mastering chemistry nbs gurukul · 2019-11-13 · mastering chemistry 2 nbs gurukul, scf- 5 market...

MASTERING CHEMISTRY 1

NBS GURUKUL, SCF- 5 MARKET GTB NAGAR, JALANDHAR. PH.NO. 5072918, 2976308, 98141-02308

NBS GURUKUL TARGET X+1-2020

CHEMISTRY SOME BASIC CONCEPTS OF CHEMISTRY

TEST - 2: COMPETITION TEST MAY 10, 2019

Prof. Adarsh Bhatti M.Sc. (Gold Medalist)

1. The crystalline salt, Na2SO4.x H2O on heating loses 55.9% of its weight. The formula of the crystalline salt is a) Na2SO4 . 5H2O (b) Na2SO4 . 7H2O (c) Na2SO4 . 10H2O (d) Na2SO4 . 6H2O Sol. (c) 100 g of crystalline salt contain H2O = 55.9 g Anhydrous salt = 100 – 55.9 = 44.1 g Molecular mass of anhydrous Na2SO4 = 2 × 23 + 32 + 4 × 16 = 142 44.1 g of anhydrous salt combine with H2O = 55.9 g

142 g of anhydrous salt combine with H2O = 1.44

9.55× 142 g = 180 g

180 u of H2O = 18

180 = 10 molecules.

2. Haemoglobin (molecular weight ( = 67200) has 0.33% iron by weight. One molecule of haemoglobin contains how many atoms of iron : a) 1 (b) 2 (c) 3 (d) 4 Sol. (d) , The number of iron atoms in haemoglobin are:

= Iron of weight Molecular × 100

Iron of Percentage ×n haemoglobi of weight Molecular

= 4= 56 × 010

0.33 × 67200

3. The most abundant elements by mass in the body of health human adult are Oxygen (61.4%);

Carbon (22.9%), Hydrogen (10.0%); and Nitrogen (2.6%). The weight which a 75 kg person would

gain if all 1H atoms are replaced by 2H atoms is

(a) 15 kg (b) 37.5 kg (c) 7.5 kg (d) 10 kg Sol. (c) Given, abundance of elements b mass oxygen = 61.4%, carbon = 22.9%, hydrogen = 10% and nitrogen = 2.6% Total weight of person = 75 kg

Mass due to 1H = 100

10×75 = 7.5 kg

1H atoms are replaced by 2H atoms,

Mass due to 2H = (7.5 2) kg

Mass gain by person = 7.5 kg 4. A drug marijuana owes its activity to tetra hydro cannabinol, which contains 70% as many carbon

atoms as hydrogen atoms and 15 times as many hydrogen atoms as oxygen atoms. The number of moles in a gram of tetrahydrocannabinol is 0.00318. Determine the molecular formula. a) C11H30O2 (b) C15H30O2 (c) C10H30O2 (d) C21H30O2 Sol. (d)

Let oxygen atom be = x part Hydrogen atom = 15 x

Carbon atom = 15 x 100

70= 10.5 x

Formula C10.5x H15x Ox

Formula weight = ( 10.5 12x ) + 15x + 16 x = 157 x



molecular wt = 00318.0

1= 314.46

x = 2 molecular formula = C21H30O2 5. Two elements X (at. Mass 16) and Y (at. Mass 14) combine to form compounds A, B and C. The

ratio of different masses of Y which combine with a fixed mass of X in A, B and C is 1 : 3 : 5. If 32 parts by mass of X combines with 84 parts by mass of Y in B, then calculate in C 16 parts by mass of X will combine with how many parts of Y ? (a) 28 (b) 42 (c) 56 (d) 70 Sol. (d) In B 32 parts of X combine with Y = 84 parts

a. 16 parts of X combine with Y = 32

1684= 42 Parts

Now No. of parts of X in both B & C are equal. Different masses of Y which combine with fixed mass of X in B & C are the ratio of 3 : 5

5

3

C inY of Mass

B inY of Mass

5

3

C inY of Mass

42

Mass of Y in C = 3

542 = 70 Parts

6. Specific volume of cylindrical virus particle is 6.02 10-2 cc/gm, whose radius and length are 7o

A

and 10o

A respectively . If NA = 6.02 1023, find molecular weight of virus ?

a) 15.4 kg/mol (b) 1.54 104 kg/mol

c) 3.08 104 kg mol (d) 3.08 103 kg/mol

Sol. (b), Volume of cylindrical virus is = r2l

= 3.14 ( 7 10-8)2 ( 10 10-8)

= 3.14 49 10-16 10-7

= 153.86 10-23 c.c. Weight of one virus particle is

= volumeSpecific

Volume

= 2× 0210.6

10 × 86.153 -23

gm

Molecular weight of virus

=02.6

10× 153.86 -21

6.02 1023 = 15.4 kg/mole

7. The specific heat of a metal is 0.16 . Its approximate atomic weight would be : a) 16 (b) 32 (c) 40 (d) 46 Sol. (C) We know from Dulong-Petits law that wt x sp. Heat = 6.4. Approximate atomic wt.

