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• SECTION 1 Sections:

Section 1

Q. 1 A solution contains Zn2+ and Cu2+ ions, each at 0.02M. If the solution is made 1 M in H3O+ and

solution is saturated with H2S, which would precipitate? K1 for H2S is 1 × 10

-7 K2 for H2s is 1 × 10

-14 Ks.p. of ZnS is 1 × 10

-23 and that of CuS is 8 × 10-37

1. CuS precipitates out

2. ZnS precipitates out

3. both precipitate

4. no precipitation occurs

0 out of 4 marks Correct Answer: 1 Solution: (1)

Q. 2 Ks.p. of CaF2 is 3.2 × 10-11M3. It is least soluble in

1. water

2. 0.1 M Ca(NH2)2

3. 0.1 M HNO3

4. 0.1 M NaF

0 out of 4 marks Correct Answer: 4 Solution: (4)

Q. 3 What is the pH of solution obtained by mixing 50 ml of 0.1 M acetic acid with 25 ml of 0.1 M NaOH? Ka of CH3COOH is 2 × 10

-5

1. 4.20

2. 4.76

3. 5.24

4. 7

0 out of 4 marks Correct Answer: 2 Solution: Ans. (2)

Q. 4 A 200.0 ml sample of H3PO3 requires 22.50 ml of 0.1250 M NaOH for complete neutralization. What is the molarity of H3PO3solution?

1. 0.014 M

2. 0.028 M

3. 0.007 M

4. 0.005 M

0 out of 4 marks Correct Answer: 3 Solution: Ans. (3)

Q. 5 `20 volume' H2O2 means

1. 20% V/V

2. 6.07 W/V

3. 20% W/V

4. 6.07 V/V

-1 out of 4 marks Correct Answer: 2 Solution: (2)

Q. 6 The molarity of a solution of H2O2 is 2. The volume strength of this solution will be

1. 11.2 volume

2. 10 volume

3. 22.4 volume

4. 5.6 volume

0 out of 4 marks (content provider) Correct Answer: 3 Solution: Ans. (3) Volume strength of H2O2 = volume of O2 released per volume of H2O2 One litre of H2O2 contains = 2 moles of H2O2 H2O2 H2O + 1/2 O2 2moles 1 mole At NTP 1 mole O2 = 22.4 l So 1l H2O2 gives 22.4 l of oxygen

Q. 7 Molarity of HCl ( ) if its density is 1.17 gm/cc is

1. 36.5

2. 32

3. 18.25

4. 63.5

-1 out of 4 marks (content provider)

Correct Answer: 2 Solution: Ans. (2)

Q. 8 What is the normality of 0.30 M H3PO4 (tribasic acid ) in the following reaction ?. H3PO4 + 2OH

- HPO4-2 + 2H2O

1. 0.30 N

2. 0.60 N

3. 0.90 N

4. 0.15 N

0 out of 4 marks (content provider) Correct Answer: 2 Solution: Ans. (2) 2 H+ ions from H3PO4 react with 2 OH- ions , hence 'n' factor for this acid base reaction for H3PO4 = 2 Normality = molarity * n = 0.3 *2 = 0.6 N

Q. 9 For the following disproportionation reaction 5Br2 + 6OH

- 5Br- + 5BrO3- + 3H2O

correct statements are:

1. equivalent weight of Br2 when it is reduced to Br

- is 80

2. equivalent weight of Br2 when it is oxidized to BrO3

- is 16

3. equivalent weight of Br2 in the net reaction is 96

4. all are correct

0 out of 4 marks (content provider) Correct Answer: 4 Solution: Ans. (4) Equivalent weight = molecular weight / 'n' factor .

When reduced to Br- ions , change in oxidation state = 1 per Br atom , since there are 2 atoms of Br , 'n' factor = 2. Equivalent weight = 160/2 = 80 When oxidized to BrO3

- , change in ox. State per Br atom = 5 , since there are 2 atoms of Br , 'n' factor = 10 Equivalent weight = 160/10 = 16 Total Equivalent weight = 80+16 = 96

Q. 10 The normality of 10 ml of a '20 V' H2O2 is

1. 1.79

2. 3.58

3. 60.68

4. 6.86

-1 out of 4 marks (content provider) Correct Answer: 2 Solution: Ans. (2) Volume strength of H2O2 = volume of O2 released per volume of H2O2 H2O2 H2O + 1/2 O2 1 l 20 l 20l = 20/22.4 moles Moles of H2O2 = 40/22.4 'n' factor =2 Number of equivalents = 3.58 moles per l = normality .

