) Residue (D) from (v) is heated on charcoal in reducing flame. It gives a magnetic substance.
Identify compounds (A) to (E) and predict all the
equations. [IIT-1980]
Sol. The fore said observations may be briefly summarised as follows :
(a) Lightgreen A solid Dil.H2SO4→
KMnO4 Pink colour disappears (b) A →∆ From the last step, one may conclude that brown residue (D) (hence also compound (A)) must be a salt of iron. Since (A) decolourises KMnO4 solution hence it should be a salt of Fe (II). The reactions involved are given below.
MnO4– + 8H+ + 5e– → Mn2+ + 4H2O From observations of (b) and (c), one concludes that compound (A) should be FeSO4 as on heating, it
SO2, gas turns dichromate solution green due to formation of green coloured sulphate of chromium (III), the different equations are,
Cr2O72– + 14H+ + 6e– → 2Cr3+ + 7H2O
5. A substance (X) is soluble in conc. HCl. When to this solution NaOH solution is added, a white precipitate is produced. This precipitate dissolves in excess of NaOH solution giving a strongly reducing solution.
Heating of (X) with sulphur gives a brown powder (Y) which is soluble in warm yellow ammonium sulphide solution. When HCl is added to the latter, a grey precipitate is produced. Heating of (X) in air gives a water soluble compound gives white gelatinous precipitate. Identify the compounds giving the reactions involved.
Sol. The reaction sequence is as follows :
(1) X Conc.HCl→ Dissolves →NaOH White ppt.
NaOH Excess →
Dissolves and the solution is strongly reducing.
(2) X
→∆
S
BrownY (NH4)2Sx→ Dissolves
→
HCl Grey ppt.
(3) X
Heat O2
→
4 2SO H conc.SolubleZ in
Fused NaOH
→
Soluble →H+
White gelatinous ppt.
(a) According to step (2), (X) appears to be tin.
) X
Sn( + 2S →
) Y ( 2
SnS SnS2 + (NH4)2Sx →
te thiostanna .
AmmNH4)2SnS3
( (NH4)2SnS3 + 2HCl →
. ppt Grey2
SnS ↓ + 2NH4Cl + H2S (b) Step (1) and (3) can also be explained, if (X) is tin
) X
Sn( + 2HCl → SnCl2 + H2
SnCl2 + 2NaOH → Sn(OH)2 + 2NaCl SnCl2 + 2NaOH →
reducing
Strongly Na2SnO2 + 2HCl (c) Sn + O2 →Fuse
(Z)2
SnO
SnO2 + 2NaOH Fuse → Na2SnO3 + H2O Na2SnO3 + 2HCl →
ppt.
gelatinous
WhiteH2SnO3↓ + 2NaCl
Brief description: cobalt is a brittle, hard, transition metal with magnetic properties similar to those of iron. Cobalt is present in meteorites. Ore deposits are found in Zaire, Morocco and Canada. Cobalt-60 (60Co) is an artificially produced isotope used as a source of γ rays (high energy radiation). Cobalt salts colour glass a beautiful deep blue colour.
Basic information about and classifications of cobalt : Name : Cobalt
Symbol : Co Atomic number : 27
Atomic weight : 58.933195 (5) Standard state : solid at 298 K Group in periodic table : 9 Group name : (none) Period in periodic table : 4 Block in periodic table: d-block Colour : lustrous, metallic, greyish tinge Classification : Metallic
Marmite, which we all eat here in England and which is what makes us English, is a source of vitamin B12, actually a compound containing cobalt. The equivalent, but altogether blander, in Australia is Vegemite. Marmite is available in the USA. Try mixing it with peanut butter.
This sample is from The Elements Collection, an attractive and safely packaged collection of the 92 naturally occurring elements that is available for sale.
ISOLATION :
Isolation: it is not normally necessary to make cobalt in the laboratory as it is available readily commercially. Many ores contain cobalt but not many are of economic importance.
These include the sulphides and arsenides linnaeite, Co3S4, cobaltite, CoAsS, and smaltite, CoAs2. Industrially, however, it is normally produced as a byproduct from the produstion of copper, nickel, and lead.
1. Show that the conics through the intersection of two rectangular hyperbolas are also rectangular hyperbolas. If A, B, C & D be the four points of intersection of these two rectangular hyperbolas, then find the orthocentre of the triangle ABC.
2. Find the area of a right angle triangle if it is known that the radius of circle inscribed in the triangle is r and that of the circumscribed circle is R.
3. Q is any point on the line x = a. If A is the point (a, 0) and QR, the bisector of the angle OQA, meets OX in R, then prove that the locus of the foot of the perpendicular from R to OQ has the equation
(x – 2a) (x2 + y2) + a2x = 0 4. Show that the equation
z4 + 2z3 + 3z2 + 4z + 5 = 0 with (z ∈ C) have no purely real as well as purely imaginary root.
5. Prove that
x x n a x x f a
0
l
+
∫
∞dx = lna
x dx a x x f a
0
+
∫
∞6. A straight line moves so that the product of the perpendiculars on it form two fixed points is a constant. Prove that the locus of the foot of the perpendiculars from each of these points upon the straight line is a circle, the same for each.
7. Prove the identity :
∫
x −0 z zx 2
e dz =
∫
x −0 4 / 4 z
x 2
2
e
e dz, deriving for the
function f(x) =
∫
x −0 z zx 2
e dz a differential equation and solving it.
8. Let α, β be the roots of a quadratic equation, such that αβ = 4 and
−1 α
α +
−1 β
β = 4 a
7 a
2 2
−
− Find the set of values of a for which α, β ∈ (1, 4)
9. Investigate the function f(x) = x5/3 – 5x2/3 for points of extremum and find the values of k such that the equation x5/3 – 5x2/3 = k has exactly one positive root.
10. Let A = {1, 2, 3, ..., 100}. If X is a subset of A containing exactly 50 elements then show that
∑
∈x ppmin = 101C51.
`tà{xÅtà|vtÄ V{tÄÄxÇzxá
This section is designed to give IIT JEE aspirants a thorough grinding & exposure to variety of possible twists and turns of problems in mathematics that would be very helpful in facing IIT JEE. Each and every problem is well thought of in order to strengthen the concepts and we hope that this section would prove a rich resource for practicing challenging problems and enhancing the preparation level of IIT JEE aspirants.
By : Shailendra Maheshwari Joint Director Academics, Career Point, Kota Solutions will be published in next issue