= 0.16

6.4 =

metal of heat sp

heat sp. × wt. At = 40

8. 4.0 g of caustic soda (mol. mass 40) contains same number of sodium ions as are present in :

a) 10.6 g of Na2CO3 (mol. mass 106) (b) 58.5 g of NaCl (formula mass 58.5) c) 100 mL of 0.5 M Na2SO4 (formula mass 142) (d) 1 gm equivalent of NaNO3 (mol. mass 85) Sol. (c) 4 g NaOH = 0.1 mol NaOH = 0.1 mol Na+ (a) 10.6 g Na2CO3 = 0.1 mol Na2CO3

= 0.2 mol Na+ (b) 58.5 g NaCl = 1 mol NaCl = 1 mol Na+

(c) 100 mL of 0.5 mol Na2SO4 = 1000

5.0× 100

= 0.05 mol Na2SO4 = 0.1 mol Na+



9. . If NA is Avogadro’s number, then number of valence electrons in 4.2 g of nitride ions (N3-) is a) 2.4 NA (b) 4.2 NA (c) 1.6 NA (d) 3.2 NA

Sol. (a) 4.2 g N-3 ions = 14

2.4mol = 0.3 mol = 0.3 × NA ions = 8 × 0.3 NA electrons = 2.4 NA (because each

ion contains 8 valence electrons ). 10. Calculate the molality of a solution of ethanol in water in which the mole fraction of ethanol is

0.040. (a) 1.31 M (b) 2.31 M (c) 3.31 M (d) 4.41 M

Ans.(b) )O(H n )OHHC(n

)OHHn(C x

252

52OHHC 32

= 0.040 (Given)

The aim is to find number of moles of ethanol in 1 L of the solution which is nearly = 1 L of water (because solution is dilute)

No. of moles in 1 L of water = 1-mol g 18

g1000 = 55.55 moles

Substituting n (H2O) = 55.55 in eqn (i) , we get 55.55 )OHH(C n

)OHHC(n

52

52

= 0.040

or 0.96 n (C2H5OH) = 55.55 × 0.040 or n (C2H5OH) = 2.31 mol Hence, molarity of the solution = 2.31 M

11. If the density of methanol is 0.793 kg L-1, what is its volume needed for making 2.5 L of its 0.25 M solution ? (a) 15.22 ml (b) 25.22 ml (c) 35.22 ml (d) 45.22 ml Ans. (b) Molar mass of methanol (CH3OH) = 32 g mol-1 = 0.032 kg mol-1

M = 1000

1000

32

1000793.0

= 24.78m Applying M1 × V1 = M2 V2 (Given solution) (Solution to be prepared)

24.78 × V1 = 0.25 × 2.5 L or V1 = 0.02522 L = 25.22 mL. 12. The density of 3 molal solution of NaOH is 1.110g mL-1. Calculate the molarity of the solution.

(a) 0.97 M (b) 1.97 M (c) 2.97 M (d) 3.97 M Sol.(c) 3 molal solution of NaOH means that 3 moles of NaOH are dissolved in 1000g of solvent.

Mass of Solution = Mass of Solvent + Mass of Solute

= 1000g + (3 40g) = 1120 g

Volume of Solution = mL110.1

1120 = 1009.00 mL

(since density of solution = 1.110g mL-1) Since 1009 mL solution contains 3 moles of NaOH

Molarity = litre in solutionin of Volume

solute of moles of Number= 1000×

1009.00

mol 3= 2.97 M

13. Calculate the percentage composition in terms of mass of a solution obtained by mixing 300g of a 25% and 400g of a 40% solution by mass. (a) 13.33 (b) 23.33 (c) 33.57 (d) 43.57 Sol. (c) Mass of solute in 300g of 25% solution = 75 g Mass of solute in 400g of 40% solution = 160g Total mass of solute = (75 + 160) g = 235 g Total mass of solution = 700 g

% of solute in final solution = 700

100 235 =

% of water in final solution = 100 – 33.5 = 66.43% 14. The system that contains the maximum number of atoms is

(a) 4.25g of NH3 (b) 8g of O2 (c) 2 g of H2 (d) 4g of He



Sol. (c)Number of atoms = mass molar

mass NA number of atoms in 1 mole

where, NA = Avogadro's number

Number of atoms in 4.25 g of NH3 = 17

25.4 NA 4 = NA

Number of atoms in 8 g of O2 = 2

N=2×N×

32

8 A

A

Number of atoms in 2 g of H2 = AA N2=2×N×

2

2

Number of atoms in 4g of He = AA N=1×N×

4

4

Thus, 2g of H2 contains the maximum number of atoms among the given. 15. How many moles of magnesium phosphate, Mg3 (PO4)2 will contain 0.25 mole of oxygen atom?