Q. 11 For the conversion of 0.240 g NaH2PO4 in a solution to monohydrogen phosphate, 21.4 ml NaOH solution is needed. The molarityof NaOH solution will be

1. 0.093

2. 0.93

3. 0.087

4. 0.083

-1 out of 4 marks (content provider) Correct Answer: 1 Solution: Ans. (1) Number of moles of NaH2PO4 = 0.24/120 0.24/120 45= 0.02 moles . For this reaction , since it goes from di to mono hydrogen , 'n' factor = 1 So number of equivalents used = 0.02 = Number of moles of NaOH Molarity of NaOH = 0.02*1000/21.4 = 0.093

Q. 12 22.7 ml of N/10 Na2CO3 solution neutralizes 10.2 ml of dilute H2SO4. Then the volume of water that

must be added to 400 ml of same H2SO4 to make it exactly is

1. 245 ml

2. 484.6 ml

3. 480 ml

4. 490.2 ml

0 out of 4 marks (content provider) Correct Answer: 4 Solution: Ans. (4) Number of equivalents must be the same , say normality of H2SO4 =N 0.1*22.7 = N*10.2 N=0.223 and number of equivalents present in 400 ml = 0.22*400 = 89 mili equivalents To make it 0.1 N 0.1*v = 89 mili equivalents V = 890 ml So additional v = 490 ml

Q. 13 50 g of 8% by mass NaOH is mixed with 100 g of 8% by mass of HCl, the resulting solution is

1. acidic

2. basic

3. neutral

4. strongly basic

-1 out of 4 marks (content provider) Correct Answer: 1 Solution: Ans. (1) 50gm solution has 8% NaOH by mass = 4 gms = 0.1 moles of NaOH 100 gm solution has 8% by mass of HCl = 8gms = 0.21 moles of HCl , Per mole of NaOH requires one mole of HCl for complete neutralization . Since HCl is present in excess , hence the solution is acidic . Answer (1)

Q. 14 How many moles of FeCr2O4 can be oxidized completely by 1 mole of KMnO4 in acidic medium?

1. 7

2. 5

3.

4.

-1 out of 4 marks (content provider) Correct Answer: 4 Solution: Ans. (4) Cr2O7

2- + 6 Fe2+ + 14H+ -->> 2 Cr3+ + 6 Fe3+ + 7 H2O Fe goes from +2 to +3 hence it's 'n' factor = 1 Cr goes from +3 to + 6 and since there are 2 atoms of Cr per molecule , it's 'n' factor = 2*3 = 6 Total 'n' factor for FeCr2O7 = 7 In KMnO4 , Mn goes from + 7 to + 2 , it's 'n' factor = +5 Since the number of equivalents be same , 1*5 = m*7 M=5/7

Q. 15 A solution of KMnO4 is reduced to MnO2. The normality of solution is 0.6. The molarity is:

1. 1.8 M

2. 0.6 M

3. 0.1 M

4. 0.2 M

0 out of 4 marks (content provider) Correct Answer: 4 Solution: Ans. (4) Mn goes from +7 in KMnO4 to +4 in MnO2 Hence change in oxidation state per molecule = + 3 = 'n' factor Normality = molarity * 'n' Molarity = 0.6/3 = 0.2

Q. 16 500 ml of 0.2M H2SO4 is mixed with 250 ml of 0.1M Ba(OH)2, the resulting normality of solution (acidic) is

1. 0.2

2. 0.1

3. 0.5

4. 0.25

0 out of 4 marks (content provider) Correct Answer: 1 Solution: Ans. (1) Normality of acids = number of moles of H+ ions generated per liter Normality of bases = number of moles of OH- ions generated per liter Total equivalents of H+ ions = 0.5*0.2*2 = 0.2 Total equivalents of OH- ions = 0.25*0.1*2 = 0.05 Total equivalents of H+ remaining = 0.15 Total volume = 0.75 l Normality = 0.15/0.75 = 0.2

Q. 17 An aqueous solution of 6.3 g oxalic acid dihydrate is made up to 250 ml. The volume of 0.1N NaOH required to completely neutralize 10 ml of this solution is

1. 40 ml

2. 20ml

3. 10 ml

4. 4 ml

-1 out of 4 marks (content provider) Correct Answer: 1 Solution: Ans. (1) Mol wt of H2C2O4.2H2O = 126 number of moles present = 6.3/126 = 0.05 moles 'n' factor for oxidation to CO2 = 2 No. of equivalents = 0.05*2 =0.1 No. of equivalents in 10 ml solution = 0.1/25 = 0.004 Number of equivalents of NaOH should be same 0.1*v= 0.004 V = 40 ml

