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7101.

Name the non-stoichiometric pointdefect responsible for colour isalkaline metal halide.

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Non-stoichiometric defect responsible for colour in alkali metal halide is metal excess (anion -vacancy) defect.

7102.

SECTION-CO,)and calculate bond order.Draw the molecular orbital diagram for oxygen molecule (O)ana

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7103.

molecular orbital of CO and NO

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7104.

2o. Which of the following does not exist on the abasis of molecular orbital theory(a) Hi(c) He(b) I(d) Li,

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According to the molecular orbital theory, in a supposed He2 molecule, both the bonding and the antibonding orbitals will have 2 electrons each. And so, the bond order (half the difference in the numbers of bonding and antibonding electrons) for such a diatomic entity is zero. That means, the He2 molecule has no existence.so ans is c option

7105.

32. Which type of 'defect has the presence ofcations in the interstitial sites ?(1) Schottky defect(2) Vacancy defect(3) Frenkel defect(4) Metal deficiency defect3633. According to molecular orbital theorywhich of the following will not be a viablemolecule?(1) He2(2) He(3) H2(4) H

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Q 32. (3) frenkel defectQ 33. (4)

7106.

molecular orbital diagram of He2

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7107.

11. Use Molecular Orbital theory to place C2, C2 C2 in order of:a) increasing bond energy and stability(b) increasing bond length

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thanku and plzz answer another question of mine of the lattice energy plzz

7108.

2. A 2-56-043k & t = -243j4Kla +61 2 lå-bl

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7109.

The electric field at a poinf near infinite thin sheetOf Charge conductor is

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Electric field at a point near aninfinite thin sheet of charged conductor is σ/2ε₀

In order to find the electric field at a point P1 on an infinite thin plane charge sheet of positive charge another point P2 is taken on the other side of the sheet, so that P1 and P2 are equidistant from the sheet. Then a cylindrical Gaussian surface is constructed through the plane which extends equally on two sides of the plane. It is seen that the electric field isuniform and does not depend on the distance from the charge sheet.

In order to find the electric field at a point P which is near but outside the conductor. The Gaussian surface is constructed and it is seen that the electric field near a plane charged conductor is twice the electric field. This is in accordance with Coulomb's theorem which states that the electric field at any point very close to the surface of a charged conductor is equal to charge density of the surface divided by free space permittivity.

7110.

The weight of a molecule of the compound C6o H122 is1.4 × 10-21 gC) 5.025 x 1023 gB) 1.09 × 10-21 gD) 16.023 x 103 g

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weight of 1 mol of C60H122 = 720 + 122 = 842g

and in 1 mol , 6×10²³ molecules are there so, weight of 1 molecule = 842/(6×10²³) = 140× 10-²³ = 1.4×10-²¹ g

7111.

Explain Rutherford's nuclear model of an atom. What are its drawbacks?ontnarticles bombard a thin sheet of Gold foil. The

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7112.

Explain density and relative density and write their formulae

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Density is the measure of mass per volume. This means that density of any solid,liquid, or gas can be found by dividing its mass in kilogram by its volume in cubic metres.density=mass/volumethe unit for density is Kg metre cube

Relative density :

The ratio between density of a substance and density of water is called relative density.

Relative density =Density of the substance/density of water at 40C

7113.

6. One atom of an element x weighs 6.643 x 1023 g. Number of moles of atoms in its 20 kg is:(a) 4(b) 40(c) 100(d) 500Iw many atoms are contained in 1 o-23I

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option D 500

7114.

Q2) Define Sublimation.

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Sublimation is the transition of a substance directly from the solid to the gas phase, without passing through the intermediate liquid phase.

where is video

7115.

7. Lihium borohydride crystallizes in an arthorhombic systemwith 4 molecules per unit cel. The unitcell dimensions are a-6.8 A, b = 4.4 A aกd c=7.2 A. If the molar mass is 21.76, then the density ofcrystals is:(A) 0.6708 g cm? (B) 1.6708 g cm(C) 2.6708 g cm-(D) of these

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7116.

. An element has a body-centred cubic (bcc) structure with a cell edge of288 pm. The density of the element is 7.2 g/cm3. How many atoms arepresent in 208 g of the element?

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Formula units per unit cell Z = 2 for BCC cubic unit cell lattice parameter a = 28 pm=288x10-8cmVolume V = a3=2.39X10-23cm3Density d = 7.2g/cm3N­A = Avogadro constant = 6.022x10²³Molecular mass M =?We know that Density d = ZM/NA Xa3,M = dxNA x a3/ZOn Substituting valuesM= 7.2g/cm3x(6.022x10²³)X (6.022x10²³)/2= 51.8gmol-151.8 g of element contains 6.022X1023208g of this element contains=?= 6.022X1023X208/51.8=2.42X1024atoms.

7117.

EXAMPLE 27. An element crystallises in a fcc latticeith cell edge of 250 pm. Calculate the density of 300 gif this element contains 2 x 10^24 atoms.

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7118.

