MOLE CONCEPT FOUNDATION BUILDER (OBJECTIVE) 1.
(D)
2.
(D)
3.
(B)
4.
(A) (Most stable isotope of carbon)
5.
(D)
6.
(C)
7.
(A) Moles of gas =
5.6 0.25 22.4
Molecular weight of gas =
7.5 30 0.25
Hence NO. 8.
(A) Molecular weight of C 60 H122 60 12 122 842. 842 Weight of a molecule = 1.39 10 21 g . 23 6.022 10
9.
(A) 1 mole contains Avogadro number of atoms.
10.
(A) 1.4 0.05. 28 Number of atoms = 0.05 2 6.02 1023 . = 6.02 1022 .
Moles of N 2
11.
(D) (A) (B) (C) (D)
12.
22.4 10 3 NA 6.022 10 23 22400 22 6.022 10 23 3.011 10 23 44 11.2 6.022 10 23 3.011 10 23 22.4 0.1 6.022 10 23 6.022 10 22
(C) Number of gms of H 2SO4 0.25 98 24.5
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13.
(D) 0.5 2 Volume of H2 in l = 0.5 22.4 11.2l .
Moles of H 2
14.
(D) 19.7 1000 100 197 Atoms of Au = 100 6.022 10 23 6.022 10 25 .
Moles of Au =
15.
(A) Mass of one molecule of CO 2
16.
44 7.3110 23 6.02 10 23
(C) Number of moles of H 2
0.224 0.01 22.4
17.
(B)
18.
WH 3 3 9g [B]
19.
In one H2O molecule: 10 proton, 8 neutrons, 10 electrons 36g 2mols Hence in 36 ml, n H2O 18g / mol Protons = 2N A 10 20 N A [C]
20.
n atoms
WN 3 14 42g
w . Hence it should be of same weight ‘W’ at.wt
[A] 21.
no. of moles = [B]
22.
23.
24.
10 3 N A 10 3 NA
wt 103 mol.wt 103 M 0 g M 0 mg
1 16 A:12 g ; B 16 8g ; C : 10 g ; D 8g 2 2 [A] A : 2.5 5N A 12.5 N A ; B :10N A ; C : 4 3N A 12N A ; D 1.8 8N A 14.4N A . Hence [D] 52 amu 13 4 amu [C]
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25.
One ion contains: 7 + 24 + 1 = 32 e total es 2 N A 32 64 NA [B]
26.
n C 0.5 6 3 [D]
27.
28.
29.
30.
wt = 36 g
28 46 36 54 ; B: ; C: ; D: 44 46 18 108 [C] A:
180 10 18 no. of es 10 10 N A 100 N A [D] n H 2O
2.48 0.01 248 nH 2O 5 0.01 molecules 0.05 N A [c] n Na 2S2O3 .5H 2O
n Ag
90 10 1 1 atom N A 5 1022 100 108 12 12
[c] 31.
n H 2O
18 333 = 9. Hence [B] 54 (96 3) (18 18)
32.
n H 2O
0.018 103 . Hence, molecules = 103 NA 18
[C]
33.
n N 3
4.2 0.3 . 14
total = 0.3 8N A 2.4N A
[A]
34.
35.
3.42 0.12 342 atom = 0.12 N A [D] n C 12 n C12H 22O11 12
8.4 0.1 84 Each contain (12 + 6 + 24) protons n MgCO3
Hence, total 0.1 42NA 2.5 1024 [B]
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36.
n total
4.4 2.24 0.2 44 22.4
molecules = 0.2N A
[B] 37.
[D]
38.
n gas
w w mol.wt. 3a
[B] 39.
40.
41.
42.
43.
558.5 10 moles 55.85 In 60 g carbon, n C 5 [A] n Fe
twice = 10 moles
Say n Mg PO n ; then n O 8n 3 4 2 0.25 8n = 0.25 n 3.125 10 2 8 [B] w w 2 : 2 2 :1 nx : ny 10 20 Hence [B]
X 46 96 180 180 X 55.9 100 [C] 25.4 8 1 1 : : 2:5 127 16 5 2 Hence I 2O5 . [C] n I : nO
44.
mol. Wt = 2 VD = 100 71 w chlorine 100 71g 100 w metal 29 g [A]
45.
(D) CaCO3 CaO CO2 1
1
1
Quantity of limes tones = wt. of one mole mole of CaCO3 = 100 kg 46.
(A) Moles of H 2S 2 11.2 Moles of SO 2 0.5 22.4
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SO 2 2H 2S 3S 2H 2O moles 1
2
given
2
0.5
3
2 3 0.5 x 1.5 1
L.R. 47.
(C) 100 1.724 58 Moles 2H 3PO4 3Mg OH 2 Mg 3 PO 4 2 6H 2 O
Moles of Mg OH 2 Moles
2 2 1.724 Given 3
3
Weight of H 3 PO 4 48.
1
6
2 1.724 98 112.6g 3
n H2O n CH3OH 2 4
wt 4 18 72g
[D] 49.
WO 3.6769 2.0769 1.6g with 2 mole X 5 mole'O ' 1.6 with ' n ' moles mole 'O ' 16 0.2 n 0.04 5 [A]
50.
Ag2CO3 2Ag 2.7 WAg 2 108 2.11g 216 60 [A]
51.
n CO2 2 n C2 H5OH 2 WCO2 2 44 88g Hence [D]
52.
53.
54.
KClO 3 KCl 3 O 2 2 48g Hence % loss in wt = 100 39.18 122.5 [C] 2 2 2 n Fe n H 2O Wiron 56 37.39 3 3 3 [A]
n CaCO3 n CaO
1.62 n CaCl2 0.0289 56
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% of CaCl2
0.0289 111 100 32.11% 10
[B] 55.
