STD XI- ENTERANCE TEST CHEMISTRY QUESTIONS:SOLUTIONS. and Cl37 17.Number of neutrons : 18,20.

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1 STD XI- ENTERANE TEST EMISTRY QUESTIONS:SOLUTIONS. 1. A is F e 3 O 4 3F e O F e 3 O Molar mass of F e 3 O 4 =168+64=232/4=58 Ans: AgNO 3 + KBr AgBr + KNO 3 1M 1M 0.25M 0.25M o.25m of AgBr=108+80=188/4=47 Ans: The element is hlorine. It exists as l17 35 Ans:02 4. Temperature is constant. P 1 =0.2 atm P 2 = 15 atm P 1 V 1 = P 2 V *V 1 =15V 2. V 1 = 15 V = = 75 Ans: O 2 10O O. M=15 Ans:15 and l37 17.Number of neutrons : 18, Ans: OONa(A) + NaO ao 4 (B) + Na 2 O 3 Parent acid of 3 OONa is 3 OO. Molar mass: 60 Ans: Al 2 O O(Bauxite) 2Al. Molar mass of Al 2 O O =138. Weight loss for 1M of Bauxite = = Weight loss for 0.5M(69 g) of Bauxite = Ans: A is Phosphorous P 2 O O 2 3 P O 4 (B) Molar mass of 3 P O 4 =98 Ans: F eso 4 F e 2 O 3 + SO 2 +SO 3 1M 0.5 M 0.5M Molar mass of SO 2 : 64 + SO /2=72 Ans: 72 1

2 11. Reaction (i) : Agl + NaNO 3 Ag + + l + Na + + NO3 All four ions freely moving in the solution. Reactions iii, v, vii, viii form insoulble hydroxides as by- products. Reactions ii,iv,vi form gaseous products along with Nal. Ans: ao 3 is responsible for temporary hardness of water. So A is a. a air ao(b) + a 3 N 2 () a 3 N O 2N 3 (D) + a(o) 2. Molar mass of N 3 :17 Ans: All except v,vi,ix. Ans: 07 2

3 STD XI- ENTERANE TEST EMISTRY QUESTIONS:SOLUTIONS. 1. A : l 2, B : 2, : l. On electrolysis of Nal + 2 O Na +, +, O, l ions formed At anode:l, O liberated l evolves as l 2, O remains in the solution At calthode: +, Na + liberated + evolves as 2, Na + remains in the solution. Al(O) 3 + 3l All O. 1M 3M 10 Molecules 30 Molecules. Ans: We need substances that produce gases.(water vapour is not gaseous at 60 0 ) 1) Al + l All ) u + l No reaction 3) Ag + l No reaction 4) MgO + l Mgl O 5) Na 2 O + 2l 2Nal + 2 O 6) KO + l Kl + 2 O 7) MnO 2 + l Mnl O + l 2 8) P bo 2 + l P bl O + l 2 9) KMnO 4 + l Kl + Mnl O + l 2 10)NaO 3 + l Nal + 2 O + O 2 Reactions 1,7,8,9,10. Ans: Ml 2 = al 2 ; M is a.(x= 4) M l 2 = Mgl 2 ; M is Mg (y = 3) x + y = 3+4= 7 Ans: X is O 2. Mg + O 2 MgO + (Y ) 3 + F e2o 3 F e(z) + 3O 20F e + 30l 2 20F el 3. 1Fe transfers 3 electrons; 20 Fe will transfer 60 electrons Ans: KI + 2 O 2 2KO + I 2. I 1 I 2 (loss of e Oxidation) O 1 O 2 (gain of e reduction) Thus reaction d is a redox reaction. 2 O 2 undergoes reduction. Ans:34 6. l: acidic; NaO NaO 3 N 4O: basic; Nal : neutral. Ans: Fullerene (Football shaped) have 6 membered rings (x=6) and 5 membered rings (y=5) Ans: P 1 T 1 = P 2 T 2 3