(a) 0.02 (b) 3.125 10-2 (c) 1.25 10-2 (d) 2.5 10-2 Sol. (b) As 8 moles of O-atoms are present in Mg3(PO4)2

0.25 mole of O-atoms are present in Mg3(PO4)2 1/8 0.25 = 3125 10-2 mol 16. The total number of electrons in 18 mL of water (density =1g mL-1 )

(a) 6.023 1025 (b) 6.023 1024 (c) 6.023 18 1023 (d) 6.023 1023

Sol. (b) In 1 mL, number of moles of H2O

17. Calculate the the volume of 1 molar NaOH solution required to convert 12 g of NaH2PO4

completely into Na3 PO4. (a) 100 (b) 150 (c) 200 (d) 250

Sol. NaH2PO4 + 2NaOH Na3PO4 + 2H2O 120 g 80 g 12 g ? 8 g

M = V

1000

M

W

2

2

1 = V

1000

40

8

V = 200 mL 18. One litre solution of N/2 HCl is heated in a beaker. It was observed that when the volume of the

solution is reduced to 600ml, 3.25g of HCl is lost. Calculate the normality of the new solution? (a) 0.358 (b) 0.50 (c) 0.685 (d) 0.85

Sol. Normality = V

1000

E

w

2

2 , g18.25 1000

1000 x 36.5 x 5.0wor

1000

1000

5.36

w5.0 2

2

When volume becomes 600 ml, 3.25g of HCl is lost.

wt. of HCl left in 600 ml = 18.25 - 3.25 = 15.00 g

Normality = 600

1000

36.5

15

V

1000

mass.Eq

w2 = 0.685 N Ans.

19. You are given one litre of 0.15 M HCl and one litre of 0.40 M HCl. What is the maximum volume of 0.25 M HCl which you can make from these solutions without adding any water ?

(a) 2.2 L L (b) 2.5 L (c) 1.5 L (d) 1.66 L Sol. Let volume of 0.40 M HCl used = xL



“ “ 0.15 M HCl used = 1 L

0.15 1 + 0.40 x = (1+x) 0.25 0.15 + 0.40 x = 0.25 + 0.25x 0.15x = 0.1

x = 15.0

1.0 = 0.66 L

1 + 0.66 = 1.66 L 20. Compute the normality of a solution obtained by mixing 100 ml of 0.1 N-HCl, 400mL of 0.2 N-

HNO3 and 500 ml of 0.125 M-H2SO4.. (a) 0.215 (b) 0.415 (c) 0.615 (d) 0.815

Sol : Eq. Wt. of H2SO4 = 49 Mol. Wt. of H2SO4 = 98

We know Molarity Mol. Wt. = Normality Eq. Wt.

0.125 98 = N 49

N = N 0.250 49

98 0.125

N = N 0.215 500400100

5000.250400 0.2 1000.1

V V V

V N VN VN

311

332211

21. Calculate the molarity of a solution of CaCl2 if on chemical analysis it is found that 500 ml of Cacl2

solution contain 1.505 1023 Cl- ions. a) 0.125 (b) 0.25 (c) 0.75 (d) 1

Sol. (b) CaCl2 2 Cl- 1 2

? 1.505 1023

23

22

10022.6

10525.7

= 0.125

M = 0.125 500

1000 = 0.25 M

22. 250 ml of 0.10 M K2SO4 solution is mixed with 250 ml of 0.20 M KCl solution. What is the concentration of K+ ions in the resulting solution ?

a) 0.1 (b) 0.2 (c) 0.3 (d)0.4

Sol. (b) K2SO4 2K+

0.10 M 2 0.10 = 0.2 M

KCl K+ 0.20M 0.20 M

M = 500

100

500

5050

250250

25020.02502.0

= 0.2 M

23. A beaker containing 20 g sugar in 100g water and another containing 10 g sugar in 100g water are placed under a bell-jar and allowed to stand until equilibrium is reached. How much water will be transferred from one beaker to another ?

a)11.3 g (b) 22.3 g (c) 33.3 g (d) 44.3 g

Sol. (c) m = 12

2

W

1000

M

W

x100

1000

34

10

x100

1000

34

20

x-100

1

x100

2

100 = 3x x = 33.3 g

24. The vapour density of a mixture of NO2 and N2O4 is 38.3 at 26.7oC. Calculate the mass of NO2 in 100g of the mixture. a) 5 g (b) 10 g (c) 15 g (d) 20 g

Sol. (d) V.D. of mixture of NO2 and N2O4 = 38.3



Mol. Wt. of mixture of NO2 and N2O4 = 38.3 2 = 76.6 (Mol. Wt. = 2 V.D.) Let NO2 present in 100g of mixture = x N2O4 present in 100g of mixture = 100 - x Mol. Wt. of NO2 = 46, Mol. Wt. of N2O4 = 92

