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QUESTION: 1

H_{2}O_{2} acts as both oxidising as well as reducing agent its product is H_{2}O, when act as oxidising agent and its product is 0_{2} when act as of reducing agent. The strength of '10P means one liter of H_{2}O_{2} solution on decomposition at S.T.P. condition liberate 10 liters of oxygen gas (H_{2}O → H_{2}O + 1/2 O_{2}) 15 gm Mn0_{4})_{2} sample containing inert impurity is completely reacting with 100 ml of '11.2 V' H_{2}O_{2}, then what will be the % purity of Ba (Mn0_{4})_{2 }in the sample? (Atomic mass Ba = 137, Mn = 55)

Solution:

QUESTION: 2

1.2 gm of carbon is burnt completely in oxygen (limited supply) to produce CO and CO_{2}. This mixture of gases is treated with solid I_{2}O_{5} (to know the amount of CO produced), the liberated iodine required 120 ml of 0.1 M hypo solution for complete titration. The % of carbon converted into CO is:

Solution:

Hence, the correct answer is Option D.

QUESTION: 3

Nitrogen (N), phosphorus (P) and Potassium (K) are the main nutrients in plant fertilizers. According to an industry convention, the numbers on the label refer to the mass % of N, P_{2}O_{5} and K_{2}O in that order calculate the N : P : K ratio of a 30 : 10 : 10, fertilizer in terms of moles of each element and express it as x: y : 1.

Solution:

Molar mass of N is 14.0067 g/mol.

Molar mass of P_{2}O_{5} is 30.973762*205 = 6349.62121 g/mol.

Molar mass of K_{2}O is 39.0983*20 = 781.966 g/mol

Ratio of masses is 30:10:10

Therefore for each 50 g of fertilizer:

For N: 30 g / 14.0067 g/mol = 30/14.0067 moles

For P_{2}O_{5}: 10 g / 6349.62121 g/mol = 10/6349.62121 moles

For K_{2}O: 10 g / 781.966 g/mol = 10/781.966 moles

Ratio of elements in terms of moles:

30/14.0067 : 10/6349.62121 : 10/781.966

30/14.0067*781.966/10 : 10/6349.62121*781.966/10 : 10/781.966*781.966/10

167.484 : 0.123 : 1.0

QUESTION: 4

In what ratio should a 15% solution of acetic acid be mixed with a 3% solution of the acid to prepare a 10% solution(all percentages are mass/mass percentages):

Solution:

QUESTION: 5

105 ml of pure water at 4°C saturated with NH_{3} gas yielded a solution of density 0.9 g/ml and containing 30% NH by mass. Find the volume of resulting NH_{3} solution.

Solution:

Given -

volume of water = 105 mL

density of solution = 0.9 g/mL

Calculation -

Since ammonia dissolved in water is 30% by weight, then using mass % formula, we can calculate the amount of ammonia in water

At 4 degree C, density of water = 0.999 g/mL

since density = mass /volume

0.999 = mass/ 105

mass of water = 104.895 g

now, let mass of ammonia be x.

30x + 3146.85 = 100x

70x = 3146.85

x = 44.955 g

Thus, ammonia dissolved in 105 mL water is 44.955 g

Therefore, total mass of solution = mass of ammonia + mass of water

Mass of solution = 44.955 + 104.895

Mass of solution = 149.85 g

So,density of solution = mass of solution / volume of solution

0.9 g/mL = 149.85 g / Volume of solution

Volume of solution = 149.85/ 0.9

Volume of solution = 166.5 mL

QUESTION: 6

The equivalent mass of H_{3}BO_{3} (M = Molar mass of H_{3}BO_{3}) in its reaction with NaOH to from Na_{2}B_{4}O_{7} is equal to –

Solution:

QUESTION: 7

X gram of pure As_{2}S_{3} is completely oxidised to respective highest oxidation states by 50 ml of 0.1 M hot acidified KMnO_{4} then x mass of As_{2}S_{3} taken is : (Molar mass of As_{2}S_{3} = 246)

Solution:

QUESTION: 8

The number of moles of ferrous oxalate oxidised by one mole of KMnO_{4} is

Solution:

QUESTION: 9

How many moles of KMnO_{4} are needed a mixture of 1 mole of each FeSO_{4} & FeC_{2}O_{4} in acidic medium

Solution:

QUESTION: 10

In the reaction

Na_{2}S_{2}O_{3} + 4C1_{2} + 5H_{2}O → Na_{2}SO_{4} + H_{2}S0_{4} + 8HCI

the equivalent weight of Na_{2} S_{2} O_{3} will be

(M= molecular weight of Na_{2}S_{2}O_{3})

Solution:

QUESTION: 11

C_{2}H_{4} + xO_{2} -> 2CO_{2} + yH_{2}O, what is the value of x + y?

