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BITSAT Chemistry Test - 6 - JEE MCQ


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30 Questions MCQ Test - BITSAT Chemistry Test - 6

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BITSAT Chemistry Test - 6 - Question 1

In the conversion of Br2 to BrO-3 the oxidation number of Br changes from

Detailed Solution for BITSAT Chemistry Test - 6 - Question 1

Oxidation number of Br in Br2 is zero because it is in elemental state,
Thus  O.N. of Br (in Br2) =0.
Let the oxidation number of Br in BrO-3 is x, oxidation number of oxygen is−2
 x+(−2×3)=−1
x+(−6)= −1
x=+6−1
x=+5
Thus, oxidation number changes from 0 to+5.

BITSAT Chemistry Test - 6 - Question 2

In the reaction, SO2 + 2H2S → 3S + 2H2O the substance that oxidised is,

Detailed Solution for BITSAT Chemistry Test - 6 - Question 2


The oxidation number of S in H2S increases from −2 to 0 in elemental sulphur and hence, H2S gets oxidized while In SO2 oxidation number of S decreases + 4 to zero. So SO2 get reduced.

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BITSAT Chemistry Test - 6 - Question 3

Oxidation involves

Detailed Solution for BITSAT Chemistry Test - 6 - Question 3

Addition of electronegative atom or removal of electropositive atom is called oxidation. Loss of Hydrogen is called Oxidation.

BITSAT Chemistry Test - 6 - Question 4

How do you account for the following observations ?
Though alkaline potassium permanganate and acidic potassium permanganate both are used as oxidants, yet in the manufacture of benzoic acid from toluene we use alcoholic potassium permanganate as an oxidant. Why ? Write a balance redox equation for the reaction.

Detailed Solution for BITSAT Chemistry Test - 6 - Question 4

Oxidation of toluene to benzoic acid in acidic medium

Oxidation of toluene to benzoic acid in basic and neutral medium

On industrial scale, alcoholic potassium permanganate is preferred to acidic or alkaline potassium permanganate because in the presence of alcohol both the reactants KMnO4 and C6H5CH3 are mixed very well and form homogeneous solution and in homogeneous medium reaction takes place faster than in heterogeneous medium. Further more in neutral medium OH- ions are produced in the reaction itself.

BITSAT Chemistry Test - 6 - Question 5

What will be the ratio of number of moles of cerric ammonium sulphate required per mole of ferrous ammonium sulphate to the number of moles of KMnO4 required per mole of ferrous ammonium sulphate?(Cerric ammonium sulphate and potassium permanganate are used as oxidising agents in acidic medium for oxidation of ferrous ammonium sulphate to ferric sulpahte.)

Detailed Solution for BITSAT Chemistry Test - 6 - Question 5

Fe2++Ce4+ ⟶Fe3++Ce3+
5Fe2++MnO4+8H+⟶5Fe3++Mn2++4H2O

BITSAT Chemistry Test - 6 - Question 6

A mixture of 0.02 mole mole of KBrO3 and 0.01 mole of KBr was treated with excess of KI and acidified. The volume of 0.1 M Na2S2O3 solution required to consume the liberated iodine will be:

Detailed Solution for BITSAT Chemistry Test - 6 - Question 6

BITSAT Chemistry Test - 6 - Question 7

A mixture of 2.3 g formic acid and 4.5 g oxalic acid is treated with concentrate H2SO4 . The evolved gaseous mixture is passed through KOH pellets. What will be the weight in grams of the remaining product at STP?

Detailed Solution for BITSAT Chemistry Test - 6 - Question 7

For gases n∝V:


CO2 is absorbed by KOH. So the remaning product is only CO. moles of CO formed from both reactions = Left mass of CO = moles × molar mass

= 2. 8g

BITSAT Chemistry Test - 6 - Question 8

An aqueous solution of 6.3 g of oxalic acid dihydrate(molar mass = 126.07 g/mol) is made up to 250 mL. The volume of 0.1 N NaOH required to completely neutralize 10 mL of this solution is -

Detailed Solution for BITSAT Chemistry Test - 6 - Question 8

Normality = 
Equivalent weight of oxalic acid dihydrate = 63
Normality = 
Now using N1V1=N2V2

BITSAT Chemistry Test - 6 - Question 9

20 ml of 0.2 M NaOH (aq) solution is mixed with 35 ml of 0.1 M NaOH (aq) solution and the resultant solution is diluted to 100 ml . 40 ml of this diluted solution reacted with 10% impure sample of oxalic acid (H2C2O4). The weight of the impure sample is

Detailed Solution for BITSAT Chemistry Test - 6 - Question 9


Milli-equivalent of NaOH = Milli-equivalent of H2C2O4
Let the weight of the impure sample be x gram

x = 0.15 gram

BITSAT Chemistry Test - 6 - Question 10

Ethylene glycol is used as an antifreeze in a cold climate. Mass of ethylene glycol which should be added to 4 kg of water to prevent it from freezing at −6° will be (Kf for water1.86 K kg mol−1 , and molar mass of ethylene glycol=62 g mol−1 )

Detailed Solution for BITSAT Chemistry Test - 6 - Question 10

Depression in freezing point is given by the formula

Where,
w1 and w2 = Weight of the solvent and solute, respectively,
m2=Molecular weight of the solute.