Now, 76.6

100=

92

x100+

46

x

Solving for x, we get x = 20.1 g 25. An alloy of Iron (54.7%), nickel (45.0%) and manganese (0.3%( has a density of 8.17 g cm-3. How

many iron atoms are there in a block of alloy measuring 10.0 cm 20.0 cm 15.0 cm ?

a) 14.41 1015 (b) 14.41 1018 (c) 14.41 1021 (d) 14.41 1025

Sol. (d) Volume of the block = 10 20 15 cm3

mass of the block = 3000 cm3 8.17 g cm-3 = 24510 g

mass of Fe (54.7%) = 24510 100

7.54g

= 239.41 mol of Fe

number of Fe atoms = 239.41 N0,

= 14.41 1025 atoms 26. The reactant which is entirely consumed in reaction is known as limiting reagent. In the reaction

2A + 4B 3C + 4D. when 5 moles of A react with 6 moles of B. then calculate the moles of C

formed ?

(a) 1.5 (b) 2.5 (c) 3.5 (d) 4.5

Sol. 2A + 4B 3C + 4D According to the above equation. 2 mols of 'A' require 4 moles of 'B' for the reaction.

Hence, for 5 moles of 'A', the moles of 'B' required = 4 mol of A Aofmol 2

B ofmol 4

But we have only 5 moles of 'B' hence, 'B' is the limiting reagent. So amount. Since 4 moles of 'B' give 3 mols of 'C' . Hence 6 moles of 'B' will give

6 mole of B C ofmol 5.4=B of mol4

C ofmol 3

27. Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction given

below :

CaCO3(s) + 2HCl(aq) CaCl2(aq) + CO2(g) + H2O(l) What mass of CaCl2 will be formed when 250 mL of 0.76 M HCl reacts with 1000g of CaCO3 ? (a) 5.5 g (b) 10.5 g (c) 15.5 g (d) 20.5 g

Sol. Number of moles of HCl = 250 mL 1000

M76.0=0.19 mol

Mass of CaCO3 = 1000 g

Number of moles of CaCO3 = g100

g1000= 10 mol

According to given equation 1 mol of CaCO3(s) requires 2 mol of HCl (aq). Hence, for the reacton of 10 mol of CaCO3(s) number of moles of HCl required would be :

10 mol CaCO3 (aq) HCl mol 20 )(CaCO mol 1

(aq) HCl mol 2

3

s

But we have only 0.19 mol HCl (aq), hence, HCl (aq) is limiting reagent. So amount of CaCl2 formed will depend on the amount of HCl available. Since, 2 mol HCl (aq) forms 1 mol of Cacl2, therefore, 0.19 mol of HCl (aq) would give :

0.19 mol HCl (aq) (aq) HCl mol2

(aq) CaClmol 1 2= 0.095 mol

or 0.095 molar mass of CaCl2 = 0.095 111 = 10.54 g



28. 2.76g of silver carbonate on being strongly heated yields a residue weighing

(a) 3.54 g (b) 3.0 g (c) 1.36g (d) 2.16 g Sol. (d)

29. A mixture of formic acid and oxalic acid is heated with concentrated H2SO4. The gas produced is

collected and on its treatment with KOH solution, the volume of the solution decreases by 1/6th. Calculate the molar ratio of the two acids in the original mixture. a) 1:1 (b) 2:1 (c) 3:1 (d) 4:1

Sol. (d) Let x moles of formic acid and y moles of oxalic acid are heated. Conc. H2 SO4

HCOOH H2O + CO x mol x mol

Conc. H2SO4

(COOH)2 H2O + CO + CO2 y mol y mol y mol Total moles of gaseous mixture = moles of CO + moles of CO2 KOH absorbs only CO2 i.e. volume occupied by y moles. Under same conditions, ratios of volumes will be proportional to molar ratio of gases

6

1

2y x

y

gasesboth of Moles

CO of 2

Moles

6y = x + 2y or 4y = x

or 4=y

x

COOH

Ratio of HCOOH: = 4 : 1 COOH 30. Gastric juice contains about 3.0 g of HCl per litre. If a person produces about 2.5 litre of gastric

juice per day, how many antacid tablets each containing 400 mg of Al(OH)3 are needed to neutralise all the HCl produced in one day ? a) 10 (b) 12 (c) 14 (d) 16

Sol . (c) HCl and Al(OH)3 react as

3HCl + Al (OH)3 AlCl3 + 3H2O

36.5 3 g 78 g

Amount of HCl produced in a day = 2.5 3 = 7.5 g

Now 36.5 3 g of HCl require Al(OH)3 = 78g

7.5 g of HCl will required Al(OH)3 = 3×5.36

78 7.5 = 5.34 g

Number of tablets required = 4.0

34.5= 13.35 = 14 tablets.