Solution:

The balance combustion equation is C_{2}H_{4} + 3O_{2} -> 2CO_{2} + 2H_{2}O, => x = 3, y = 2, => x + y = 5.

QUESTION: 12

The following equations are balanced atomwise and chargewise.

(i) Cr_{2}O_{7}^{2-} + 8H+ + 2H_{2}O_{2} → 2Cr^{3}+ + 7H_{2}O + 2O_{2}

(ii) Cr_{2}O_{7}^{2-} + 8H+ + 5H_{2}O_{2}→ 2Cr^{3+} + 9H_{2}O + 4O_{2}

(iii) Cr_{2}O_{7}^{2-} + 8H+ + 7H_{2}O_{2}→ 2Cr+ + 11H_{2}O + 5O_{2}

The precise equationl equations representing the oxidation of H_{2}O_{2} is/are

Solution:

The correct answer is option A

Cr_{2}O_{7}^{2-} converts into Cr^{3+} in acidic medium I.e. in H+ medium.

First balance the Cr atom on both sides and then Oxygen atom. H^{+} is in excess due to acidic medium.

Add H^{+} as +ve charge to balance the charge on both sides.

QUESTION: 13

An excess of NaOH was added to 100 mL of a ferric chloride solution. This caused the precipitation of 1,425 g of Fe(OH)_{3}. Calculate the normality of the ferric chloride solution

Solution:

QUESTION: 14

In the reaction CrO_{5} + H_{2}SO_{4} →Cr_{2}(SO_{4})_{3} + H_{2}O + O_{2} one mole of CrO_{5} will liberate how many moles of O_{2}

Solution:

**The balanced chemical reaction is,**

**From the balanced chemical reaction, we conclude that**

As, 4 moles of CrO_{5} react to give 7 moles of O_{2}

So, 1 mole of CrO_{5} react to give 7/4 moles of O_{2}

QUESTION: 15

0.4g of a polybasic acid H_{n}A (all the hydrogens are acidic) requires 0.5g of NaOH for complete neutralization. The number of replaceable hydrogen atoms in the acid and the molecular weight of 'A' would be : (Molecular weight of the acid is 96 gms.)

Solution:

QUESTION: 16

A solution of Na_{2}S_{2}O_{3} is standardized iodimetrically against 0.1262 g of Mr% This process requires 45 mL of the Na_{2} S_{2}0_{3} solution. What is the strength of the Na_{2}S_{2}O_{3}?

Solution:

QUESTION: 17

25.0 g of FeSO_{4}.7H_{2}O was dissolved in water containing dilute H_{2}SO_{4}, and the volume was made up to 1.0 L. 25.0 mL of this solution required 20 mL of an N/10 KMnO_{4} solution for complete oxidation. The percentage of FeSO_{4}^{.} 7H_{2}O in the acid solution is

Solution:

QUESTION: 18

1.0 mol of Fe reacts completely with 0.65 mol of O_{2} to give a mixture of only Fe0 and Fe_{2}O_{3}. The mole ratio of ferrous oxide to ferric oxide is

Solution:

2Fe + O_{2} →→ 2FeO

4Fe + 3O_{2} →→2Fe_{2}O_{3}

Let x moles of Fe reacts to produce FeO

Thus amount of Fe reacted to produce Fe_{2}O_{3} = 1-x Moles

Now

2 Moles of Fe reacts to produce FeO with = 1 moles of O_{2}

1 moles of Fe reacts to produce FeO with = 1/2 moles of O_{2}

x moles of Fe to produce FeO with = 0.5x moles of O_{2}

4 moles of Fe reacts to produce Fe_{2}O_{3} with = 3 moles of O_{2}

1-x moles of Fe reacts to produce Fe_{2}O_{3} = (3/4) x (1-x) Moles of O_{2}

EQUATION :

