Mole Concept

Mole Concept Atomic weight Q.1

One atomic mass unit in kilogram is (A) 1/NA (B) 12 / NA

(C) 1/1000 NA

(D) 1000 / NA

Q.2

The number of g-atoms of nitrogen in its 7 gm is equal to number of g-atoms in (A) 6 gm Mg (B) 28 gm Fe (C) 30 gm Ca (D) 20 gm Hg

Q.3

2 isotopes of an element are present in 1 : 2 ratio of number, having mass number M and (M + 0.5) respectively. The mean mass number of element will be (A) 3M + 1 (B) 1.5 M + 0.5 (C) 0.5 M + 0.5 (D) None of these

Q.4

A sample of nitrogen contains 90% N14 and 10% N15 isotopes. The average number of neutron per atom is (A) 14.10 (B) 7.05 (C) 7.10 (D) 14.20

Q.5

From 2 mg calcium, 1.2 × 1019 atoms are removed. The number of g-atoms of calcium left is (A) 5 × 10–5 (B) 2 × 10–5 (C) 3 × 10–5 (D) 5 × 10–6

Q.6

Three isotopes of an element have mass numbers M, (M+1) and (M+2). If the mean mass number is (M + 0.5), then which of the following ratio of number may be accepted for M, (M+1) and (M+2) in the order (A) 1 : 1 : 1 (B) 4 : 1 : 1 (C) 3 : 2 : 1 (D) 2 : 1 : 1 Molecular weight

Q.7

The number of electron in 3.1 mg NO3¯ is (A) 32 (B) 1.6 × 10–3

(C) 9.6 × 1020

(D) 9.6 × 1023

Q.8

The number of neutrons in 0.45 g water, assuming that all the hydrogen atoms are H1 atoms and all the oxygen atoms are O16 atoms, is (A) 8 (B) 0.2 (C) 1.2 × 1023 (D) 4.8 × 1024

Q.9

A gaseous mixture contains CO2 (g) and N2O (g) in a 2 : 5 ratio by mass. The ratio of the number of molecules of CO2 (g) and N2O (g) is (A) 5 : 2 (B) 2 : 5 (C) 1 : 2 (D) 5 : 4

Q.10 Number of oxygen atoms present in 1.61 gm of Na2SO4.10H2O is equal to number of atoms in (A) 3.0 gm C2H6 (B) 18.0 gm C6H12O6 (C) 0.98 gm H2SO4 (D) 8.0 gm SO3 Q.11

Which of the following contain largest number of carbon atoms? (A) 15 gm ethane, C2H6 (B) 40.2 gm sodium oxalate, Na2C2O4

[1]

(C) 72 gm glucose, C6H12O6

(D) 35 gm pentene, C5H10

Q.12 If the mass of proton is doubled and that of neutron is halved, the molecular weight of CO2, consisting only C12 and O16 atoms, will (A) not change (B) increase by 25% (C) decrease by 25% (D) increase by 50% Q.13 The shape of Tobacco Mosaic Virus (TMC) is cylindrical, having length and diameter 3000 Å and 170 Å, respectively. The density of the virus is 0.08 gm/ml. The molecular weight of TMC is (A) 3.28 (B) 5.44 × 10–24 (C) 5.44 × 10–18 (D) 3.28 × 106 Q.14 If molecular weight of glucose-1-phosphate is 260 and its density is 1.5 g/ml. What is the average volume occupied by 1 molecule of this compound? (A) 43 × 10–23 ml (B) 0.67 ml (C) 0.17 × 1023 ml (D) 29 × 10–23 ml Q.15 The number of hydrogen atoms in 0.9 gm glucose, C6H12O6, is same as (A) 0.048 gm hydrazine, N2H4 (B) 0.17 gm ammonia, NH3 (C) 0.30 gm ethane, C2H6 (D) 0.03 gm hydrogen, H2 Q.16 The number of g-molecules of oxygen in 6.02 × 1024 CO molecules is (A) 1 g-molecule (B) 0.5 g-molecule (C) 5 g-molecule (D) 10 g-molecule Concept of mole Q.17 0.36 gm of a triatomic gas, X3, occupies 168 ml at 1 atm & 273 K. The atomic weight of X is (A) 48 (B) 16 (C) 24 (D) 12 Q.18 What time, it would take to spend Avogadro’s number of rupees at the rate of 10 lac rupees per second? (A) 6.023 × 1017 sec (B) 1.909 × 1010 year 14 (C) 1.673 × 10 hour (D) all of these Q.19 Ethanol, C2H5OH, is the substance commonly called alcohol. The density of liquid alcohol is 0.7893 g/ ml at 293 K. If 1.2 mole of ethanol are needed for a particular experiment, what volume of ethanol should be measured out? (A) 55 ml (B) 58 ml (C) 70 ml (D) 79 ml Q.20 Which of the following will occupy greater volume under the similar conditions of pressure and temperature? (A) 6 gm oxygen (B) 0.98 gm hydrogen (C) 5.25 gm nitrogen (D) 1.32 gm helium Q.21 n mol of N2 and 0.05 mol of Ar are enclosed in a vessel of capacity 2 L at 1 atm and 27°C. Find n (R = 0.082 L atm mol K–1) (A) 0.30 (B) 0.10 (C) 0.03 (D) 0.06 Q.22 The volume of water at 277 K is 18 ml. One ml of water contains 20 drops. The number of molecules in one drop of water will be (A) 1.07 × 1021 (B) 1.67 × 1021 (C) 2.67 × 1021 (D) 1.67 × 1020