1. The crystalline salt, Na2SO4.x H2O on heating loses 55.9% of its weight. The formula of the crystalline salt is a) Na2SO4 . 5H2O (b) Na2SO4 . 7H2O (c) Na2SO4 . 10H2O (d) Na2SO4 . 6H2O

2. Haemoglobin (molecular weight ( = 67200) has 0.33% iron by weight. One molecule of haemoglobin contains how many atoms of iron : a) 1 (b) 2 (c) 3 (d) 4

3. The most abundant elements by mass in the body of health human adult are Oxygen (61.4%);

Carbon (22.9%), Hydrogen (10.0%); and Nitrogen (2.6%). The weight which a 75 kg person would

gain if all 1H atoms are replaced by 2H atoms is

(a) 15 kg (b) 37.5 kg (c) 7.5 kg (d) 10 kg 4. A drug marijuana owes its activity to tetra hydro cannabinol, which contains 70% as many carbon

atoms as hydrogen atoms and 15 times as many hydrogen atoms as oxygen atoms. The number of moles in a gram of tetrahydrocannabinol is 0.00318. Determine the molecular formula. a) C11H30O2 (b) C15H30O2 (c) C10H30O2 (d) C21H30O2

5. Two elements X (at. Mass 16) and Y (at. Mass 14) combine to form compounds A, B and C. The ratio of different masses of Y which combine with a fixed mass of X in A, B and C is 1 : 3 : 5. If 32 parts by mass of X combines with 84 parts by mass of Y in B, then calculate in C 16 parts by mass of X will combine with how many parts of Y ? (a) 28 (b) 42 (c) 56 (d) 70

6. Specific volume of cylindrical virus particle is 6.02 10-2 cc/gm, whose radius and length are 7o

A

and 10o

A respectively . If NA = 6.02 1023, find molecular weight of virus ?

a) 15.4 kg/mol (b) 1.54 104 kg/mol

c) 3.08 104 kg mol (d) 3.08 103 kg/mol 7. The specific heat of a metal is 0.16 . Its approximate atomic weight would be :

a) 16 (b) 32 (c) 40 (d) 46 8. 4.0 g of caustic soda (mol. mass 40) contains same number of sodium ions as are present in :

a) 10.6 g of Na2CO3 (mol. mass 106) (b) 58.5 g of NaCl (formula mass 58.5) c) 100 mL of 0.5 M Na2SO4 (formula mass 142) (d) 1 gm equivalent of NaNO3 (mol. mass 85)

9. If NA is Avogadro’s number, then number of valence electrons in 4.2 g of nitride ions (N3-) is a) 2.4 NA (b) 4.2 NA (c) 1.6 NA (d) 3.2 NA

10. Calculate the molality of a solution of ethanol in water in which the mole fraction of ethanol is 0.040. (a) 1.31 M (b) 2.31 M (c) 3.31 M (d) 4.41 M

11. If the density of methanol is 0.793 kg L-1, what is its volume needed for making 2.5 L of its 0.25 M solution ? (a) 15.22 ml (b) 25.22 ml (c) 35.22 ml (d) 45.22 ml

12. The density of 3 molal solution of NaOH is 1.110g mL-1. Calculate the molarity of the solution.

(a) 0.97 M (b) 1.97 M (c) 2.97 M (d) 3.97 M 13. Calculate the percentage composition in terms of mass of a solution obtained by mixing 300g of a

25% and 400g of a 40% solution by mass. (a) 13.33 (b) 23.33 (c) 33.57 (d) 43.57

14. The system that contains the maximum number of atoms is

(a) 4.25g of NH3 (b) 8g of O2 (c) 2 g of H2 (d) 4g of He



15. How many moles of magnesium phosphate, Mg3 (PO4)2 will contain 0.25 mole of oxygen atom?

(a) 0.02 (b) 3.125 10-2 (c) 1.25 10-2 (d) 2.5 10-2 16. The total number of electrons in 18 mL of water (density =1g mL-1 )

(a) 6.023 1025 (b) 6.023 1024 (c) 6.023 18 1023 (d) 6.023 1023

17. Calculate the the volume of 1 molar NaOH solution required to convert 12 g of NaH2PO4 completely into Na3 PO4. (a) 100 (b) 150 (c) 200 (d) 250

18. One litre solution of N/2 HCl is heated in a beaker. It was observed that when the volume of the solution is reduced to 600ml, 3.25g of HCl is lost. Calculate the normality of the new solution? (a) 0.358 (b) 0.50 (c) 0.685 (d) 0.85

19. You are given one litre of 0.15 M HCl and one litre of 0.40 M HCl. What is the maximum volume of 0.25 M HCl which you can make from these solutions without adding any water ?

