Solution Iit Jee

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Transcript of Solution Iit Jee

2 PIE IIT JEE PAPER I & II

(SOLUTION 2010)

IIT-JEE 2010 SOLUTION - 2

SOLUTIONS TO IITJEE 2010 (PAPER 1)PART1 : CHEMISTRYSECTIONI

Straight Objective Type This section contains 8 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1. In the reactionHBr OCH3 the product are

(A)

Br

OCH3 and H 2

(B)

Br and CH 3Br

(C) 1. (D)O

Br and CH 3OH

(D)

OH and CH 3Br

HBr CH3 Br

O H

: Br CH3

OH + CH 3Br

2.

Plots showing the variation of the rate constant (k) with temperature (T) are given below. The plot that follows Arrhenius equation is

(A)

k

(B)T

k

T

(C)

k

(D)T

k

T

2.

(A) From Arrhenius equation: K = Ae Ea / RT , K will increase exponentially with temperature. Hence (A).PIE EDUCATION, Corporate Office: 44A/2, Sarvapriya Vihar, New Delhi - 16. Ph: 01141828585. Fax: 01126520144 Head OfficeAXIS 117/N/581, Kakadeo Kanpur Visit us : www.pieeducation.com

3 PIE IIT JEE PAPER I & II

(SOLUTION 2010)

IIT-JEE 2010 SOLUTION - 3

3.

3.

The species which by definition has ZERO standard molar enthalpy of formation at 298 K is (B) Cl2(g) (A) Br2(g) (D) CH4(g) (C) H2O(g) (B) The standard molar enthalpy of formation of an element is zero at 298K. Br2 although an element, exists in the liquid state at 298K. Hence (B) The ionization isomer of [Cr(H2O)4Cl(NO2)]Cl is (B) [Cr(H2O)4Cl2](NO2) (A) [Cr(H2O)4(O2N)]Cl2 (D) [Cr(H2O)4Cl2(NO2)]H2O (C) [Cr(H2O)4Cl(ONO)]Cl (B) [Cr(H2O)4Cl(NO2)]Cl [Cr(H2O)4Cl(NO2)]+ + Cl [Cr(H2O)4Cl2]NO2 [Cr(H2O)Cl2]+ + NO2 The correct structure of ethylenediaminetetraacetic acid (EDTA) isHOOC CH2 N HOOC HOOC CH2 CH2 N HOOC CH2 CH2 CH2 N CH2 COOH CH CH N CH2 COOH CH2 COOH CH2 COOH HOOC COOH N HOOC HOOC CH2CH2 N COOH COOH CH2 CH2 H N CH CH N H HOOC CH2 CH2 COOH

4.

4.

5.

(A)

(B)

(C) 5.

(D)

(C) EDTA isHOOC HOOC CH2 N CH2 CH2 CH2 N CH2 COOH CH2 COOH

Hence (C)6.

6.

The bond energy (in kcal mol1) of C C single bond is approximately (A) 1 (B) 10 (C) 100 (D) 1000 (C) The bond energy of C C single bond is 100 kcals mol1. Hence (C). Other data are absurd. The synthesis of 3-octyne is achieved by adding a bromoalkane into a mixture of sodium amide and an alkyne. The bromoalkane and alkyne respectively are (A) BrCH2CH2CH2CH2CH3 and CH3CH2CCH (B) BrCH2CH2CH3 and CH3CH2CH2CCH (C) BrCH2CH2CH2CH2CH3 and CH3CCH (D) BrCH2CH2CH2CH3 and CH3CH2CCH (D)CH3 CH 2 CH 2 CH 2 Br NaNH 2 CH3CH2C C H CH 3 CH 2 C C : Na NH3 NaBr +

7.

7.