Q. 18 50 ml of an aqueous solution of H2O2 treated with an excess of KI solution in presence of dilute H2SO4. The liberated I2 required 20 ml of 0.05M Na2S2O3 solutions for complete reaction. The strength of H2O2 is

1. 0.224V

2. 0.112V

3. 0.056V

4. None of these

0 out of 4 marks (content provider) Correct Answer: 2 Solution: Ans. (2) 'n' factor for Na2S2O3 = 1

Number of equivalents of Na2S2O3 = number of equivalents of H2O2 20*0.05*1 = 50*N N= 0.02 'n' factor for H2O2 = 2 Molarity for H2O2 = 0.02/2 = 0.01 Volume strength of H2O2 = volume of O2 released per volume of H2O2 One litre of H2O2 contains = 0.01 moles of H2O2 H2O2 H2O + 1/2 O2 This gives 0.005 moles of O2 = 0.112 l = vol strength of H2O2

Q. 19 0.2 mol of HCl and 0.1 mol of barium chloride were dissolved in water to produce a 500 ml. solution. The molarity of the Cl- is

1. 0.06M

2. 0.09M

3. 0.12M

4. 0.80M

-1 out of 4 marks (content provider) Correct Answer: 4 Solution: Ans. (4) 0.2 moles of HCl = gives 0.2 moles 0f Cl- ions 0.1 mole of BaCl2 = gives 0.2 moles of Cl- ions so we have a total of 0.4 moles of Cl-ions volume = 0.5 l Molarity = 0.4/0.5 = 0.8 M

Q. 20 0.7g of Na2CO3×xH2O were dissolved in water and the volume was made to 100 ml, 20 ml of this solution required 19.8 m of N/10 HCl for complete neutralization. The value of x is:

1. 7

2. 3

3. 2

4. 5

0 out of 4 marks (content provider) Correct Answer: 3 Solution: Ans. (3) say total molecular weight = m 0.7/5 = 0.14 gm required 19.8*0.1 meq of NaOH = 1.98 meq Number of moles of Na2CO3.xH2O = 0.14/m 'n' factor = 2 0.14*2/m = 1.98/1000 M = 142 = 106+18x X = 2

Q. 21 The hydrated salt Na2SO4×nH2O undergoes 55.9% loss in weight on heating and becomes anhydrous. The value of n will be

1. 5

2. 3

3. 7

4. 10

0 out of 4 marks (content provider) Correct Answer: 4 Solution: Ans. (4) say total weight of H20 in a mole = x then x/(x + 132) =0.559 solving this x = 180 So degree of hydration = 10

Q. 22 What volume of 0.1 M H2SO4 will be required to produce 17.0g of H2S by the reaction 5H2SO4 + 8NaI 4Na2SO4 + 4I2 + H2S + 4H2O?

1. 70.0 L

2. 50.0L

3. 25.0 L

4. 5.0 L

0 out of 4 marks (content provider) Correct Answer: 3 Solution: Ans. (3) 17gn H2S = 0.5 moles H2S This will require 5*0.5 = 2.5 moles of H2SO4 Say volume = v 0.1*v = 2.5 V = 25 l

Q. 23 How many grams of copper will be replaced in 2 L of a 1.50 M CuSO4 solution if the latter is made to react with 27.0g of aluminium? (Cu = 63.5, Al = 27.0)

1. 190.50g

2. 95.25g

3. 31.75g

4. 10.65g

0 out of 4 marks (content provider) Correct Answer: 2 Solution: Ans. (2) 3CuSO4 + 2Al Al2(SO4)3 + 3 Cu 3 moles 1 mole ? Al is the limiting reagent , 1.5 moles of Cu will be obtained 1.5 mole = 95.25 gms of Cu .

Q. 24 An aqueous solution of urea containing 18g urea in 1500cm3 of solution has a density of 1.502

g/cm3 of solution. If the molecular weight of urea is 60, then the molality of solution is

1. 0.2

2. 0.192

3. 0.064

4. 1.2

0 out of 4 marks (content provider) Correct Answer: 2 Solution: Ans. (2) 18gm urea = 18/60 = 0.3 moles 1.5 l of solution has a mass of 1.5*1.5 = 2.25 kgs 2.25 kgs has 0.3 moles 1 kg has 0.3/2.25 = 0.13 So molality = 0.13 moles/kg

Q. 25 The solubility of Fe(OH)3 is x mol L-1. Its Ksp would be

1. 9x3

2. 3x4

3. 27x4

4. 9x4

4 out of 4 marks (content provider) Correct Answer: 3 Solution: Ans. (3) Fe(OH)3 Fe3+ + 3 OH- X x 3x Ksp = [x][3x]3 Ksp = 27x4

Q. 26 Ksp of H2S is 1 × 10-22. [S2-] in a buffer of pH 6 is :

1. 1 × 10-16 M

2. 1 × 10-12 M

3. 1 × 10-10 M

4. 1 × 10-8 M.