(D)
3BaCl 2 2Na 3PO4 Ba 3 PO4 2 6NaCl Moles 3 2 1 6 1 0.2 0.5 0.2 0.1 2 (L.R.) 56.
Ca OH 2 H 2SO 4 CaSO 4 2H 2O 0.5
0.2 LR
n CaSO4 n Ca OH 0.2 2
[A] 57.
A 2B C 3D 5
8 LR
nB n 4 ; n D 3 B 12 2 2 Hence [B] nC
n : 4Al 58.
3C
2400 200 12
1350 50 27
Al4C3
L.R given 4Al 144 w 1800g given 50 W [D]
59.
2A 2B B 2C 3C 4D
60.
61.
2 2 4 n D nA 2 1 3 32 3 [D] Mol.wt. 0.8 28 0.2 32 28.8 M VD 14.4 2 [C] Dcl2 wrt air
Dcl2 Dair
M cl2 M air
71 29
Hence [A]
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62.
30.4 14 16x 14 x 2 100 M 46 oxide 1.44 MO 2 32
Say NO X . Then
Doxide wrt O2 [B]
63.
molality
n w solvent
1000 urea : NH 2 C NH 2 || O
18
60 1000 0.192 1500 1.052 18 [B]
64.
65.
66.
1 n Molarity 1000 98 1000 0.01 V mL 1000 [D]
Al3 20 0.2 2 0.2M 40 [A]
1 mole KMnO4 5 moles FeSO4 V 0.01 50 0.01 V 10mL [D]
67.
100 4 n H 0.001 2 2 10 1000
no. of H 2 104 N A 1.2 1020 [B] 68.
3 molal 3 mole NaOH in 1000g solvent 120 1000 vol 1009mL d 1.11 n 3 Molarity 1000 2.97 V mL 1.009 [A]
69.
(B)
2.65 1000 0.1 M. 106 250 0.1 10 After dilution of 10 mL solution = 0.001M 1000 n NaCl 1 70. X NaCl 0.0177 n NaCl n H 2O 1 1000 18 [A] CENTERS: MUMBAI / DELHI / AKOLA / KOLKATA / LUCKNOW / NASHIK / GOA / PUNE # 7
Molarity of NO 2 CO 3
GET EQUIPPED FOR JEE MAIN
1.
KClO3 KCl 3 O 2 2 3 2Al O Al 2O 3 2 2 nO n Al2O3 2 1 3 2 [A]
2.
Consider 1 L solution 29 d 1000 H2SO4 3.6 98 100 d = 1.22 g/mL [A]
N2
3. n:
3H2
10
, n O2 3
2
2NH3
15 LR
10 5 n N2 5
n H2 0
10 moles
[A]
Fe2 SO 4 3 3BaCl2 3BaSO 4 2FeCl3
4.
n:
1
? n BaCl2 3
n FeCl3 2
2
1
n BaCl2 2 3 0.75moles 2
[C] 5.
3CuSO4 2Al Al2 SO4 3 3Cu displaces
54g Al 192 g Cu displaces 27g Al 96g [C]
6.
CH4
2O2
CO2
– 5 8 5–4=1 – 4 n CO2 4 ; nCH 4 (remaining) = 1
2H2O – 8
[A] 7.
X C10H X nO 2 10CO 2 H 2O 2 X Hence, n 10 4 X with 1 mole C10 H X 10 moles 4 with 2.5 moles 32.5 moles
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x 32.5 1 13 4 2.5 x 13 10 4 12 [C] i.e. 10
8.
9.
10.
with
100g CaCO3 2mole HCl with 25L g 0.75 M HCl 1000 0.9375g [D] 2.125 V L Molarity 143.5 2.125 1000 Molarity 0.59 143.5 25 [B] with 2X 3 16g Oxygen n AgCl n Cl n HCl
with
1g 0.16g Oxygen 3 16 X 150 0.16 2 [D] 11.
12.
3 2Al O2 Al2O3 2 1 n: n 2 3 with 2 27g Al mole O2 2 with 1 mole 2 2 27 18g 3 [D] n BaSO4 nSO2 nS (POAC on S)
13.
8 1 32 4
[D]
n NaBr n1 , n KBr n 2 say 0.97 n AgBr n Br n1 n 2 0.00516 108 80 Also, n1 103 n 2 119 0.560
0.56 103 0.00516 0.00178 16 WKBr 119n 2 0.212 g [B] 16g 31.2 A : nH 4 4 ; B : nH 4 1.64 16g 76 n2
14.
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C : n H 22
34.2 36 2.2 ; D : n H 12 2.4 342 180
Hence [A] 15.
Total atoms
= 200 0.05 NA 1020 NA
0.05 NA 31022 [C] 16.
Mol. Wt of A 2 B3 150 96 246
For 5 mol, 246 5 g = 1.23 kg
[C]
17.
18. 19.
20.
21.
200 144 6.43 N A ; C N A 3 9N A 342 48 D : 2.5 3N A 7.5N A . Hence [A]
A :10N A ; B :11
[D] obvious 1 1 1 1 A : 3N A ; B : 26N A ; C : 8N A ; D : 2N A 44 114 30 26 Hence [A]
9.2 2 n 1 n 0.4 46 [C]
n CO2 n , say. Then n O 2n
22.
wt 0.4 30 12g
8 1 n 16 4
[D]
A : 0.2 14 g 2.8g ; B :
3 1023 6 1023
12 g 6g ; C : 32 g ; D : 7 g.
Hence [A] 23.
[D]
1 gram molecule: 44 g 1 molecule of CO 2 = 44 amu
24.
n H n 2 2n 4 10n
n C 2n 1 2n n C : n H 1: 5 [A] 25.