5Volume of oxygen at STP required to burn completely 22 g of propane, C,H,?(A) 22.7 L(B) 45.4L5.75 L(D) 11.351

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Balanced equation of Combustion of propane isC3H8 + 5O2 = 3CO2 + 4H2O

At NTP44g of propane requires 5*22.4L of oxygen for complete combustion22g of propane requires 22*5*22.4/44 = 56L

Volume of oxygen required for complete combustion is 56L

7119.

calculate the number of atoms in 8 gram of iron

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7120.

An element has atomic mass 93 g/mol and density 11-5 g/cmº. If edge length of itsunit cell is 300 p.m. identify the types of unit cell.

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Number of atoms per unit cell, z = d × a³ × NA / M --> (1)

Here, d = 11.5 g cm⁻³ , M = 93 g mol⁻¹ , NA = 6.022 × 10²³ mol⁻¹

a = 300 pm = 300 × 10⁻¹⁰ cm = 3 × 10⁻⁸ cm

Now, Substituting these values in the expression (1), we get

z = 11.5 g cm⁻³ × (3 × 10⁻⁸ cm)³ × 6.022 × 10²³ mol⁻¹ / 93 g mol⁻¹

FINAL RESULT = 2.01

As there are 2 atoms of the element present unit cell, therefore, the cubic unit cell must be body centered

7121.

The edge length of a face centred cubic cell ofan ionic substance is 508 pm. If the radius ofthe cation is 110 pm, the radius of the anion is:17.

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For an fcc solid, edge length ,a=2(r++r−)a=2(r++r−)

⇒2(110+r−)=508

⇒220+2r−=508

⇒2r−=508−220

⇒2r−=288

∴r−=144pm

7122.

14. Unit cell of crystalline sodium hydrogen dincetate(molecular mass 142) has density of 1.4 g ml. It'sunit cell contains 24 molecules. Calculate the edgelength of the unit cell. (Ans. 15.93 x 108 cm)

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15.93 × 10`8 cm is the answers

7123.

What volume of air containing 21% oxygen by volume is required to completely burn 1 kg of carboncontaining 100% combustible substances?

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7124.

Silvermetalcrystallizeswitha face centred cubic lattice. The length of unit cell is found to be 4.077x10-8 cm. Calculate atomic radius and density of silver.

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7125.

Copper crystallises in a face-centred cubic latticewith a unit cell length of 361 pm. What is the radiusof copper atom in pm?(1) 108(3) 157(2) 128(4) 181

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7126.

The metal calcium crystallines in a face-centredcubic unit cell with a= 0.556 nm. Calculate thedensity of the metal if(i) it contains 0.2% Frenkel defect,(ii) it contains 0.1% Schottky defect.

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7127.

LB, C, D 30at> E e gsere aos devdes S 4,1, 11,7 30609 9 oSG0,). 863 gm0 b). wODS उन्ठ०, ©). wOW wo d). olrd P ¢)- polrd TESD.). 553 pH Devs o), 584 B wBire 8o Fraved? (AS1)

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7128.

bo(0)to SCH,HOW bogCH3 – CH – CH – CH,BE

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1-bromo 3-methyl butane

common name- isopentyl bromide

7129.

A certain compound contains Calcium,Carbonand Nitrogen in the mass ratio, 20: 6:14The empirical formula of the compound is

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7130.

s beeupied by the nucleus.The mass-charge ratio for A' ions is 1.97 x 10' kg C', Calculate the mass of A atom.Calculate the Coma eCo

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Mass/Charge of A+= (1.97x10^-7kg/charge)

Charge of an electron= 1.6 x 10^-19

Thus Mass of 'A' atom = (1.6 x 10^-19)(1.97x10^-7)= 3.16 x10^-26Kg

7131.

omL 7%& ombrtionn 0%!’%’44 ८ it jfi//m’/%% 2.॥0%... (04-90 (027W)y 0: | 50; कक हा (5): 50५1 ८0% a8 ज

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This isomerism is Ionisation Isomerism.

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7132.

110g of a mixture of Caco, and Na.co.ignition suffered a loss in weight of 2.29. Themass ratio of Caco, and Na,co, is1) 1:1 2) 1:1.4 3) 1.4:1 4) 1.75:1

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Answer:1)1:1Explanation:Na2CO3 is thermally stable and does not decompose on ignition but CaCO3 decompse to give CO2.

CaCO3 → CaO + CO2

1mole i.e 100gm of CaCO3 gives 44 gm of CO2.

Wt.of CaCO3 giving 2.2 gm of CO2 =( 2.2X100/44)=5gm

Wt.of CaCO3 =5gm,

Wt.of Na2CO3=10-5=5gm,

There foe ratio of wts=1:1

the right answer is (1)1:1

7133.

5859606263+4/-1Single ChoiceWhich one of the following is an example forheterogeneous catalysis?А250, +O, NO ~250;BC2H2011 +H20 Die C,H,Os+CH20CN2+O, PL™2NOCHỊCOOCH3 + NaOH=CH2COONa + CH20Submit

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7134.