[2]

Q.23 How many mole of electron weight one kilogram? (A)

1 × 108 9.108  6.023

(B)

1 ×1031 9.108

(C)

1 × 1054 9.108

(D) 6.023 × 1023

Q.24 112.0 ml of NO2 at 1atm & 273 K was liquefied, the density of the liquid being 1.15 gm/ml. Calculate the volume of and the number of molecules in the liquid NO2. (A) 0.10 ml and 3.01 × 1022 (B) 0.20 ml and 3.01 × 1021 (C) 0.20 ml and 6.02 × 1023 (D) 0.40 ml and 6.02 × 1021 Fill in the blank: Q.25 For the reaction, 2x + 3y + 4z  5w Initially if 1 mole of x, 3 mole of y and 4 mole of z is taken. If 1.25 mole of w is obtained then % yield of this reaction is _______. Q.26 The vapour density of a mixture of gas A (Molecular mass = 40) and gas B (Molecular mass = 80) is 25. Then mole % of gas B in the mixture would be ________. Q.27 For the reaction 2A + 3B + 5C  3D Initially if 2 mole of A, 4 mole of B and 6 mole of C is taken. With 25% yield, moles of D which can be produced are _________. Single Correct: Q.28 One gram of the silver salt of an organic dibasic acid yields, on strong heating, 0.5934 g of silver. If the weight percentage of carbon in it 8 times the weight percentage of hydrogen and one-half the weight percentage of oxygen, determine the molecular formula of the acid. [Atomic weight of Ag = 108] (A) C4H6O4 (B) C4H6O6 (C) C2H6O2 (D) C5H10O5 Q.29 Mass of sucrose C12H22O11 produced by mixing 84 gm of carbon, 12 gm of hydrogen and 56 lit. O2 at 1 atm & 273 K according to given reaction, is C(s) + H2(g) + O2 (g)  C12H22O11(s) (A) 138.5 (B) 155.5 (C) 172.5 (D) 199.5 Q.30 40 gm of a carbonate of an alkali metal or alkaline earth metal containing some inert impurities was made to react with excess HCl solution. The liberated CO2 occupied 12.315 lit. at 1 atm & 300 K. The correct option is (A) Mass of impurity is 1 gm and metal is Be (B) Mass of impurity is 3 gm and metal is Li (C) Mass of impurity is 5 gm and metal is Be (D) Mass of impurity is 2 gm and metal is Mg Q.31 The percentage by mole of NO2 in a mixture of NO2(g) and NO(g) having average molecular mass 34 is : (A) 25% (B) 20% (C) 40% (D) 75%

[3]

Q.32 The minimum mass of mixture of A2 and B4 required to produce at least 1 kg of each product is : (Given At. mass of 'A' = 10 ; At. mass of 'B' = 120)