4 P 1 = 1.8 atoms P 2 = 2.1 atoms T 1 = 300 T 2 =? T 2 = = 350K. = 77 c 1.8 Ans: Molecular weight of SO 3 : 80; NaO: 40 Each molecule of SO 3 is double in weight of NaO. Number of molecules in 100g. SO 3. = Number of molecules in 50g. NaO. OR 80 gms of SO 3 : molecules 100g molecules 40 gm of NaO : molecules molecules =50 g NaO Ans: % of 720=180 g. 2Al + 2NaO 2NaAlO O. 2M(54 g) 3M(6 g). 180g 20gms.(10 Moles) Ans: MgO 3 MgO + O 2 84g 40g 21g 10g. 21g of pure MgO 3 i.e 10% of 210 gms. Ans: NaO SO 4 Na 2 SO O + 2O 2 2M 1M 0.6M 0.3M we need 0.3M 2 SO 4. for complete reaction. Given 900 ml 3M. solution for 0.3 M we need 90 ml solution Ans: ; Molecular weight of 3 8 = 44 Ans: 44 4

5 14. 2KlO 3 2Kl + 3O gms 96 gms.? gms = gm of KlO 3 is decomposed = x 4.90;x= Ans: 20 STD XI- ENTERANE TEST EMISTRY QUESTIONS:SOLUTIONS. 15. A is Aluminium. Al 2 (O 3 ) 3 + l = All 3 (B) + 2 O + O 2 Al 2 O 3 + NaO = 2NaAlO 2 () + 2 O Molecular weight of NaAlO 2 :82 Ans: OO + 3l 2 = l 3 OO + 3l. 1M(60 gm) 1M(163.5 gm) (T) T-100.5=63 Ans: Butane: 4 10 : = 4 : 10 = 48 gm : 10 gm = 24 gm : 5 gm. ence 5 g associated with 24 g carbon. Ans: (O 3 ) 2 = +4 O 6 ; (P O 4 ) 3 = P +5 O 8 P = 5 = 4 P = 01 Ans: There are 14 elements in the actinide series. Actinium (89) to Lawrencium (103).These are in 7th period, 3rd group. Radium is 7th period, 2nd group. So the atomic number of Ra=103-15=88 Ans: : 3 = 2 + Br 2 = 3 Br 2 Br( 3 6 Br 2 ). Mol wt of 3 6 Br 2. = 202 = Ans: Sample A is 90% m/m S(g) contains 90 5 gms of acid 100 Sample B is 10 % m/m 5 g contains 5 10 gms. = 0.5 gms of acid. 100 Resultant solution : 70% m/m = 70 S s =15 Ans: If the mass of BaO is taken as X gms, mass of ao will be (X -2.5)gms BaO + ao SO 4 BaSO 4 + aso O 5

6 153g 56g 233g 136g Xg (2.5-X)g 1.523Xg 2.43(2.5-X)g 1.523X gm +2.43(2.5-X)gm =4.713g; X=1.5g % of X (BaO) in the mixture= /2.5 = 60 Ans: % pure Bauxite ( Al 2 O O) = 69 gm in 1 Kg = 0.5 mol. Al 2 O 3 + 2NaO 2NaAlO O 0.5mole 1mole(40g) Ans: Third member of the homologous series of O; 2 5 O will have molecular weight , as the difference in mol wt of any two members in homologous series is 14. Ans: Volume of propane in the mixture =40 10/100=4 lit Volume of butane in the mixture = 10-4 = 6 lit 3 8(g) (g) 3O 2(g) O ( g) 1*4 : 5*4 : 3*4 : 4* (g) (g) 8O 2 (g) O(g) 2*3 : 13*3 : 8*3 : 10*3 Total volume of carbon dioxide gas added to the atmosphere: 12 lit (from propane)+24 lit (from butane)=36 lit Ans: XO 2(g) gas bleaches blue litmus so it is sulphur dioxide gas. aso 3 ao + SO 2 1.5M 1.5M (96 gm) Ans: 96 6