1) 0.5x + 3/4(1−x) =0.65 moles of O_{2}

Thus

x = 0.4 moles

2 Moles of Fe produces = 2 mole of FeO

0.4 moles of Fe produces = 0.4 moles of FeO

4 moles of Fe produces = 2 moles of Fe_{2}O_{3}

(1-0.4 )moles of Fe produces = (1/2) x 0.6 Moles = 0.3 moles of Fe_{2}O_{3}

Thus

FeO/Fe_{2}O_{3} = 0.4/0.3 = 4/3

QUESTION: 19

25 mL of a solution containing HCf and H_{2}S0_{4} required 10 mL of a 1 N NaOH solution for neutralization. 20 mL of the same acid mixture on being treated with an excess of AgNO_{3} gives 0.1425 g of AgCf. The normality of the HC and the normality of the H_{2}S0_{4} are respectively

Solution:

QUESTION: 20

If 10 gm of V_{2}O_{5} is dissolved in acid and is reduced to V^{2+} by zinc metal, how many mole of I_{2} could be reduced by the resulting solution if it is further oxidised to VO^{2+ }ions ?

[Assume no change in state of Zn^{2}+ions] (V = 51, 0 = 16, I = 127) :

Solution:

QUESTION: 21

0.70 g of mixture (NH_{4})_{2} SO_{4} was boiled with 100 mL of 0.2 N NaOH solution till all the NH_{3}(g) evolved and get dissolved in solution itself. The remaining solution was diluted to 250 mL. 25 mL of this solution was neutralized using 10 mL of a 0.1 N H_{2}SO_{4} solution. The percentage purity of the (NH_{4})_{2} SO sample is

Solution:

QUESTION: 22

A mixed solution of potassium hydroxide and sodium carbonate required 15 mL. of an N/20 HCI solution when titrated with phenolphthalein as an indicator. But the same amount of the solution when titrated with methyl orange as an indicator required 25 mL of the same acid. The amount of KOH present in the solution is

Solution:

QUESTION: 23

The percentage of copper in a copper(II) salt can be determined by using a thiosulphate titration. 0.305 gm of a copper(II) salt was dissolved in water and added to, an excess of potassium iodide solution liberating iodine according to the following equation

2Cu^{2} (aq) + 4I^{–} (aq) 2CuI(s) + I_{2}(aq)

The iodine liberated required 24.5cm^{3} of a 0.100 mole dm^{-3} solution of sodium thiosulphate

2S_{2}0_{3}^{2-} (aq) + I_{2}(aq) → 2I^{–} (aq) + S_{4}0_{6}^{2-} (aq)

the percentage of copper, by mass in the copper (ll) salt is. [Atomic mass of copper = 63.5]

Solution:

QUESTION: 24

KIO_{3} reacts with KI to liberate iodine and liberated Iodine is titrated with standard hypo solution, The reactions are

1. IO_{3}^{–} + I^{–} →I_{2} ** (valency factor = 5/3)**

2. I_{2} + S_{2}O_{3}^{2–} → S_{4}O_{6}^{2–} + I^{– }**(valency factor = 2)**

meq of hypo = meq of I_{2} = meq of IO_{3}^{–} + meq of I^{–}

IO_{3}^{–} react with I^{–} ⇒ meq of IO_{3}^{–} = meq of I^{–}

**Statement-1 : **meq of hypo = 2 x meq of IO_{3}^{–}

**Statement-2 : **valency factor of I_{2} in both the equation are different therefore we cannot equate milliequivalents in sequence

Solution:

The correct answer is option A

Milliequivalent of IO_{3}^{−} is equal to milliequivalent of I^{−}.

Milliequivalent of hypo is equal to milliequivalent of IO_{3}^{−} plus milliequivalent of I^{−}.

The milliequivalent of hypo is twice the milliequivalent of IO_{3}^{−}. It is also twice the milliequivalent of I^{−}.

QUESTION: 25

In the reaction:

2 C_{2}H_{6} + 7 O_{2} → 4 CO_{2} + 6 H_{2}O

how many moles of CO_{2} are formed when 1 mole of O_{2} is consumed?

Solution:

1 mole O_{2} x (4 mole CO_{2} / 7 mole O_{2} ) = 4/7 mole CO_{2}

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