 2AB2 + 4A2B 5A2 + 2B4  (A) 2120 gm (B) 1060 gm

(C) 560 gm

(D) 1660 gm

Q.33 74 gm of a sample on complete combustion gives 132 gm CO2 and 54 gm of H2O. The molecular formula of the compound may be (A) C5H12 (B) C4H10O (C) C3H6O2 (D) C3H7O2 Q.34 An iodized salt contains 0.5 % of NaI. A person consumes 3 gm of salt everyday. The number of iodide ions going into his body everyday is (A) 10–4 (B) 6.02 ×10–4 (C) 6.02 × 1019 (D) 6.02 × 1023 Q.35 The mass of CO2 produced from 620 gm mixture of C2H4O2 & O2, prepared to produce maximum energy is (Combustion reaction is exothermic) (A) 413.33 gm (B) 593.04 gm (C) 440 gm (D) 320 gm Q.36 In the quantitative determination of nitrogen, N2 gas liberated from 0.42 gm of a sample of organic compound was collected over water. If the volume of N2 gas collected was

100 ml at total pressure 11

860 mm Hg at 250 K, % by mass of nitrogen in the organic compound is [Aq. tension at 250 K is 24 mm Hg and R = 0.08 L atm mol–1 K–1 ] (A)

10 % 3

(B)

5 % 3

(C)

20 % 3

(D)

100 % 3

Q.37 The mass of P4O10 produced if 440 gm of P4S3 is mixed with 384 gm of O2 is P4S3 + O2  P4O10 + SO2 (A) 568 gm (B) 426 gm (C) 284 gm (D) 396 gm Q.38 Calculate percentage change in Mavg of the mixture, if PCl5 undergo 50% decomposition. PCl5  PCl3 + Cl2 (A) 50% (B) 66.66 % (C) 33.33 % (D) Zero Q.39 The mass of Mg3N2 produced if 48 gm of Mg metal is reacted with 34 gm NH3 gas is Mg + NH3  Mg3N2 + H2 (A)

200 3

(B)

100 3

(C)

400 3

(D)

150 3

Q.40 The number of carbon atoms present in a signature, if a signature written by carbon pencil weights 1.2 × 10–3 g is (A) 12.04 × 1020 (B) 6.02 × 1019 (C) 3.01 × 1019 (D) 6.02 × 1020 Q.41 The average atomic mass of a mixture containing 79 mole % of 24Mg and remaining 21 mole % of 25Mg and 26Mg , is 24.31. % mole of 26Mg is (A) 5 (B) 20 (C) 10 (D) 15

[4]

Assertion Reason: Q.42 Statement -1 : 2A + 3B  C 4/3 moles of 'C' are always produced when 3 moles of 'A' & 4 moles of 'B' are added. Statement -2 : 'B' is the liming reactant for the given data. (A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1. (B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1. (C) Statement-1 is false, statement-2 is true. (D) Statement-1 is true, statement-2 is false. More than one correct: Q.43 Given following series of reactions: (I) NH3 + O2  NO + H2O (II) NO + O2  NO2 (III) NO2 + H2O  HNO3 + HNO2 (IV) HNO2  HNO3 + NO + H2O Select the correct option(s): (A) Moles of HNO3 obtained is half of moles of Ammonia used if HNO2 is not used to produce HNO3 by reation (IV) (B)

100 % more HNO3 will be produced if HNO2 is used to produce HNO3 by reaction (IV) than if 6 HNO2 is not used to produce HNO3 by reaction (IV)

1 th of total HNO3 is produced by reaction (IV) 4 (D) Moles of NO produced in reaction (IV) is 50% of moles of total HNO3 produced.

(C) If HNO2 is used to produce HNO3 then

Comprehension: Q.44 A 4.925 g sample of a mixture of CuCl2 and CuBr2 was dissolved in water and mixed thoroughly with a 5.74 g portion of AgCl. After the reaction the solid, a mixture of AgCl and AgBr, was filtered, washed, and dried. Its mass was found to be 6.63 g. (a)

(b)

(c)

(d)

% By mass of CuBr2 in original mixture is (A) 2.24 (B) 74.5

(C) 45.3

(D) None

% By mass of Cu in original mixture is (A) 38.68 (B) 19.05

(C) 3.86

(D) None

% by mole of AgBr in dried precipate is (A) 25 (B) 50

(C) 75

(D) 60

No. of moles of Clr ion present in the solution after precipitation are (A) 0.06 (B) 0.02 (C) 0.04

(D) None

[5]

Q.45 NaBr, used to produce AgBr for use in photography can be self prepared as follows : Fe + Br2  FeBr2 ....(i) FeBr2 + Br2  Fe3Br8 ....(ii) (not balanced) Fe3Br8 + Na2CO3  NaBr + CO2 + Fe3O4 ....(iii) (not balanced) (a)

