RIESGO IMPORTANTE RIESGO INTOLERABLE
2. PLIEGO DE CONDICIONES TÉCNICAS
+ + 4I– →
2 24
HgI
+ + +
0
Hg
2. Calculate mol of Ca(OH)2 required to carry out following conversion taking one mol in each case : (a) COOH
COOH into COO COO Ca (b) H3PO4 into CaHPO4
(c) NH4Cl into NH3
(d) NaHCO3 into CaCO3
Sol. (a) COOH
COOH is a dibasic acid COOH
COOH + Ca(OH)2 COO COO Ca
1 mo l 1 mo l
Ca(OH)2 required = 1 mol
(b) H3PO4 + Ca(OH)2 → CaHPO4 + 2H2O 1 mol of H3PO4 ≡ 2H+ neutralised by 1 mol of Ca(OH)2
Ca(OH)2 required = 1 mol
(c) 2NH4Cl + Ca(OH)2 → CaCl2 + 2NH3 + 2H2O 2 mol NH4Cl ≡ 1 mol Ca(OH)2
1 mol NH4Cl ≡ 0.5 mol Ca(OH)2
(d) 2NaHCO3 + Ca(OH)2 → Na2CO3 + CaCO3 + 2H2O
2 mol NaHCO3 ≡ 1 mol Ca(OH)2 1 mol NaHCO3 ≡ 0.5 mol Ca(OH)2
3. A colourless salt (A), soluble in water, gives a mixture of three gases (B), (C) and (D) along with water vapours. Gas (B) is blue towards litmus paper, gas (C) red and gas (D) is neutral. Gas (B) is also obtained when (A) is heated with NaOH and gives brown ppt. with K2HgI4. Solution thus obtained gives white ppt. (E) with CaCl2 solution in presence of CH3COOH. Precipatete (E) decolorises MnO−4/H+. Gas(C) turns lime water milky while gas (D) burns with blue flame and is fatal when inhaled. Identify (A) to (D) and explain chemical reactions.
Sol. Gas (B) gives brown ppt. with K2HgI4
⇒ gas (B) is NH3 ⇒ gas (A) has NH4+ (C) turns lime water milky
⇒ gas (C) can be SO2 or CO2
Gas (D) is also obtained along with (C). Gas (D) burns with blue flame and is fatal when inhaled
⇒ gas (D) is CO ⇒ gas (C) is CO2
⇒ (A) has C2O42–
It is confirmed by the fact that CaCl2 gives white ppt.
CaC2O4(E) which decolourises MnO4–/H+
⇒ (A) is (NH4)2C2O4
Explanation :
(NH4)2 C2O4→∆ 2NH3 + CO2 + CO +H2O (A) (B) (C) (D) (B) is blue towards litmus (basic) (C) is red toward litmus (acidic) (D) is neutral
(NH4)2C2O4+2NaOH→∆ Na2C2O4+ ) B ( NH
2 3 +2H2O
Na2C2O4+ CaCl2 →
) E ( . ppt White
O
CaC2 4 ↓ + 2NaCl
UNDERSTANDING
Inorganic Chemistry
NH3 + K2HgI4 → O Hg
Hg NH2I brown ppt (Iodide of Million’s base) 2MnO4– +16H++5C2O42– → 10CO2+2Mn2++ H2O
violet colourless
4. A solution of a salt (A) when treated with calculated quantity of sodium hydroxide gave a green coloured ppt (B), which dissolve in excess of NaOH. (B) acts as a weak base and loses water on heating to give a green powder (C). The green powder is used as refractory material. When (C) is fused with an alkali in presence of air or oxidising agent, a yellow coloured solution (D) is obtained. Identify the compounds from (A) to (D)
-Sol. The compound (A) is chromic salt. The chemical reactions are as under -
(i) With calculated quantity of sodium hydroxide - CrCl3 + 3NaOH → Cr(OH)3 + 3NaCl
green ppt (B)
(ii) In excess of sodium hydroxide, soluble NaCrO2 is formed
Cr(OH)3 + NaOH → NaCrO2 + 2H2O
(sod. chromite)
(iii) Since Cr(OH)3 contains -OH group, so it will act as a base. On heating it will lose water to give Cr2O3
powder (C)
2Cr(OH)3 → Cr2O3 + 3H2O
(C)
(iv) On fusing Cr2O3 with an alkali in presence of oxygen or oxidising agent, a yellow soluble chromate will be formed -
2Cr2O3 +8NaOH + 3O2 → 4Na2CrO4 + 4H2O yellow soln. (D)
5. Two moles of an anhydrous ester (A) are condensed in presence of sodium ethoxide to give a β-keto ester (B) and ethanol. On heating in an acidic solution compound (B) gives ethanol and a β-keto acid (C).