Mixture analysis

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Mixture analysis is a very active research topic in statistics and machine learning, with new developments in methodology .

7135.

Write uses of Fullerene.

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Fullerenes may be used for drug delivery systems in the body, in lubricants and as catalysts. The tube fullerenes are callednanotubes.

7136.

57585960616263+4Single ChoiceIn the adsorption of a gas on solid, Freundlichisotherm is obeyed. The slope of the plot is zero.Then the extent of adsorption is :Directly proportional to the pressureof the gasDirectly proportional to the squareroot of the pressure of the gasInversely proportional to the squareroot of the pressure of the gasIndependent of the pressure of thegas

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in the adsorption of a gas on solid Freundlich isotherms is Obeyed the slope of the plot is zero then the extend of adsorptionis A.directly proportional to the pressure of thegas

7137.

Uses of Ammona Gas

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Ammonia is used as a refrigerant gas, for purification of water supplies, and in the manufacture of plastics, explosives, textiles, pesticides,dyesand other chemicals. It is found in many household and industrial-strength cleaning solutions.

7138.

)Write uses of Ethanol

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Ethanolisusedin medical wipes and most common antibacterial hand sanitizer gels as an antiseptic.Ethanol kills organisms by denaturing their proteins and dissolving their lipids and is effective against most bacteria and fungi, and many viruses.

7139.

The density (in gm/ml) of the stainless steol spherical Dalls used tn ball-bearings รกre 102atoms are present in each ball of dlamuter 1 cm 11 lhe balls contain 84% iron by m ass ? T

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How many iron atoms are present in a stainless steel ball bearing having a radius of 0.254 cm ?The stainless steel contains 85.6% Fe by weight and has a density of 7.75g/cm cube.

this is an answer to similar question.

7140.

Example 5Find output voltage Vo2VVoü1K02M-1V

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7141.

basic concepts of chemistry H2 why 2 is used

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H has 1 electron less than Nobel gas configuration so it shares it's electron with one more H to attain stability

but oxygen also have O2 why?

and other gases

7142.

0/45, / The volume of water that must be added to a mi( 48-/ 2 MHCl to obtain 0.25 M solution of HCL is - T(1) 750 ml (2) 100 ml (3) 200 m¢ (4)300m(Cl and 750 ml of

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7143.

53. Find out the volume of 98% w/w H2SO4 (density1.8 gm/ml), must be diluted to prepare 12.0 litres of 2.4 M sulphuric acidsolution. I

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7144.

(b)State Huckel's rule and illustrate it with the help of a suitable example.

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7145.

1 What are the necessary conditions for any system to be aromatic?

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7146.

Q. 14, whether H(-) ions is presentin basic solution? If yes, then why thesolution is basic ?

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When a base is dissolved in water then it will produce H⁺ ions as wells as OH⁻ ions. if H⁺ ion concentration is more than OH⁻ ion then the solution is acidic. And they show basic character if OH⁻ ion concentration is more than H⁺ ion concentration.That means they are basic due to presence of more OH- ions.

7147.

List the basic conditions required for an electric current to flow. Draw an electric circuitshowing basic electric components used.

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basic conditions required for an electric current to flow is a voltage source and resistance this resistance can be considered as a form of electric appliances just like a bulb or a heater Anything that have sort of resistance and after that there should be a switch for that the switch will work as the conditions of on and off if we switch off then no current would be flowing there and if we switch on the current will be flowing through the appliances

7148.

1. Calculate the Normality of 500 ml of 0.4 M sulphuric acid solution?

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There is a relation b/w Normality and Molarity :

For acids.

Normality = Molarity xBasicity.

wherebasicityis the number ofH+ions a molecule of an acid can give.

Here, Normality is 0.4 and No. Of H+ ions in H2SO4 soln. is 2

Now, Put the values in to the formula,

Normality = Molarity xBasicity.

0.4 = molarity × 2 (no. Of H+ ions)

Molarity = O.4/2 =0.2 Moles per Litre

7149.

12. 25 ml of3.0MHNO, are mixed with 75 ml o4.0MHINO, the molarity of the final mixture would be

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Answer:- Molarity of the mixture solution is 3.75M.

Solution:- Volume is an additive property so the volume of the mixture would be the sum of the volumes of two solutions of the acid being added together.

We could easily solve this problem using dilution equation:

M1V1 + M2V2 = MV

Where, M and V are the molarity and volume of the mixture.

Let's plug in the values in the equation:

3.0(25) + 4.0(75) = M(25+75)

75 + 300 = M(100)

375 = M(100)

M = 375/100

M = 3.75

So, the molarity of the mixture solution is 3.75M.

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7150.

10. How does Huckel rule help to decide the aromaticcharacter of a compound?

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Erich Huckel is a German chemist and physicist. He proposed a theory to determine the characteristics properties of the aromatic compound (containing at least one benzene ring) .

The rule states that if a cyclic, planer molecules containing (4n + 2 pi) electrons it has a aromatic properties. This rule is known as Huckel Rule.

Where, n = no. Of carbon atoms in compound pi = no. Of pi electrons