Mass of iron required to produce 2.06 × 103 kg NaBr (A) 420 gm (B) 420 kg (C) 4.2 × 105 kg

(D) 4.2 × 108 gm

(b)

If the yield of (ii) is 60% & (iii) reaction is 70% then mass of iron required to produce 2.06 × 103 kg NaBr (A) 105 kg (B) 105 gm (C) 103 kg (D) None

(c)

If yield of (iii) reaction is 90% then mole of CO2 formed when 2.06 × 103 gm NaBr is formed (A) 20 (B) 10 (C) 40 (D) None

Q.46 Preparation of cobalt Metaborate involves the following steps of reactions: (i)

Ca2B6O11 + Na2CO3 (aq) Boiled  CaCO3 (insoluble) + Na2B4O7 + 2NaBO2

(ii)

 Na2B4O7  NaBO2 + B2O3

 CoO + B2O3  Co(BO2)2. Mass of Ca2B6O11 in kg required to produce 14.5 kg of Co(BO2)2 , assuming 100% yield of each reaction is (A) 32.2 (B) 40 (C) 28.2 (D) 30

(iii) (a)

(b)

If the yield of reaction (i), (ii) & (iii) is 60%,

200 % & 32.2 % respectively, then mass of Ca2B6O111 in 3

kg required to produce 14.5 kg of Co(BO2)2 is (A) 250 (B) 200 (C) 190

(D) 150

Q.47 Water is added to 3.52 grams of UF6. The products are 3.08 grams of a solid [containing only U, O and F] and 0.8 gram of a gas only. The gas [containing fluorine and hydrogen only], contains 95 % by mass fluorine. [Assume that the empirical formula is same as molecular formula.] (a) The empirical formula of the gas is (A) HF2 (B) H2F (C) HF (D) HF3 (b)

(c)

The empirical formula of the solid product is (A) UF2O2 (B) UFO2

(C) UF2O

(D) UFO

The percentage of fluorine of the original compound which is converted into gaseous compound is (A) 66.66 % (B) 33.33 % (C) 50 % (D) 89.9 %

[6]

Match the column : Q.48 Match the following : Column I (A) C12 H22 O11 (B) CH3COOH (C) Ca(OH)2 (D) CH3COOC2 H5

(P) (Q) (R) (S)

Q.49 Match the following: Column I (A) Gram atom present in one atom

Column II 8 NA atoms / mol 14 × NA atoms / mol 45 NA atoms / mol 2 NA of O-atoms / mol

(P)

Column II 2 NA

(B)

NA gram atom contains atom

(Q)

1 NA

(C)

No. of protons in 1 gm molecule of H2

(R)

4 NA

(D)

No. of electrons added to 32 gm O atom to convert it into O2–

(S)

NA2

Q.50 One type of artifical diamond (commonly called YAG for yttrium aluminium garnet) can be represented by the formula Y3Al5O12.[Y = 89, Al =27] Column I Column II Element Weight percentage (A) Y (P) 22.73% (B)

Al

(Q)

32.32%

(C)

O

(R)

44.95%

[7]

ANSWER KEY Q.1

C

Q.2

B

Q.3

Q.8

C

Q.9

B

Q.15 C

D

Q.4

C

Q.5

C

Q.6

B

Q.10 C

Q.11

D

Q.12 B

Q.13 D

Q.14 D

Q.16 C

Q.17 B

Q.18 D

Q.19 C

Q.20 B

Q.21 C

Q.22 B

Q.23 A

Q.24 B

Q.25 50%

Q.28 B

Q.29 B

Q.30 B

Q.31 A

Q.32 A

Q.33 C

Q.34 C

Q.35 C

Q.36 A

Q.37 B

Q.38 C

Q.39 A

Q.40 B

Q.41 C

Q.42 C

Q.43 A,C,D

Q.26 25%

Q.44 (a) C, (b) A, (c) B, (d) A

Q.45 (a) B, (b) C, (c) B

Q.46 (a) A, (b) A

Q.47 (a) C, (b) A, (c) A

Q.7

C

Q.27 0.75

Q.48 (A) R, (B) P,S (C) S (D) Q,S Q.49 (A) Q (B) S (C) P (D) R Q.50 (A) R, (B) P, (C) Q

[8]