Total charge = 1 N A 3e 3N A e coulomb Hence [D]
69.98 Mol.wt 2112 mol.wt 360 100 [D] CENTERS: MUMBAI / DELHI / AKOLA / KOLKATA / LUCKNOW / NASHIK / GOA / PUNE # 10 26.
12g H 2O
8% H 2O
27.
45g silica 45g silica 43g others 43g others 100goriginal ' w 'grams 8 % of w = water i.e. 92 % of w = silica others 92 Hence, w 88g w 95.65 100 45 % of silica 100 47% 95.65 [D]
28.
M3N2 . 28 % nitrogen 28 3M 28 28 M 24 100 [C]
29.
0.014% mol.wt 2 at.wt of N 0.014 i.e. M 2 14 28 100 2800 M 2 105 14 103 [D]
30.
(A) Average atomic mass =
90 20 21x 22 10 x 100
20.11
x = 9% 31.
(B)
32.
(C) 6.023 10 23 1 6.023 10 23 3.01 10 22 Moles of HCl = 0.05 6.02 1023 3.01 10 22 HCl 0.05 6.02 10 23
Moles of Ca OH 2
Ca OH 2 2HCl CaCl 2 2H 2 O 1
2
1
0.05
1 0.05 1 0.025 2
L.R. 33. (A) CENTERS: MUMBAI / DELHI / AKOLA / KOLKATA / LUCKNOW / NASHIK / GOA / PUNE # 11
1.595 0.01 1595 Weight of solvent = 100 – 1.595 = 98.505 98.505 Volumes of solvent 82 10 3 L 1.2 1000 0.01 Molarity 0.12M 82 103 (B) 28 (A) atoms of O 2 6.022 10 23 ~ 3 10 23 32 3 (B) atoms of Be 6.022 10 23 ~ 2 10 23 9 8 (C) atoms of C 6.022 10 23 4 10 23 12 19 (D) atoms of F2 6.022 10 23 1 10 23 19 (C) X Y X Y 20 80 : 1 : 2 XY2 10 200
Moles of CuSO 4
34.
35.
36.
(C) Auogaduos hypothesis
37.
(A) 3 2.68 0.00335 24 100 Number of magnesium atoms = 0.00335 6.022 10 23 2.011021 atoms. (A) 25 10 3 Moles of comphon 0.164 10 3 10 12 16 16 Number of atoms 0.164 103 6.022 10 23 . 9.9 1019 (D) Moles of e 52 2 54 .
Moles of magnesium =
38.
39.
40.
(B) Moles of Ag =
1 . 107
1 107 2 107 2 32 Mass of Ag 2S 1.1495 107 2 1.1495 Mass of ore required = 100 85.78g 1.34
Moles of Ag 2S required =
41.
(D) Moles of Al = 27
27
1
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2Al 2NaOH 2H 2 O 2NaAlO 2 3H 2 Moles 2 2 2 2 3 31 Given 1 excess 1.5 2 (L.R.) Vol. of H 2 evolved = 1.5 22.4 33.6 L . WINDOW TO JEE MAIN 1. (A)
n solub le Vso lub le Lt
Molarity
Vsolution is affected by Temperatuere. 2.
(C)
n Fe
560 10 56
No. of atoms = 10 NA In 70 g of N In 20 g of H 3.
(A)
5.
(B)
70 N A 5 NA 14 20 no. of atoms = N A 20 N A 1 no. of atoms =
4.
(D)
6.02 10 20 NA Molarity 0.01 0.1 6.
(C)
7.
(C) V=1L
Wtotal 11.021000 1020g nsoluble = 2.05
Wtotal
352.8 100 1216.55g 29 = 1020 – 123 = 897 g
molality
2.05 2.28 0.897
8.
(B)
9.
(B) V = 1L nsoluble = 3.6 wsoluble = 3.6 98 352.8
352.8 100 1216.55g 29 1216.55 density 1000 w total
= 1.22 g/ml
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10.
(C)
14.
(A)
11.
(B)
12.
(C)
13.
(C)
weight N A species atomic weight In 4 g of hydrogen 4 Number of atoms N A 2 4N A 2 [Here species = 2 because hydrogen is present as H2] In 71 g of chlorine = 2NA 71 Number of atoms = N A 2 2N A 71 In 127 g of iodine, 127 Number of atoms = N A 2 2N A 127 In 48 g of magnesium, 48 Number of atoms = N A 1 2N A 24 [Here Mg is present as Mg so species = 1] Thus, the number of atoms are largest in 4 g of hydrogen. Number of atoms
15.
(b) Heavy water is D2O In it, Number of p 1 2 8 10 Number of e 1 2 8 10 Number of n 0 1 2 8 10 (D have1 n0 because it is actually, 1H2)
16.
(d) 18 g H2O contains 2 g H 0.72 g H2O contains 0.08 g H. 44 g CO2 contains 12 g C 3.08 g CO2 contains 0.84 g C
C:H
0.84 0.08 : 0.07 : 0.08 7 :8 12 1
Empirical formula = C7H8
17.
(c) 3 M solution means 3 moles of solute (NaCl) are present in 1000 L of solution. Mass of solution = volume of solution density
1000 1.252 = 1252 g Mass of solute = No. of mole molar mass of NaCl
358.5g = 175.5 g Mass of solvent = (1252 – 175.5)g = 1076.5 g = 1.076 kg
Molality
moles of solute mass of solvent in kg
3 2.79m 1.076
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18.
(a)
M1V1 M 2V2 V1 V2 10 2 200 0.5 200 10 20 100 210 120 0.57 M 210
Final concentration, M
FOUNDATION BUILDER (SUBJECTIVE) 1.
2.
X 2X X y given . For B : Y 20 40 20 1.6 0.1moles 6 10 22 molecules 16 Each molecule has (6 + 4) = 10e s n CH 4
total e s 6 1023 3.
n H 2O
18g 1mole 18g mol
1 molecule has (2 + 8) = 10 e s 1 mole contains 10N A electrons. 4.