(a) 2.2 L L (b) 2.5 L (c) 1.5 L (d) 1.66 L 20. Compute the normality of a solution obtained by mixing 100 ml of 0.1 N-HCl, 400mL of 0.2 N-

HNO3 and 500 ml of 0.125 M-H2SO4.. (a) 0.215 (b) 0.415 (c) 0.615 (d) 0.815

21. Calculate the molarity of a solution of CaCl2 if on chemical analysis it is found that 500 ml of Cacl2

solution contain 1.505 1023 Cl- ions. a) 0.125 (b) 0.25 (c) 0.75 (d) 1

22. 250 ml of 0.10 M K2SO4 solution is mixed with 250 ml of 0.20 M KCl solution. What is the concentration of K+ ions in the resulting solution ?

a) 0.1 (b) 0.2 (c) 0.3 (d)0.4 23. A beaker containing 20 g sugar in 100g water and another containing 10 g sugar in 100g water are

placed under a bell-jar and allowed to stand until equilibrium is reached. How much water will be transferred from one beaker to another ?

a)11.3 g (b) 22.3 g (c) 33.3 g (d) 44.3 g 24. The vapour density of a mixture of NO2 and N2O4 is 38.3 at 26.7oC. Calculate the mass of NO2 in

100g of the mixture. a) 5 g (b) 10 g (c) 15 g (d) 20 g

25. An alloy of Iron (54.7%), nickel (45.0%) and manganese (0.3%( has a density of 8.17 g cm-3. How

many iron atoms are there in a block of alloy measuring 10.0 cm 20.0 cm 15.0 cm ?

a) 14.41 1015 (b) 14.41 1018 (c) 14.41 1021 (d) 14.41 1025 26. The reactant which is entirely consumed in reaction is known as limiting reagent. In the reaction

2A + 4B 3C + 4D. when 5 moles of A react with 6 moles of B. then calculate the moles of C

formed ?

(a) 1.5 (b) 2.5 (c) 3.5 (d) 4.5 27. Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction given

below :

CaCO3(s) + 2HCl(aq) CaCl2(aq) + CO2(g) + H2O(l) What mass of CaCl2 will be formed when 250 mL of 0.76 M HCl reacts with 1000g of CaCO3 ? (a) 5.5 g (b) 10.5 g (c) 15.5 g (d) 20.5 g

28. 2.76g of silver carbonate on being strongly heated yields a residue weighing

(a) 3.54 g (b) 3.0 g (c) 1.36g (d) 2.16 g 29. A mixture of formic acid and oxalic acid is heated with concentrated H2SO4. The gas produced is

collected and on its treatment with KOH solution, the volume of the solution decreases by 1/6th. Calculate the molar ratio of the two acids in the original mixture. a) 1:1 (b) 2:1 (c) 3:1 (d) 4:1

30. Gastric juice contains about 3.0 g of HCl per litre. If a person produces about 2.5 litre of gastric juice per day, how many antacid tablets each containing 400 mg of Al(OH)3 are needed to neutralise all the HCl produced in one day ? a) 10 (b) 12 (c) 14 (d) 16







1. Sol. (c) 100 g of crystalline salt contain H2O = 55.9 g Anhydrous salt = 100 – 55.9 = 44.1 g Molecular mass of anhydrous Na2SO4 = 2 × 23 + 32 + 4 × 16 = 142 44.1 g of anhydrous salt combine with H2O = 55.9 g

142 g of anhydrous salt combine with H2O = 1.44

9.55× 142 g = 180 g

181 u of H2O = 18

180 = 10 molecules.

2. Sol. (d) , The number of iron atoms in haemoglobin are:

= Iron of weight Molecular × 100

Iron of Percentage ×n haemoglobi of weight Molecular

= 4= 56 × 010

0.33 × 67200

3. Sol. (c) Given, abundance of elements b mass oxygen = 61.4%, carbon = 22.9%, hydrogen = 10% and

nitrogen = 2.6%

Total weight of person = 75 kg

Mass due to 1H = 100

10×75 = 7.5 kg

1H atoms are replaced by 2H atoms,

Mass due to 2H = (7.5 2) kg

Mass gain by person = 7.5 kg 4. Sol. (d) Let oxygen atom be = x part

Hydrogen atom = 15 x

Carbon atom = 15 x 100

70= 10.5 x

Formula C10.5x H15x Ox

Formula weight = ( 10.5 12x ) + 15x + 16 x = 157 x

molecular wt = 00318.0

1= 314.46

x = 2 molecular formula = C21H30O2 5. Sol. (d) In B 32 parts of X combine with Y = 84 parts

b. 16 parts of X combine with Y = 32

1684= 42 Parts

Now No. of parts of X in both B & C are equal. Different masses of Y which combine with fixed mass of X in B & C are the ratio of 3 : 5