(C) on decarboxylation gives (D) of molecular formula C3H6O. Compound (D) reacts with sodamide to give a sodium salt (E), which on heating with CH3I gives (F), C4H8O, which reacts with phenyl hydrazine but not with Fehling reagent. (F) on heating with I2
and NaOH gives yellow precipitate of CHI3 and sodium propionate. Compound (D) also gives iodoform, but sodium salt of acetic acid. The sodium salt of acetic acid on acidification gives acetic acid which on heating with C2H5OH in presence of conc.
H2SO4 gives the original ester (A). What are (A) to (F) ?
Sol. (i) Acetic acid on heating with C2H5OH gives original compound (A).
CH3COOH + C2H5OH
∆
→
H2SO4
) A
( 2 5
3COOC H CH
+ H2O (ii) CH3COOC2H5 (A) on heating with C2H5ONa
undergoes Claisen condensation to give (B), which is aceto acetic ester.
CH3CO OC2H5 + H CH2COOC2H5 C2H5ONa Reflux (A)
+ C2H5OH + CH3COCH2COOC2H5
(B)
(iii) (B) on heating in acidic solution gives (C) and ethyl alcohol.
) B
(2 2 5
3COCH COOC H
CH + HOH H→+
) C ( 2
3COCH COOH
CH + C2H5OH
(iv) (C) on decarboxylation gives acetone (D).
) C ( 2
3COCH COOH
CH −CO2
→∆
) D
( 3
3COCH CH
(v) (D) reacts with NaNH2 to form sodium salt (E), which on heating with CH3I gives butanone (F).
) D
( 3
3COCH
CH + NaNH2
NH3
−
→∆
) E
( 2
3COCH Na CH
NaI –
I CH →3
) F
( 2 3
3COCH CH
CH
(vi)
) F
( 2 3
3COCH CH
CH + 3I2 + 4NaOH →∆ CHI3 + CH3CH2COONa + 3NaI + 3H2O (vii)
) D
( 3
3COCH
CH + 3I2 + 4NaOH →∆ CHI3 + CH3COONa + 3NaI + 3H2O CH3COONa HCl → CH3COOH + NaCl Thus, (A) CH3COOC2H5
(B) CH3COCH2COOC2H5
(C) CH3COCH2COOH (D) CH3COCH3
(E) CH3COCH2Na (F) CH3COCH2CH3
1. For complex numbers z1 = x1 + iy1 and z2 = x2 + iy2
we write z1 ∩ z2, if x1 ≤ x2 and y1 ≤ y2. The for all complex numbers z with 1 ∩ z, we have
z z +
− 1
1 ∩ 0, Justify the result.
2. AP and BQ are fixed parallel tangents to a circle, and a tangent at any point C cuts them at P and Q respectively. Show that CP.CQ is independent of the position of C on the circle.
3. Let f(x) = ax2 + bx + c & g(x) = cx2 + bx + a, such that |f(0)| ≤ 1, |f(1)| ≤ 1 and |f(–1)| ≤ 1. Prove that
|f(x)| ≤ 5/4 and |g(x)| ≤ 2
4. A straight line is drawn throguh the origin and parallel to the tangent to the curve
a y a x+ 2− 2
= ln
+ −
y y a
a 2 2
at an arbitrary
point M. Show that the locus of the points P of intersection of this straight line and the straight line parallel to the x-axis and passing through the point M is a circle.
5. Show that
∑
=n − +
r r r
n r r
C C
0( 2) 2 =
+ +
odd is n If 2, n
1
even is n If 1, n
1
6. Let In =
∫
01 −tan 1xdx
xn , then expression In in terms of In–2.
7. If f
+ 3 y
x =
3 ) ( ) (
2+ f x + f y
for all real x and y. If f ´(2) = 2, then f(2) is - Passage :
Let Z denotes the set of integers. Let p be a prime number and let z1 ≡ {0, 1}. Let f : z → z and g : z → z1 are two functions defined as follows :
f(n) = pn; if n ∈ z and
g(n) = 1; if n is a perfect square
= 0, otherwise.
8. g(f(x)) is - (A) many one into (B) many one onto (C) one one onto (D) one one into 9. f(g(x)) = p has -
(A) no real root
(B) at least one real root (C) infinity many roots (D) exactly one real root 10. g(f(x)) is –
(A) non periodic function (B) odd function
(C) even function (D) None of these
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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
11 Set
1. g(x) = sin x ; 0 ≤ x < π/2
Hence g(x) is continuous and differentiable in (0,∞) 2. x choose (at least one from each) 4 cones :
4 – 1C3 – 1 = 3C2 = 3 ways
so required ways (either 2 or 3 different flavours)
= 60 + 6C2 3 = 60 + solve them together to find pt. D
t
similary use eq. of external angle bisector line AE
⇒ r= p
Compare real parts & get (a) Compare imaginary. parts & get (b)
6. Let Ei = the event that originator will not receive a letter in the ith stage.
Originator sands letters to two persons so in 1st stage he will not get letter.
Prob. that letter sent by 1st received is not received by originator is
1 1
similarly prob. that letter sent by 2nd receipiant is not received by originator is
1