O2 :10 e,8 protons,8 neutrons per ion. in1mole:10N A e,8 N A protons,8N A neutrons
5.
Atomic mass = NA mass of oneatom
6 1023 6.64 1023 g 40 g wt 1 wt of one atom 3.98 10 23 2.5 1022
6.
no. of atoms =
7.
removed 1021 44amu 7.35 102 g CO2 ,remaining 200 73.5 126.5mg n CO2
8.
126.5 103 0.002875 44
1mole N3 charge NA 3e 2.88 105 C
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9.
3.2 103 2 10 4 moles 64 n S n Na 2S2O3 .5H 2O 2 2 5 103 10 2 n O nSO 2 2
n O : nS 10.
11.
104 102
0.01
n O 3 n NaNO3 2 n NO2 1 m 10m 2 0.03 0.333 0.363 6 1 n N n NaNO3 n NO 2 10 103 6 = 0.01 + 0.166 = 0.176 23 6 10 t s 6 1017 s 6 10 6 1017 t hr 1.67 1014 3600 1.67 1014 t yr 1.9 1010 years 24 365
12.
atomic wt = 6.644 1023 6 1023 = 40 g/mol 40 1000g n 1000 moles 40 g / mol
13.
nC
106 g 12g / mol
No. of atoms = n C 6 1023 5 1016 14.
15.
16.
r = 0.1 inch = 0.254 cm 85.6 Fe ball 100 ball Vball density 4 3 0.254 7.75 0.532 g 3 85.6 Fe 0.532 g 0.455g 100 0.455 n Fe and no. of atoms = 4.9 1021 56
0.086 starch = wt of 1 atom = 31 g 100 3100 starch 3.6 104 0.086 VNH 3 n NH 3 22.4 L
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17.
18.
3.4 22.4 4.48L 17
PV 1 1 0.04464 RT 0.0821 273 n moleculs n O2 N A 2.69 1022 n O2
O3 O2 600mL Vml 600 V mL
V 600 V 32 1g 48 22400 22400 V = 200 mL 19.
Element
% (with in 100 g)
K
40.2
Mn
26.8
P
33
no. of (in 100 g) atom 40.2 1.03 39 26.8 0.48 55 33 1.06 31
ratio 2 1 2
K 2Mn P2 20.
Say
nO n
n H 15n 70 And n C 15n 10.5n 100 C10.5H15O or C 21H30O 2 is empirical formula 1 Mol. Wt 314 0.00318 C21H30O2 Then
21.
weight
9.03 10 20 0.311g weight
6.02 1023 mol.wt. mol.wt 207.33g 131.3 19n 207.3 n 4 22.
23.
58.97 102 59.9 n C 5 100 13.81 H 102 14.08 n H 14 100 27.42 N 102 27.97 n N 2 100 C5H14 N 2 C
C
12 12 CO2 0.9482 0.2586 44 44
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n C 0.02155 2 H H 2O 0.02154 18 n H 0.02154 n C : n H 1:1 CH 24.
12 cylinder 100 cylinder r 2 h density Co
3.14 6.25 10 8.2 1610.7
Co 1 12 n Co 1610.7 3.28 58.9 58.9 100
no.of atoms 3.28 6 10 23
1.98 1024 25.
Mol. Wt = wt of 1 mole mix = 2VD = 76.6 (x mol. NO2 + (1 – x) mol. N 2O4 ) = 76.6 g x 46 1 x 92 76.6
15.4 n NO2 in 1 mole = 0.335 46 100 n mix in100 g 76.6 x
n NO 2 in 100 0.335 n mix
= 0.437 26.
molality
n solvent
1000
Consider 1L of solvent
C2 H5OH mol.wt 46
n=8 solvent 1.025 1000 8 46 657
molality
27.
8 1000 12.18 657
2NaHCO3 Na 2CO3 CO2 H2O
Na 2CO3 no effect loses
2 84g NaHCO3 62g CO2 H 2O loses
g 0.124
0.124 168 0.336 g 62 0.336 100 % of NaHCO3 16.8% 2 and Na 2CO3 100 16.8 83.2%
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28. C 74.27
74.27 6.1892 12 7.79 7.79 1 12.99 0.928 14 4.95 0.309 16
H 7.79 N 12.99 O
%
29.
4.95
6.1892 20 0.309 7.79 25 0.309 0.918 3 0.309 0.309 1 0.309
C20 H 25 N3O 20 of C atoms 100 49 40.816 %
2CH3CHO O 2 2CH 3COOH 20g 10g : 20 10 0.45 0.31 n: 44 32 L.R (A) n CH3COOH n CH3CHO 0.45
CH3COOH 27.27g. 10 20 44 0.852 32 2 n O2 32 2.727g
(B) n O2 left
O2
(C) % yeild
23.8 100 87.2% 27.3
30.
3 20 4 40 8 50 n CH n A 3.2 2 100 2 100 3 100
31.
n CH 4 n1 and n C2H 4 n 2 ,say
now, n1 16 n 2 28 5g 14.5 also, n CO2 n1 2n 2 0.33 44 n1 0.193 and n 2 0.068
CH 4 100 16n1 100 60%. 5 5 %C2 H 4 40% %CH 4
32.
POAC on carbon n C n K 2CO3 1 n K Zn 2
moles of product
2 Fe
CN 6 2
n K2CO 3 12
12
0.0166
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33.
n Cu n Cu NO3 .3H 2 O (POAC on Cu ) 2
10 product 63.5 124 54 63.5 38.03g 34.
AgNO3 NaCl 5.77 4.77 n n 170 58.5 0.03394 0.08 L.R. n AgCl n AgNO3 0.03394
AgCl
NaNO3
AgCl 0.03394 143.5 4.87g
35.