-1 out of 4 marks (content provider) Correct Answer: 3 Solution: Ans. (3) Ksp =[H+]2[S2-] Since[ H+ ] = 10-6 Putting the values , [S2-] = 10-10

Q. 27 Which of the following solutions will have pH of 4.74:

1. 100 mL of 1 M CH3COOH (pKa = 4.74) at the equivalent point using NaOH

2. 50 mL of 1 M CH3COONa + 25 mL of 1 M HCl

3. 50 mL of 1 M CH3COOH + 25 mL of 1 M NaOH

4. Both (1) and (3).

0 out of 4 marks (content provider) Correct Answer: 4 Solution: Ans. (4) At equivalent point ph = pKa And when 50 mL of 1 M CH3COOH + 25 mL of 1 M NaOH are used , salt acid buffer is formed And ph = pKa + log ([salt]/[acid ]) 25 ml * 1 M of salt is formed and the same amount of acid is left . So again ph = pKa = 4.74

Q. 28 pKa (CH3COOH) is 4.74. x mol of lead acetate and 0.1 mol of acetic acid in one L solution make a solution of pH = 5.04. Hence, x is :

1. 0.2

2. 0.05

3. 0.1

4. 0.02.

0 out of 4 marks (content provider) Correct Answer: 3 Solution: Ans. (3) ph = pKa + log ([salt]/[acid ]) 5.04 = 4.74 + log (x/0.1) 0.3 = log(x/0.1) X = 0.2

Q. 29 pH of a saturated solution of Ba(OH)2 is 12. Hence, Ksp of Ba(OH)2 is :

1. 5 × 10-7 M3

2. 5 × 10-4 M2

3. 1 × 10-6 M3

4. 4 × 10-6 M3.

0 out of 4 marks (content provider) Correct Answer: 1 Solution: Ans. (1) Ba(OH)2 Ba2+ + 2 OH- s/2 s now [H+] = 10-12 so [OH-] = 10 -2 = s Ba2+ = s/2 = 5*10-3 Ksp = (s/2)(s)2 = 5*10-7 M3

Q. 30 For a sparingly soluble salt AP Bq the relationship of it's solubility product Ksp with its solubility S is given by:-

1. KSP = S

P + q pp qq

2. KSP = S

p + q pq qp

3. KSP = S

pq pp qq

4. KSP = S

pq pp + q qp + q

0 out of 4 marks (content provider) Correct Answer: 1 Solution: Ans. (1) ApBq pA + qB S ps qs Ksp = (ps)p(qs)q = ppqq sp+q

Q. 31 KSP of CuS, Ag2S and HgS are 10-31, 10-44 and 10-54 respectively. Select the correct order for their

solubility in water

1. Ag2S > HgS > CuS

2. HgS > CuS > Ag2S

3. HgS > Ag2S > CuS

4. Ag2S > CuS > HgS

0 out of 4 marks (content provider) Correct Answer: 4 Solution: Ans. (4) Extent of solubility is inversely proportional to solubility product Hence ,

Q. 32 A gas 'X' at 1 atm is bubbled through a solution containing a mixture of 1 MY- and 1 MZ- at 25oC. If

the reduction potential of Z > Y > X, then

1. Y will oxidize X and not Z

2. Y will oxidize Z and not X

3. Y will oxidize both Z and X

4. Y will reduce both X and Z

0 out of 4 marks (content provider) Correct Answer: 1 Solution: Ans. (1) Since the reduction potential of X is less than that of Y , it will get reduced easily by Y , which gets oxidized .