PIE EDUCATION, Corporate Office: 44A/2, Sarvapriya Vihar, New Delhi - 16. Ph: 01141828585. Fax: 01126520144 Head OfficeAXIS 117/N/581, Kakadeo Kanpur Visit us : www.pieeducation.com

4 PIE IIT JEE PAPER I & II

(SOLUTION 2010)

IIT-JEE 2010 SOLUTION - 4

CH 3 CH 2 CH 2 CH 2 C C CH 2 CH 33 Octyne

Hence (D).8.

The correct statement about the following disaccharide isCH2OH H O H (a) OH H OH H OH H HOH2C H O (b) HO CH2OH OH H H

OCH2CH2O

8.

(A) Ring (a) is pyranose with -glycosidic link (B) Ring (a) is furanose with -glycosidic link (C) Ring (b) is furanose with -glycosidic link (D) Ring (b) is pyranose with -glycosidic lilnk (A) Ring (a) is pyranose with -glycosidic link but ring (b) is furanose with -glycosidic linkage. Hence (A)SECTIONII

9.

Multiple Correct Answer Type This section contains 5 multiple correct questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONE OR MORE may be correct. In the Newman projection for 2,2-dimethylbuaneX H3C CH3

H

Y

H

9.

X and Y can respectively be (A) H and H (C) C2H5 and H (B, D)CH3 H3C CH2 C CH32,2 Dimethyl buane

(B) H and C2H5 (D) CH3 and CH3

CH3

PIE EDUCATION, Corporate Office: 44A/2, Sarvapriya Vihar, New Delhi - 16. Ph: 01141828585. Fax: 01126520144 Head OfficeAXIS 117/N/581, Kakadeo Kanpur Visit us : www.pieeducation.com

5 PIE IIT JEE PAPER I & II

(SOLUTION 2010)

IIT-JEE 2010 SOLUTION - 5

H H3C CH3 H3C

CH3 CH3

H

C 2H5 (I)

H

H

CH3 (II)

H

I. Considering rotation about C1 C2 bond one of the conformers will be as represented by the Newman projection formula (I). Hence (B). Also II. Considering rotation about C2C3 bond, one of the conformers can be represented by the above Newman projection formula (II). Hence (D)10.

10.

The reagent(s) used for softening the temporary hardness of water is (are) (B) Ca(OH)2 (A) Ca3(PO4)2 (D) NaOCl (C) Na2CO3 (B, C) Temporary hardness is due to dissolved Ca(HCO3)2 Ca(HCO3)2 + Ca(OH)2 2CaCO3 + 2H2O Ca(HCO3)2 + Na2CO3 CaCO3 + 2NaHCO3 Hence (B) and (C) Among the following, the intensive property is (properties are) (A) molar conductivity (B) electromotive force (C) resistance (D) heat capacity (A, B) By decreasing or increasing the amount of an electrolytic solution, neither molar conductance nor EMF is going to change. Aqueous solutions of HNO3, KOH, CH3COOH, and CH3COONa of identical concentrations are provided. The pair(s) of solutions which form a buffer upon mixing is(are) (B) KOH and CH3COONa (A) HNO3 and CH3COOH (D) CH3COOH and CH3COONa (C) HNO3 and CH3COONa (C, D)OH

11.

11.

12.

12.

13.

In the reaction

NaOH(aq)/ Br2 the intermediate(s) is(are)

PIE EDUCATION, Corporate Office: 44A/2, Sarvapriya Vihar, New Delhi - 16. Ph: 01141828585. Fax: 01126520144 Head OfficeAXIS 117/N/581, Kakadeo Kanpur Visit us : www.pieeducation.com

6 PIE IIT JEE PAPER I & II

(SOLUTION 2010)

IIT-JEE 2010 SOLUTION - 6

O Br

O

(A)Br O

(B)Br O Br

(C)Br

(D)Br

13.