CaCO3 CaO CO 2 n1
n1
MgCO3 MgO CO2 n2
n1 100 n 2 84 1.84 n1 56 n 2 40 0.96 n1 0.01 n 2 0.01 %CaCO3 36.
0.01 100 100 54.35% 1.84
Cl2 2KOH KCl KClO H 2 O
3KCl O 2KCl KClO
3
1 1 3 n Cl2 n KClO4 n Cl2 1 3 4 4 1385 n Cl2 4 40 39 35.5 64
Cl 2 40 71 2840g
37.
POAC on Cl (eventually on completion) n Cl2 2 n KCl 1 n KClO4 n Cl 142 2 2 4 0.5 71 4 3.5moles 1g KClO3 n 2 moles
n KCl
38.
n1 moles
KClO 4 KCl
KCl O 2
3 146.8 n O2 n 1 n1 0.00437 2 22400 n2
1g n 0.00379 39 35.5 48 1
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3 n 2 0.00284 4 1g O 2 0.79029g
n KClO4
resiude
0.00284 39 35.5 64 100 0.79 49.789%
%KClO 4
39.
CH 3 x AlCl y CH 4 yCl Cl AgNO3 AgCl NO3 1. n CH4 x.n CH3
x
AlCly
0.222 0.643 x. 16 15x 27 35.5y
2. n AgCl n Cl y.n CH3
x
AlCl y
0.996 0.643 y. 108 35.5 15x 27 35.5y
x 1.99 2 y 0.643 2y 0.222 in1, 16 30y 27 35.5y 1 2
y 1 and x 2 40.
3 KClO3 KCl O2 2 6.125g Zn 2HCl ZnCl 2 H 2
H 2 1 O2 H2 O 2 in (1), 3 n O2 n KClO3 0.075 2 in (3), n H 2 2 n O 2 0.15 in (2), n Zn n H 2 0.15 41.
42.
(1) (2) (3)
Zn 0.15 65.3 9.795g (A): B, (B): A, 7 7 (C): n C n B . 2 2 C O 2 CO, CO 2 n1
n2
POAC on C 12 n C n1 n 2 1 12 POAC on O : n O n1 2n 2
20 1.25 16
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n 2 1.25 1 0.25 and n1 0.75 n CO : n CO2 n1 : n 2 3 :1 43.
2NaOH H 2SO 4 Na 2SO4 2H 2O n NaOH 15 1 1 4 n H 2SO 4 7.5 10 2 1000 10 2 H2SO4 strength 1000 VH2SO4 mL
6.125g / L 44.
Molarity
n 10 103 1000 103 0.1M V mL 100
in gram / L 0.1 39 16 1 5.6g / L
45.
100 4 n SO2 n H 2SO4 0.001M 10 4 1000 no. of ion n SO2 n A 6 1019 4
46.
n CuSO 4 .5H 2 O n Cu 2 0.5 0.01 5 10 3
weight n mol.wt 5 103 249.5 1.2475g 47.
M1V1 M 2 V2 M 3V3 50 0.5 75 0.25 M final M 3 50 75 0.35Molar
48.
Molality
49.
n I2
n
1000 solvent 3 30 1000 0.4molal 250
n I2 n C6 H 6
0.2
Say, we have 1 mole mix. Then, n I2 0.2 and n C6H 6 0.8 molality
50.
n I2
C6 H6
1000
0.2 1000 3.205m. 0.8 78
Consider 1L solution. t solution 1000 1.06 10609.
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10 solution 106g 100 n KCl Molailty 1000 Vsolution (mL) 106 74.5 1000 1.4228M 1000
KCl
51.
30%NH3 . 70%water. 70 i.e. solution water 105g 100 100 i.e. solution 150 150g 70 150 Vsolution 166.67mL density 0.9
52.
Consider 1L of solution, solution 1.025 1000 1025g
n ethanol M V 8 1 8moles ethanol 8 46 368 n molality ethanol 1000 solvent 8 1000 1025 368 12.176molal
53.
2SO 2 O 2 2SO3
nSO 2 nSO3 nSO3 5 54.
55.
56.
4FeS2 11O2 2Fe2O3 8SO2 600 800 5 23 120 32 1 So moles of Fe2O3 2.5 2 146 n NH3 n HCl 4 36.5 Wt NH3 4 17 68g 2H 2 O 2 2H 2O 6 29 3 0.90625 2 32 LR
wt H 2O formed 0.90625 2 18 32.625 g wt H 2 left 3 0.90625 2 2 2.l325g 57.
245 w 3 2 95 58.5
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58.
w 226 g AgCl AgBr n1
n2
n1 n 2 108 60.94 %Ag 100 n1 143.5 n 2 188 100 n 1 0.31955 n2 n1 35.5 100 n n1 35.5 %Cl 100 2 n1 143.5 n 2 188 n1 143.5 188 n2 4.856% %Br 100 60.94 4.856 34.2%
59.
CO2 H2O unbalanced COOH2 H SO 2
4
POAC n C n CO2 1 n COOH 2
10 2 2 90 9
2 V CO2 22.4L 4.977L 9 60.
acid is H3A. salt is Ag3A
1moleAg3A 3mole Ag n Ag 0.37 108 n Ag3 A 0.00114 3 3 0.607 0.00114 mol.wt of Ag3 A mol. wt 108 3 A 531
A 207
wt of H3A 210 GET EQUIPPED FOR JEE ADVANCE ONE OPTION CORRECT 1.
N2 3H 2 say, wt :14x 3x 14x x 3x to 28 2 2 x 3x tt y 3y 2 2 NH 3 was 40% by mol.