Q. 33 The dissociation constants of two weak acids HA1 and HA2 are 3 × 10-4 and 1.8 × 10-5, respectively.

The relative strength of the acids will be

1. 1 : 4

2. 4 : 1

3. 1 : 16

4. 16 : 1

0 out of 4 marks (content provider) Correct Answer: 1 Solution: Ans. (1) Left

Q. 34 The hydrogen ion concentration and pH of the solution made by mixing 100 ml of 1M HNO3 with 100 ml of 0.8 M KOH, are

1. [H+] = 0.1, pH = 1

2. [H+] = 0.01, pH = 2

3. [H+] = 1 × 10-12, pH = 12

4. [H+] = 1 × 10-7, pH = 7

0 out of 4 marks (content provider) Correct Answer: 1

Solution: Ans. 1 100 ml of 1 M HNO3 , gives 0.1 moles of H+ And 100 ml of 0.8 M KOH , gives 0.08 moles of OH- Total volume = 0.2 l So 0.08 moles of H+ will be fully consumed and salt will be formed . while 0.02 moles of H+ will remain , so concentration of H+ ion = 0.02/0.2 = 0.1 M And Ph = -log[H+] = 1

Q. 35 Which of the following solution will have pH close to 1.0?

1. 100 ml of M/10 HCl + 100 ml of M/10 NaOH

2. 55 ml of M/10 HCl + 45 ml of M/10 NaOH

3. 10 ml of M/10 HCl + 90 ml of M/10 NaOH

4. 75 ml of M/5 HCl + 25 ml of M/5 NaOH

0 out of 4 marks (content provider) Correct Answer: 4 Solution: Ans. (4) In 75 ml of M/5 HCl + 25 ml of M/5 NaOH Moles of [H+] = 75/5 = 15 milli moles Moles of [OH-]= 25/5 = 5 milli moles Number of moles of H+ remaining = 10 milli moles Total volume = 100 ml SO remaining H+ ion concentration = 10 /100 = 0.1 M And Ph = 1

Q. 36 The correct order of increasing [H3O+] in the following aqueous solution is

1. 0.01 M H2S < 0.01 M H2SO4 < 0.01 M NaCl < 0.01 M NaNO2

2. 0.01 M NaCl < 0.01 M NaNO2 < 0.01 M H2S < 0.01 M H2SO4

3. 0.01 M NaNO2 < 0.01M NaCl < 0.01 M H2S < 0.01 M H2SO4

4. 0.01 M H2S , 0.01 M NaCl < 0.01 M NaNO2 < 0.01 M H2SO4

-1 out of 4 marks (content provider) Correct Answer: 3 Solution: Ans. (3) H2SO4 is the strongest acid , hence it has the highest H+ ion concentration , and after that there is H2S , among the salts , NaCl is the salt of the stronger acid (HCl), than NaNO2 , hence NaCl has a higher H+ ion concentration .

Q. 37 The vapour pressure of a solution of a non-volatile solvent (A) in solvent (B) is 95% of vapour pressure of solvent at the same temperature. If MB = 0.3 MA where MB and MA are molecular weights of B and A respectively the weight ratio of solvent and solution are

1. 0.15

2. 0.57

3. 0.2

4. 0.4

0 out of 4 marks (content provider) Correct Answer: 2 Solution: Ans. (2) Pa = Pa*(Xa) 0.95 = Xa = (Wa/Ma) / [(Wa/Ma)+ (Wb/Mb)] Substituting the values , we get Wa/Wb = 0.57

Q. 38 At 4oC an osmotic pressure rise of 2.6 mm of protein solution was observed which was prepared

by dissolving 0.75 g protein in 125 cm3 of an aqueous solution. If density of solution is 1.0 gm/cm3 the mol wt of protein will be:

1. 9.4 × 105

2. 5.4 × 105

3. 5.4 × 1010

4. 9.4 × 1010

0 out of 4 marks (content provider) Correct Answer: 2 Solution: Ans. (2)

Q. 39 The vapour pressure of a solution of 5 g of non electrolyte in 100 g of water at a particular temperature is 2985 Nm-2. The vapour pressure of pure water at that temperature is 3000 Nm-2. The molecular weight of the solute is

1. 180

2. 90

3. 270

4. 200

0 out of 4 marks (content provider) Correct Answer: 1 Solution: Ans. (1)

Q. 40 Three sparingly soluble salts A2X, AX and AX3 have the same solubility product. Their solubilities will be in the order

1. AX3 > AX > A2X

2. AX3 > A2X > AX

3. AX > AX3 > A2X

4. AX > A2X > AX3

-1 out of 4 marks (content provider) Correct Answer: 4 Solution: Ans. (4) For

ApBq pA + qB S ps qs Ksp = (ps)p(qs)q = ppqq sp+q For AX ..... P=1 , q= 1 Ksp =s2

S= Ksp For AX2 ..... P =1 , q = 2 Ksp =4s3

S = ( 4Ksp )3 So the value of s keeps on decreasing hence solubility decreases as , q/p ratio increases .

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