5

3

C inY of Mass

B inY of Mass

5

3

C inY of Mass

42



Mass of Y in C = 3

542 = 70 Parts

6. Sol. (b), Volume of cylindrical virus is = r2l

= 3.14 ( 7 10-8)2 ( 10 10-8)

= 3.14 49 10-16 10-7

= 153.86 10-23 c.c. Weight of one virus particle is

= volumeSpecific

Volume

= 2× 0210.6

10 × 86.153 -23

gm

Molecular weight of virus

=02.6

10× 153.86 -21

6.02 1023 = 15.4 kg/mole

7. Sol. (C) We know from Dulong-Petits law that wt x sp. Heat = 6.4. Approximate atomic wt.

= 0.16

6.4 =

metal of heat sp

heat sp. × wt. At = 40

8. Sol. (c) 4 g NaOH = 0.1 mol NaOH = 0.1 mol Na+ (d) 10.6 g Na2CO3 = 0.1 mol Na2CO3

= 0.2 mol Na+ (e) 58.5 g NaCl = 1 mol NaCl = 1 mol Na+

(f) 100 mL of 0.5 mol Na2SO4 = 1000

5.0× 100

= 0.05 mol Na2SO4 = 0.1 mol Na+

9. Sol. (a) 4.2 g N-3 ions = 14

2.4mol = 0.3 mol = 0.3 × NA ions = 8 × 0.3 NA electrons = 2.4 NA (because each

ion contains 8 valence electrons ).

10. Ans.(b) )O(H n )OHHC(n

)OHHn(C x

252

52OHHC 32

= 0.040 (Given)

The aim is to find number of moles of ethanol in 1 L of the solution which is nearly = 1 L of water (because solution is dilute)

No. of moles in 1 L of water = 1-mol g 18

g1000 = 55.55 moles

Substituting n (H2O) = 55.55 in eqn (i) , we get 55.55 )OHH(C n

)OHHC(n

52

52

= 0.040

or 0.96 n (C2H5OH) = 55.55 × 0.040 or n (C2H5OH) = 2.31 mol Hence, molarity of the solution = 2.31 M

11. Ans. (b) Molar mass of methanol (CH3OH) = 32 g mol-1 = 0.032 kg mol-1

M = 1000

1000

32

1000793.0

= 24.78m Applying M1 × V1 = M2 V2 (Given solution) (Solution to be prepared)

24.78 × V1 = 0.25 × 2.5 L or V1 = 0.02522 L = 25.22 mL. 12. Sol.(c) 3 molal solution of NaOH means that 3 moles of NaOH are dissolved in 1000g of solvent.

Mass of Solution = Mass of Solvent + Mass of Solute

= 1000g + (3 40g) = 1120 g

Volume of Solution = mL110.1

1120 = 1009.00 mL

(since density of solution = 1.110g mL-1) Since 1009 mL solution contains 3 moles of NaOH



Molarity = litre in solutionin of Volume

solute of moles of Number= 1000×

1009.00

mol 3= 2.97 M

13. Sol. (c) Mass of solute in 300g of 25% solution = 75 g Mass of solute in 400g of 40% solution = 160g Total mass of solute = (75 + 160) g = 235 g Total mass of solution = 700 g

% of solute in final solution = 700

100 235 =

% of water in final solution = 100 – 33.5 = 66.43%

14. Sol. (c)Number of atoms = mass molar

mass NA number of atoms in 1 mole

where, NA = Avogadro's number

Number of atoms in 4.25 g of NH3 = 17

25.4 NA 4 = NA

Number of atoms in 8 g of O2 = 2

N=2×N×

32

8 A

A

Number of atoms in 2 g of H2 = AA N2=2×N×2

2

Number of atoms in 4g of He = AA N=1×N×4

4

Thus, 2g of H2 contains the maximum number of atoms among the given. 15. Sol. (b) As 8 moles of O-atoms are present in Mg3(PO4)2

0.25 mole of O-atoms are present in Mg3(PO4)2 1/8 0.25 = 3125 10-2 mol 16. Sol. (b) In 1 mL, number of moles of H2O

17. Sol.(c) NaH2PO4 + 2NaOH Na3PO4 + 2H2O 120 g 80 g 12 g ? 8 g

M = V

1000

M

W

2

2

1 = V

1000

40

8

V = 200 mL

18. Sol.(c) Normality = V

1000

E

w

2

2 , g18.25 1000

1000 x 36.5 x 5.0wor

1000

1000

5.36

w5.0 2

2

When volume becomes 600 ml, 3.25g of HCl is lost.

wt. of HCl left in 600 ml = 18.25 - 3.25 = 15.00 g

Normality = 600

1000

36.5

15

V

1000

mass.Eq

w2 = 0.685 N Ans.