2NH 3
2y
40 x 3x 3y 2y y 100 2 2 2x x 5y 2x 2y y 3.5 7 y
i.e. 2y
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X N2
x y 1 x 2 y 2 2x 2y 2 x y 1 1.75 1 0.75 2 2.5 5 0.15
(A) 2.
n(A) n(B) M A n(A) 1.4g and M B n(B) 0.8 M 0.8 B 0.57. M A 1.4 (C)
3.
with 3.2g metal 0.4g oxygen 64g metal g oxygen
64 0.4 8g 3.2 128g metal with 16g O i.e. M 2O (B)
4.
with 4M 96g O. sin ce X 4 O 6
with 5.72g 4.28g O 5.72 6 16 MX 32 4 4.28 (A)
5.
224 0.01 mol.wt 22400 wt 1 mol.wt 100 n 0.01 n
3 at.wt. 100 at.wt 33.3g 33.3 mass of one atom 5.53 1023 g 6.02 10 23 (C)
6.
7.
PV 2 0.35 3.123 102 mol.wt RT 0.0821 273 1 i.e. 3.123 102 2 At.wt at.wt 16 wt of one atom . NA NA (C) n
I 2 Cl2 ICl, ICl3 n1
n2
POAC on I CENTERS: MUMBAI / DELHI / AKOLA / KOLKATA / LUCKNOW / NASHIK / GOA / PUNE # 25
n(I)
25.4 n1 n 2 127
POAC on Cl 14.2 n(Cl) n1 3n 2 35.5 n1 : n 2 1:1 (A)
8.
FeSO 4 : n1 SO 42 n1 and Fe 2 n1 Fe 2 SO 4 3 : n 2 SO 42 3n 2 and Fe 3 2n 2
n1 3n 2 given
n1 3 n2
Fe 2 n n n 1 1 2 3: 2 3 Fe 2n 2 2 (D) 9.
0.36M : V1 say and 0.15M : V2say M V M 2 V2 M final 0.24 1 1 V1 V2 36V1 0.15V2 0.24 V1 V2 V 0.36 1 0.15 V2 or 0.24 V1 1 V2 V V 0.36 1 0.15 0.24 1 V2 V2 V 0.09 3 1 V2 0.12 4 (D)
10.
0.24
At mass N A mass of an atom
6 1023 3.98 1023 24g (C) 11.
Fe 2 Fe CN 6
Fe 3 56 7 C 6 12 3 (C) 12.
obvious (D)
13.
obvious (B)
14.
1gatom 1mole of atom 14g.
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(A) 15.
Na 2CO3 2HCl 2NaCl H 2O CO 2 n HCl 2 n Na 2CO3 VHCl M HCl
V3 2
1.431 V 9mL. 106
(B)
16.
17.
They must have same mol. wt. (C)
V2micron sphere V
0
20 Asphere
4 2 10 6 3 3 109 3 9 4 2 10 3
(A) 18.
19.
3 KClO3 KCl O2 2 n o2 0.1 2 n KClO3 3 3 30 2 2 2 122.5 % purity 30 100 81.66% 10 (B) V ml m 1 0.65 6.5 10 4 moles 1000 1000 BaCl 2 .2H 2 O 137 71 36 6.5 10 4 0.1586g n
BaCl2 137 71 6.5 10 4 0.1352g (A)
20.
1.36 V 200 2.4 1.24 500 V 102.941mL (B)
11.5
100 M C6 H5CH3 9.31g 71
21.
t
22.
CuSO 4 .5H 2O n1 , MgSO 4 .7H 2 O n 2 total t 5g and anhydrous 3g 249.5n1 246n 2 5
M C6H5COOK
and 159.5n1 120n 2 3 on solving, n1 0.0149 and n 2 0.0052 CuSO 4 .7H2 O. 3.729g
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%by t
3.72 100 74.4% 5
(C) 23.
C 7 H 6 O 3 C 4 H 6 O 3 C9 H 8 O 4 C 2 H 4 O 2 : 2g 4g 0.01449 n : 0.0144 0.039 L.R
theoretical yield 0.01449 M C9H 8O 4 2.69 %yeild 80.76% (A) 24.
25.
26.
2XI3 3Cl 2 2XCl3 3I2 0.5 0.236 n XI3 n XCl3 M 381 M 106.5 M 138.88 139 (B) 2 2 no.of molecules 500cm 0.21nm n V Molarity NA 6 1023 n i.e.V 2.395 10 5 L. 4.24 256 (B)
in 10 mL CuCl2 n1 , and CuBr2 n 2 n AgBr 2n 2 and n AgCl 2n1 2n 1 143.5 2n 2 188 0.9065g
and 2n 1 2n 2 188 1.005g then n1 0.00115 and cuBr2 0.35g 25% and 58%
(A) 27.
n XH 4 2n and n X2 H6 n, say
n x n XH4 n X2H6 2 4n 5 4n and 2n X 4 n. 2X 6 5.628 X 5 5 i.e. X 4 2X 6 5.628 2X 4X 5 10 5 7.5 or, 5.628 2 X 2 X 17.5 or X 27.86 28 0.628 (A) i.e.
28.
M AgNO3
0.0125 39 80 0.0105M 1mL
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42.5 n AgNO3 1000 0.00446 n NaBr 2n Na 2SO 4
0.0105 n AgNO3
n1
n2
n1 2n 2 0.00446
also, n1 103 n 2 142 2 5 t of 1 29.
th
5
portion
Let acid be HA Salt: BaA 2 .2H2O
BaA2 H 2SO4 BaSO4 2HA
4.29 21.64 0.477 137 2A 36 1000 A 121 HA 122
30.
total moles =n (say) 0.15 n moles of CH 3COOH 0.15n 60 0.85 n 18 30 30 n 1.234 9 15.3 n NaOH n CH3COOH 0.15 n 0.18519 VNaOH 18.5 L B
31.