(A, C)OH O O H O Br Br

OH H2 O

Br Br

Br Br

O Br Br Br H+

O BrBr Br

O BrH +

O Br

Br

H Br

Br

H

Br

SECTIONIII

Linked Comprehension Type This section contains 2 paragraphs. Based upon the first paragraph 3 multiple choice questions and based upon the second paragraph 2 multiple choice questions have to be answered. Each of these questions has four choices A), B), C), D) out of which ONLY ONE is correct. Paragraph for Questions 14 to 16 Copper is the most noble of the first row transition metals and occurs in small deposits in several countries. Ores of copper include chalcanthite (CuSO4 5H2O), atacamite (Cu2Cl(OH)3), cuprite (Cu2O), copper glance (Cu2S) and malachite (Cu2(OH)2CO3). However, 80% of the world copper production comes from the ore chalcopyrite (CuFeS2). The extraction of copper from chalcopyrite involves partial roasting, removal of iron and self-reduction.PIE EDUCATION, Corporate Office: 44A/2, Sarvapriya Vihar, New Delhi - 16. Ph: 01141828585. Fax: 01126520144 Head OfficeAXIS 117/N/581, Kakadeo Kanpur Visit us : www.pieeducation.com

7 PIE IIT JEE PAPER I & II

(SOLUTION 2010)

IIT-JEE 2010 SOLUTION - 7

14.

14.15.

Partial roasting of chalcopyrite produces (A) Cu2S and FeO (C) CuS and Fe2O3 (B) Iron is removed from chalcopyrite as (A) FeO (C) Fe2O3 (D) In self-reduction, the reducing species is (A) S (C) S2 (C)

(B) Cu2O and FeO (D) Cu2O and Fe2O3

15.16.

(B) FeS (D) FeSiO3

16.

(B) O2 (D) SO2

Paragraph for Questions 17 to 18 The concentration of potassium ions inside a biological cell is at least twenty times higher than the outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. A simple model for such a concentration cell involving a metal M is: M(s) | M+ (aq; 0.05 molar) ||M+ (aq; 1 molar) | M(s) For the above electrolytic cell the magnitude of the cell potential |Ecell| = 70 mV. 17.

17.18.

For the above cell (A) Ecell < 0; G > 0 (B) Ecell > 0; G < 0 (C) Ecell < 0; G > 0 (D) Ecell > 0; G < 0 (B) Ecell = 0.0591 log 0.05 = +ve as Ecell is +ve so G is less than zero. If the 0.05 molar solution of M+ is replaced by a 0.0025 molar M+ solution, then the magnitude of the cell potential would be (A) 35 mV (B) 70mV (C) 140mV (D) 700 mV (C) As 0.0591 log 0.05 = 70 mV 0.0591 log 0.0025 = 140 mVSECTIONIV

18.

Integer Type This section contains TEN questions. The answer to each question is a single digit integer ranging from 0 to 9. The correct digit below the question number in the ORS is to be bubbled. 19. 19.

The value of n in the molecular formula BenAl2Si6O is 3 The value of n in the BenAl2Si6O18 is 3.PIE EDUCATION, Corporate Office: 44A/2, Sarvapriya Vihar, New Delhi - 16. Ph: 01141828585. Fax: 01126520144 Head OfficeAXIS 117/N/581, Kakadeo Kanpur Visit us : www.pieeducation.com

8 PIE IIT JEE PAPER I & II

(SOLUTION 2010)

IIT-JEE 2010 SOLUTION - 8

20.

The total number of basic groups in the following form of lysine isH3N O CH2 CH2 CH2 CH2 CH C H2N O

20.21. 21.

2 In lysine NH2 and COO are basic groups.

Based on VSEPR theory, the number of 90 degree F-Br-F angles in BrF5 is 0 In BrF5 the shape is square pyramidal type but due to one lone pair no F-Br-F bond will be of 90 angle, it is around 85. Amongst the following, the total number of compounds whose aqueous solution turns red litmus paper blue is (NH4)2C2O4 NaCl Zn(NO3)2 KCN K2SO4 K2CO3 NH4NO3 LiCN FeCl33 The aqueous solutions of KCN, K2CO3 and LiCN are alkalin