BITSAT Chemistry Test - 6 - Question 11

Depression in freezing point of 0.322 molal aqueous solution of glucose will be

Detailed Solution for BITSAT Chemistry Test - 6 - Question 11

Depression in freezing point (ΔTf)
ΔTf=Kfm
 =1.86×0.322
=0.5989 K

BITSAT Chemistry Test - 6 - Question 12

Liquids A and B form an ideal solution. Vapour pressure of pure A and B are 7 x 103 and 12 x 103 Pa respectively. Find the vapour composition of A and B if mole fraction of 'A' in liquid solution is 0.4.

Detailed Solution for BITSAT Chemistry Test - 6 - Question 12

Since, 
According to Raoult's Law: 


Now, 

BITSAT Chemistry Test - 6 - Question 13

The freezing point of an aqueous solution is - 0.186°C. The Ebullioscopic constant of the solvent is Kb = 0.52°C kg mol-1 and Cryoscopic constant of the solvent is Kf = 1.86°C kg mol-1. find the increase in boiling point.

Detailed Solution for BITSAT Chemistry Test - 6 - Question 13


m=0⋅10

=0⋅052 K

BITSAT Chemistry Test - 6 - Question 14

The following salts were dissolved in water to get an aqueous solution of mentioned concentration. Which one of the following will have maximum freezing point?

Detailed Solution for BITSAT Chemistry Test - 6 - Question 14


Lesser the ΔTf more the freezing point of solution.
In all other cases concentration is more than this. so, i×m will also be higher than this.

BITSAT Chemistry Test - 6 - Question 15

Two elements, A and B, form compounds having molecular formulae AB and AB4 when dissolved in 20g of C6H6, 1.0g AB2 lowers the freezing point by 2.3K whereas 1.0g of AB4 lowers it by 1.3K. The molal depression freezing point constant for benzene is 5.1 kg mol-1. The atomic masses of A and B (in g mol-1) respectively are (assume AB2 and AB4 are non-ionisable)

Detailed Solution for BITSAT Chemistry Test - 6 - Question 15

Suppose, the atomic masses of A and B are x and y respectively.
Molar mass of AB2 = x + 2y





Similarly, for the 2nd case,


Solving equations (i) and (ii), we get,
x=25.69; y=42.64
Therefore, A = 26 and B =42.64

BITSAT Chemistry Test - 6 - Question 16

The freezing point (in°C) of solution having 0.1 g of K3 [Fe (CN)6] (molecular weight 329)in 100 g of water (Kf = 1.86 K kg mol- 1) is

Detailed Solution for BITSAT Chemistry Test - 6 - Question 16

Van't Hoff's factor (i)= 
Molality = 



(As freezing point of water is 0°C)

BITSAT Chemistry Test - 6 - Question 17

A 5.25% solution of a substance is isotonic with 1.5% solution of urea (molar mass = 60 g mol-1) in the same solvent. If the densities of both the solutions are assumed to be equal to 1.0 g cm-3, molar mass of the substance will be:

Detailed Solution for BITSAT Chemistry Test - 6 - Question 17

Isotonic : Two solutions having equal osmotic pressure at a given temperature are called isotonic solutions.
Osmotic pressure is directly proportional to molarity (C) of the solution at a given temperature.
Mathematically, it can be calculated as given below. 
π = CRT
Since, the two solutions are isotonic,

BITSAT Chemistry Test - 6 - Question 18

When 0.2g of acetic acid is added to 20g of benzene, then the freezing point of benzene decreases by 0.45°C. If acetic acid associates to form a dimer in benzene, then find the percentage association of acetic acid in benzene.
(Kf for benzene=5.12 K kg mol−1)

Detailed Solution for BITSAT Chemistry Test - 6 - Question 18

BITSAT Chemistry Test - 6 - Question 19

Molal elevation constant of a liquid is

Detailed Solution for BITSAT Chemistry Test - 6 - Question 19

The formula for elevation in boiling point is ΔTb = Kb x m where, Kb = molal elevation constant and m = molality.
Molality can be written as where, w = weight of the solute, W = weight of the solvent and GMM =
GMM = Gram molecular mass of the solute. When m = 1, which means 1 mole of a solute in 1 Kg of a solvent, ΔTb = kb where,

Tb = Boiling point of the solvent and
Lv = Latent heat of vapourisation.