C2 H 6 3.5O2 n1 moles
2CO2 3 H 2O
C2 H 4 3O2 2CO2 2 H 2O n2 moles
PV 1 40 1.218 RT 0.0821 400 130 also 3.5 n1 3n2 nO2 4.06 32 n1 0.817 n1 n2
n2 0.401 %C2 H 4 33% and C2 H6 67% A 32. Al K S O
no. of atom 0.3889 0.388 0.775 3.1
% 10.5 15.1 24.8 49.6
ratio 1 1 2 8
2
33.
Vmolecule
3 100 A 300 A 4
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1.299 1024 mol. Wt N A Vmol desnity
6 1023 1.299 1024 1.2 103 kg m3 939 kg (B) More than one correct 1.
3 moles in 1L 1250 g
wNa2S2O3 3 46 64 48 474 474 100 37.92% 1250 3 3 (B) x 0.065 1250 474 46.11 3 18 n 1000 (C) molality of Na wsolvent 3 2 1000 7.73 1250 474 (A) % by weight
2.
mol. wt = wt of 22.4 L=28.896 g mol.wt VD 14.48 2 (A) and (B)
3.
[A] : 32 g
[ B]
1 64 32 g 2
[ D] : 32 g
Ca NO3 2 Na2C2O4 CaC2O4 2 NaNO3
4. millimoles :
6 3
3 LR
3
6
[A],[C],[D] 5.
n
1.12 0.02 56 0.02 100 2 g
CaO n CaCO3
wCaCO3
wCaCl2 0.02 111 2.22 g wNaCl 2.22 g
[ A, C]
6.
nNaCl 100 m moles ; n HCl 300 m moles
nCaCl2 200 m moles 200 Ca 2 , 400 Cl
cation 600 3 anions 800 4 Cl 800 2 M 400 CENTERS: MUMBAI / DELHI / AKOLA / KOLKATA / LUCKNOW / NASHIK / GOA / PUNE # 30
A , C
7.
Obvious A,B,D nH 4
8.
3 mole NH 3
wH 3 14 42 g
WN 3 14 42 g molecule 3 N a 18 1023 atoms 4 3 N A 72 1023 [ A], [B]. [C],[D] 9.
Obvious : [A], [B]
10.
[B],[C]: obvious others depend on volume
14.
Hence [C], [D] 14 Mol. Wt 22.4 28 11.2
Match the following 1.
2.
13 12 100 38.33% 407 6 (II) wt % of H 100 1.47% 407 (III) wt of H: wt of Cl 6 : 6 35.5 (IV) mo. of C: O =13:2 (I) wt % of C
(a)
wSO2 Wo2
(P) (A) (C) (E)
2s
(b) d 10 5 2 g cc sp. gr 2 s (c) M 2VD 32 Q (d) molecular
3.
132 3 44
at anons = 9(R)
20 (a) Al 3 0.04 M 400 H 40 0.084 500 Total =0.12 M Cl 60 40 0.2 M 500 (P), (S) 20 (b) K 0.2M 100
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Cl 20 0.2M 100 (S) 12 (c) K 0.12 M 100 [ P], [Q] SO42 6 0.06 M 100 (d) wH 2 SO4 200
24.5 49 n H 2 SO4 1 2 100
H 1 1000 5 M 200 SO42 1 2 1000 2.5M 200 [R] 4.
(A) VSO2 11.2 L
wSO2 32 g 1 2 NA 2 (B) n H 2 1 2 VH 2 11.2 L total atoms
wH2 1 g , , total atoms N A P (C) no. of atoms 0.5 3 N A 1.5 N A [P], [Q], [R] (D) 1moleO2 V 22.4 L Atoms 12 1023 wt 32 g
[S]
COMPREHENSION TYPE Passage 1 1.
wt of 1 atom 1amu 1.66 1024 g. (C)
2.
n S n H 2SO4 100 wt 3200g. (A)
3.
3.4 M s 2 32 M 1882.3 (B) 100
4.
C O 2 n C n C n O2 1
VO2
20 Vair 22.4C Vair 112L (B) 100
Passage – 2 1.
Consider 1 L.
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n KOH 6.9 KOH 6.9 56 386.4 30 solu 386.4 solu 1288g 100 d 1.2889g / mL. (A) 134 PV 0.2 2 M H2SO4 2 n NH3 1000 RT 0.0821 303 M H 2SO 4 0.06 (C)
2.
1600 0.205 0.2 V 40mL (A) 1600 V
3.
n H 2S
n H2SO4
5 34 5 1 5 34 V 0.2 5 V 25L (A)
4.
n H2SO4
Passage – 3
18g 3 1023 g (D) 23 6 10
1.
m H2O
2.
Avogadro’s law. (A)
3.
obvious Mass is 16amu. (C)
4.
obvious (A) Passage – 4
1.
AgNO3 NaCl AgCl NaNO3 5.77 4.77 n: n AgCl 0.0339 170 58.5 wt 4.88g 0.0339 0.081 L.R (A)
2.
H 2SO4 0.12 98 11.7g (A)
INTEGER
1.
0.5 mole N3 .N 3 has 10e . 5moles.
2.
n
CO2
132 3 44
nC 3
3.
MCl X : say. mol.wt M 106.5
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n Cl n MClX X n Cl n Ag
0.22x M 106.5x
0.51 3 103 170
6.4 112s (Dulong petite’s law) 0.57 0.22 x 3 103 x 3 112 106.5x M
8 2800 4 100 56
4.
n Fe
5.
x 5 20 2 2.6 5x 40 2.6x 52 x 20
6.
n 1 Cn H 2n 2 n O 2 nCO 2 n 1 H 2O 2 n 1 n 2 7 4n 3n 2 7n or n 2 n 4
1440 2 mol.wt 60 12
7.
n
8.
5 solu 0.3g solu 6g 100
9. 10.