7 STD XI- ENTERANE TEST EMISTRY QUESTIONS:SOLUTIONS. 13. Atomic number 15: electronic configuration: 2, 8, 5. ; group 15 Ans: Mg + N 2 Mg 3 N 2 (X) 3M 1M 1M (40 molecules) Mg 3 N 2 + 6l 3Mgl 2 + 2N 3 (Y ) 1M 6M 3M 2M. 40 Molecules 80 Molecules Ans: (orrection in the Question paper : 73 grams of dry l) B is Na 2Na O (xs) 2NaO ( aq) + 2 2M(46 g) 2M (80g.) NaO + l Nal + 2 O. 1M 1M 1M 1M. 2M 2M :Solution is neutral; p=7(x) (80g) (73g) 6X=42 Ans: A is Sulphur (S) S + O 2 SO 2 (B) SO 2 + O 2 SO 3 () SO O 2 SO 4 (D) S + F e F es (E) F es + 2 SO 4 F eso S(F ) (E) (D) Molecular weight of 2 S (34) + Atomic weight of S (32) Ans: A is N 4 l. N 4 l + NaO Nal + 2 O + N 3 N 4 l + AgNO 3 Agl + N 4 NO 3 white ppt Molecular weight of A = =53.5 XY.5 Ans: Bal 2 + Na 2 SO 4 BaSO 4 +2Nal. + 20MNa 2 SO 4 (unreacted) 1M 1M 1M 2M 10M 30M 10M (T) 20M Ans: O 2 (X) + 2 2O lessair + 2 O (g) O (g) + 2 (g) Sum of molecular weights of O (28) + 2 (2) =30 Ans: 30 7

8 20. (75% of 800)= 600g (6M) ao 3 ao 3 ao + O 2 1M 1M 1M. 100g. 44 g (6M) (6M) O 2 + Na 2 O 2 Na 2 O O 2 2O 2 + 2Na 2 O 2 2Na 2 O 3 + O 2. 1M 0.5M 6M 3M 3M O 2 = = Ans: ao 3 ao + O 2 l ao + 2 O a(o) 2 2 aol2 + 2 O ao 3 al 2 1M(100g) 1M(111g) 0.5M(50g) 0.5M(63.5) Ans: 50 g molecules = 3M O 2 molecules 3M O 2 molecules = 2 moles of O 3 molecules = = Ans: It is Iodine Ans: O + l l(a) + l. 4 9 l + NaO 4 9 O(B) + Nal O Ans: 09 8

9 STD XI- ENTERANE TEST EMISTRY QUESTIONS:SOLUTIONS. 14. ao + 2 O a(o) 2 ; a(o) 2 is X a(o) 2 + 2NO 3 a(no 3 ) O. a(no 3 ) 2 is Y 1M 2M 1M 2M 24 Molecules 48 Molecules Ans: Y is a(no 3 ) 2 Ans: Loss of electrons means oxidation. i.e. se in +ve valency. ere it is F e 2+ F e 3+ (I) (II) (II) must be F e 2 O 3. 6F e 2 O l 12F el O. 6 F e 2 O 3 gives 12 F el 3 i.e. 12*3 = 36 l ions. Ans: X is Al 2 (O 3 ) 3 Al 2 (O 3 ) 3 Al2 O 3 + 3O 2 X Y Al 2 (O 3 ) 3 has 9 oxygen atoms. Ans: a(o) 2 + 2O 2 excess a(o 3 ) 2. (Z) Valency of O 3 = -1 Ans: Mg + N 2 Mg3 N O 3Mg(O) 2 + 2N 3 3M 1M 1M 3M 2M 1.5M 0.5M 1M Ans: Mg(O) 2 + 2NO 3 Mg(NO 3 ) O. 1M 1M. 3M 3M. It will be a neutral solution p=07. Ans: SO S 3S O 1M 2M 3M 10 Molecules+20 Molecules=30 atoms. Ans: Nal in 2 O Na +, l, +, O + evolves at cathode,l at anode. NaO is formed. X = NaO. 2NaO + O 2 Na 2 O O. Na 2 O O found in nature. Ans. 10 9

10 23. B is l 2 a(o) 2 + l 2 al.ol(d) + 2 O. Ans: Ethanol Acetaldehyde Acetic Acid O O O 1 (-O) single bond.no (-O)Single bond 1 (-O) single bond x=1 y = 0 z =1 x + y + z =2 Ans: ON 2, Next homolgue 3 7 ON 2. Difference in Molecular Weight of homologues ( 2 ) = 12+2=14 Ans: X is Al and Y is F e 2 O 3 2Al (s) + F e 2 O 3(s) 2F e (s) + Al 2 O 3(s) Fe Molar mass = 56 Ans: 56 O 10

Some Basic Concepts of Chemistry

Q 1. Some Basic Concepts of Chemistry What weight of AgCI will be precipitated when a solution containing 4.77 g of NaCI is added to a solution of 5.77 g of AgNO 3? (IIT JEE 1978 3 Marks) Q 2. One gram