BITSAT Chemistry Test - 6 - Question 20

The required energy for photoelectric effect for few metals is as listed below:
Li - 2.4 eV; Mg - 3.7 eV; Ag - 4.3 eV; Fe - 4.7 eV
If light of wavelength of 400 nm falls on these metals, which of these metals will show photoelectric effect?

Detailed Solution for BITSAT Chemistry Test - 6 - Question 20

h = 6.63 x 10-34, c = 3 x 108 m/s, λ = 400 x 10-9 m
E = hc/ λ = 3.105 eV (1 eV = 1.6 x 10-19 J)
So, only lithium will show the photoelectric effect (3.105 > 2.4).

BITSAT Chemistry Test - 6 - Question 21

Electron emits radiation with a wavelength of 4900 x10-10 m from a certain electronic state and is lying in the Balmer series in the spectrum of hydrogen atom. Find the electronic state of the electron. Take approximate value and RH = 1.097 x 107m-1.

Detailed Solution for BITSAT Chemistry Test - 6 - Question 21

Given:

According to the question,
We know that,

Since the radiation is in the Balmer series (Visible region)
Therefore, n1=2

Hence, this is the correct option.

BITSAT Chemistry Test - 6 - Question 22

In SF4, what is the hybridisation of sulphur?

Detailed Solution for BITSAT Chemistry Test - 6 - Question 22

In SF4,

'S' has one lone pair and is involved in 4 bonds.
Therefore, the sulphur atom is sp3d hybridised and the trigonal bipyramidal geometry of the molecule is distorted to see-saw due to the presence of the lone pair of electrons.

BITSAT Chemistry Test - 6 - Question 23

The t1/2 values for the decomposition of CH3CHO at constant temperature and at initial pressure of 340 mm and 170 mm of Hg were 410 and 820 s, respectively. The order of the reaction is

Detailed Solution for BITSAT Chemistry Test - 6 - Question 23

BITSAT Chemistry Test - 6 - Question 24

Given 
Find the potential for the cell:

Detailed Solution for BITSAT Chemistry Test - 6 - Question 24

From the given representation of cell, Ecellcan be found as follows


Hence, this is the correct option.

BITSAT Chemistry Test - 6 - Question 25

For the given reaction the first order rate constant for the decomposition of Ammonium Nitrite is 2.60x 10-5 s-1 at 700 K. What will be the value of rate constant of the reaction at 900 K when the energy of activation is 150 kJ mol-1?

Detailed Solution for BITSAT Chemistry Test - 6 - Question 25

Given:

According to the question,

Hence, this is the correct option.

BITSAT Chemistry Test - 6 - Question 26

Which of the following equations is/are correct for energy of electron in nth orbit?

Detailed Solution for BITSAT Chemistry Test - 6 - Question 26

We know that

Therefore option (C) is correct.

BITSAT Chemistry Test - 6 - Question 27

Which of the following has/have bond order zero and show(s) diamagnetic behaviour?

I. Ne2
II. He2
III. N2
IV. O2

Detailed Solution for BITSAT Chemistry Test - 6 - Question 27

We know that,Total number of electrons in He2 = 4 is the MO configuration.
Now,  Bond order

As there is no unpaired electron, hence, it is diamagnetic in nature. Similarly, Ne2 has a bond order of zero and behave diamagnetic in nature. Hence, this is the correct option.

BITSAT Chemistry Test - 6 - Question 28

One mole of gas is heated at constant volume and temperature is raised from 298 to 308K. The heat supplied to the gas is 200 J. Which of the following is correct?

Detailed Solution for BITSAT Chemistry Test - 6 - Question 28

We know that
dU = Q + W
Here, volume of the system is not changing, so work done is zero.
So, dU = Q = 200 J

BITSAT Chemistry Test - 6 - Question 29

The ka for HOCl Is 2.6 x 10-8.

What will be the value of equilibrium constant for the given reaction?

(Take kw = 1 x 10-14)

Detailed Solution for BITSAT Chemistry Test - 6 - Question 29

We know that,

Hence, this is the correct option.

BITSAT Chemistry Test - 6 - Question 30

At 300 K, the conductivity of saturated solution of KCl in water is 2.8 x 10-6 Scm-1. Its solubility product at 300 K (given λ(K+) = 73 S cm2 mol-1, λ(Cl-) = 67 Scm2 mol-1) is

Detailed Solution for BITSAT Chemistry Test - 6 - Question 30

Given:
λ(K+) = 73 S cm2 mol-1, λ∞(Cl-) = 67 Scm2 mol-1
Therefore,

Hence,

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