19. Sol. (d) Let volume of 0.40 M HCl used = xL “ “ 0.15 M HCl used = 1 L

0.15 1 + 0.40 x = (1+x) 0.25 0.15 + 0.40 x = 0.25 + 0.25x 0.15x = 0.1



x = 15.0

1.0 = 0.66 L

1 + 0.66 = 1.66 L 20. Sol (a) Eq. Wt. of H2SO4 = 49 Mol. Wt. of H2SO4 = 98

We know Molarity Mol. Wt. = Normality Eq. Wt.

0.125 98 = N 49

N = N 0.250 49

98 0.125

N = N 0.215 500400100

5000.250400 0.2 1000.1

V V V

V N VN VN

311

332211

21. Sol. (b) CaCl2 2 Cl- 1 2

? 1.505 1023

23

22

10022.6

10525.7

= 0.125

M = 0.125 500

1000 = 0.25 M

22. Sol. (b) K2SO4 2K+

0.10 M 2 0.10 = 0.2 M

KCl K+ 0.20M 0.20 M

M = 500

100

500

5050

250250

25020.02502.0

= 0.2 M

23. Sol. (c) m = 12

2

W

1000

M

W

x100

1000

34

10

x100

1000

34

20

x-100

1

x100

2

100 = 3x x = 33.3 g

24. Sol. (d) V.D. of mixture of NO2 and N2O4 = 38.3

Mol. Wt. of mixture of NO2 and N2O4 = 38.3 2 = 76.6 (Mol. Wt. = 2 V.D.) Let NO2 present in 100g of mixture = x N2O4 present in 100g of mixture = 100 - x Mol. Wt. of NO2 = 46, Mol. Wt. of N2O4 = 92

Now, 76.6

100=

92

x100+

46

x

Solving for x, we get x = 20.1 g

25. Sol. (d) Volume of the block = 10 20 15 cm3

mass of the block = 3000 cm3 8.17 g cm-3 = 24510 g

mass of Fe (54.7%) = 24510 100

7.54g

= 239.41 mol of Fe

number of Fe atoms = 239.41 N0, = 14.41 1025 atoms

26. Sol.(d) 2A + 4B 3C + 4D

According to the above equation. 2 mols of 'A' require 4 moles of 'B' for the reaction.

Hence, for 5 moles of 'A', the moles of 'B' required = 4 mol of A Aofmol 2

B ofmol 4

But we have only 5 moles of 'B' hence, 'B' is the limiting reagent. So amount. Since 4 moles of 'B' give 3 mols of 'C' . Hence 6 moles of 'B' will give



6 mole of B C ofmol 5.4=B of mol4

C ofmol 3

27. Sol.(b) Number of moles of HCl = 250 mL 1000

M76.0=0.19 mol

Mass of CaCO3 = 1000 g

Number of moles of CaCO3 = g100

g1000= 10 mol

According to given equation 1 mol of CaCO3(s) requires 2 mol of HCl (aq). Hence, for the reacton of 10 mol of CaCO3(s) number of moles of HCl required would be :

10 mol CaCO3 (aq) HCl mol 20 )(CaCO mol 1

(aq) HCl mol 2

3

s

But we have only 0.19 mol HCl (aq), hence, HCl (aq) is limiting reagent. So amount of CaCl2 formed will depend on the amount of HCl available. Since, 2 mol HCl (aq) forms 1 mol of Cacl2, therefore, 0.19 mol of HCl (aq) would give :

0.19 mol HCl (aq) (aq) HCl mol2

(aq) CaClmol 1 2= 0.095 mol

or 0.095 molar mass of CaCl2 = 0.095 111 = 10.54 g 28. Sol. (d)

29. Sol. (d) Let x moles of formic acid and y moles of oxalic acid are heated. Conc. H2 SO4

HCOOH H2O + CO x mol x mol

Conc. H2SO4

(COOH)2 H2O + CO + CO2 y mol y mol y mol Total moles of gaseous mixture = moles of CO + moles of CO2 KOH absorbs only CO2 i.e. volume occupied by y moles. Under same conditions, ratios of volumes will be proportional to molar ratio of gases

6

1

2y x

y

gasesboth of Moles

CO of 2

Moles

6y = x + 2y or 4y = x

or 4=y

x

COOH

Ratio of HCOOH: = 4 : 1 COOH 30. Sol . (c) HCl and Al(OH)3 react as

3HCl + Al (OH)3 AlCl3 + 3H2O

36.5 3 g 78 g

Amount of HCl produced in a day = 2.5 3 = 7.5 g

Now 36.5 3 g of HCl require Al(OH)3 = 78g

7.5 g of HCl will required Al(OH)3 = 3×5.36

78 7.5 = 5.34 g

Number of tablets required = 4.0

34.5= 13.35 = 14 tablets.

mastering chemistry nbs gurukul · 2019-11-13 · mastering chemistry 2 nbs gurukul, scf- 5 market...

Documents