0.25 89600 Fe n 56 n 4 100 CO2 C6 H10O5 1g algae 2g algae 1g strach
1g strach POAC on carbon n CO2 1 n (C6H10O5 )n 6n 1 1 6n 162n 27 1 27 time 8 4.7 10 3 EXPERTISE ATTAINERS 1.
POAC on Co
n Co3O4 3 n Co 1 n Co
0.2125 3 Co n Co 59 0.156 g 177 64
n PPt 1 n Co ppt n ppt mol.wt 1.52g
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2.
90 0.5g Fe Fe2O3 n1 100 10 0.5g Fe Fe3O 4 n 2 100 0.45 n1 2 n1 0.04 56 0.05 n2 3 n 2 0.0003 56 wt of mix= 160 n1 232 n 2 0.71g (a)
(b) 0.5g Fe Fe 2O 3 n
0.5 n 4.46 103 56 Fe2 O3 0.7142 g n2
3.
(i) n AgNO3 n AgCl n NaCl n HCl n1 n2 2.567 n1 n 2 0.0179 143.5 (ii) NaCl is not affected 1.341 n Cl n AgCl n 2 143.5 n 2 0.009345 n1 0.0856
Now, n1 58.5 n 2 M 1gram 0.5 M 53.5 0.009345 4.
y z C x H y Clz O2 xCO 2 H 2O Cl2 2 2 0.22 0.195 n ………(1) x CO2 44 12x y 35.5Z y n 0.22 0.0804 …..(2) H 2O 18 12x y 35.5 z 2 768 37.24 0.12 PV 760 1000 0.0012 n RT 0.0821 382 12x y 35.5z Solving, x 2; y 4 and z 2 C2H4Cl2
5.
…..(3)
Consider 1 mole mix t 55.4
N 2 , NO 2 , N 2O4 n1
n2
n3
n1 n 2 n3 1 Now, after heating,
n
NO2 n 2 2n3
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N O 2
2NO2
4
no. of moles = n1 n 2 2n 3 1 n 3
New Average mol. Wt 39.57 55.4 39.57 1 n3 n 3 0.4 Now, n1 28 n 2 46 n 3 92 55.4 28n1 46n 2 18.6
……(1)
also n1 n 2 n3 1 n1 n 2 0.6
…….(2)
Solving (1) and (2), n1 0.5 ,and n 2 0.1 5 :1: 4 n
6.
IO 3 n HSO3 n 3 5.8 0.8788 HSO 3 1 3 23 127 48
w NaHSO3 9.139 g n
I in 1st n IO3
5.8 198
IO3 n I n IO3 0.00586 1 5 M NaIO3 5.8 198 n
IO3 0.2L 200 mL M IO n
V
3
7.
AgNO3 KI KNO3 AgI 2KI KIO3 6HCl 3ICl 3KCl 3H 2O M KI VKI n KIO3 M 20 KI 2 1 2 M KI 0.3M
30
1 10
1
KI, excess n KIO3 n KI, excess 10 m mole 2 1 Original KI 50 0.3 15m mole n
Now,
KI used 5 m mole
n AgNO3 n KI used 5mmole w (AgNO 3 ) 0.85 g purity 85%
8.
Let % of boron will at. Wt. 10.0 = x Let % of boron will at. Wt. 11.01 = (100 – x) x 10.01 100 x100.01 10.81 100 x = 20%
CENTERS: MUMBAI / DELHI / AKOLA / KOLKATA / LUCKNOW / NASHIK / GOA / PUNE # 36
9.
Let NaCl = w gms kCl = (118 – w) gms POAC on Cl w 0.118 w 0.2451 M Nacl M kCl M Agcl
w 0.118 w 0.2451 58.5 74.5 74.5 143.5 w = 0.0338 gm Nacl = 0.0338 gm kcl = 0.0842 gm n nacl 5.777 104 n kcl 1.1310 3
POAC on Na; moles of Na 2O 2 = moles of NaCl 1 5.777 104 ; n Na 2O 2
1.1310 3 n k 2O 2
5.777104 Weight of Na 2O 62 0.01gm 2 1.1310 3 Weight of k 2o 94 0.1062 2 % Na2O = 3.58%. % k2o = 10.62 % 10.
CxHy
y C x H y x y x o 2 xCO 2 H 2 O 2 POAC on carbon 5 x vol.of CO 2 1 Now 1 vol. of CO2 = 10 mL (that obtained by kOH) x=2 Vol. of O2 reactionary = 15 ml Vol. of O2 reacted = 15 ml 1 ml of CxHy react with (2 + y/x) ml O2 5 ml of CxHy react with (2 + y/x) = ml 15 (given) y= 4 formula = C2H4 11.
(a) CO2 C 2CO POAC on carbon Let CO = x l CO2 = (1 – x) l x 1 2 1 x 1.6 1 2 – x = 1.6 = 1 x = 0.4 l & (1 – x)0.6 L (b) The molecular formula = M3N2 2 14 100 28 % Nitrogen = 3x 14 2
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Where x = atomic wt of metal 100 28 24 x 3 12.
monobro min ation C 2 H5 pat1 85%yeildC4 H10 55gm 90% yield
POAC on carbon x 90 85 55 2 4 1 100 100 412 101 212 61 x = 74.37 gms 74.37 V 22.4L 55.53L 30
WINDOW TO JEE ADVANCED INTEGER TYPE 2.
(4) Boltzmann constant, k
R or R k N A NA
1.380 10 23 6.023 10 23 = 8.31174 J K–1 8.312
Hence, no. of significant figures is 4 3.
(8) Mass of 1 L solvent = 0.4g mL1 103 mL = 400 g = 0.4 kg Mole of solute 3.2 So molality (m) Mass of solvent (kg) 0.4 =8m
CENTERS: MUMBAI / DELHI / AKOLA / KOLKATA / LUCKNOW / NASHIK / GOA / PUNE # 38
