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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1451. |
The number of atoms in `100 g` of an fcc crystal with density `= 10.0g cm^(-3)` and cell edge equal to `200 pm` is equal toA. `5 xx 10^(24)`B. `5 xx 10^(25)`C. `6 xx 10^(23)`D. `2 xx 10^(25)` |
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Answer» Correct Answer - A `d = (Z xx M)/(N_(A)a^(3))` `N_(A) = (4 xx 100)/(10 xx (2 xx 10^(-8))^(3)) = 5 xx 10^(24)` (here : 200 Pm `= 2 xx 10^(-8) cm`) |
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| 1452. |
The number of atoms in 100 g of a fcc crystal with denstiy = `10.0 g//cm^(3)` and cell edge equal to 200 pm is equal toA. `5xx10^(24)`B. `5xx10^(25)`C. `6xx10^(23)`D. `2xx10^(25)` |
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Answer» `rho=(Z_(eff)xxMv)/(a^(3)xx10^(-30)xxN_(A))` [For fcc,`Z_(eff)=4`/unit cell] `therefore Mw=(rhoxxa^(3)xx10^(-30)xxN_(A))/(Z_(eff))` `=((10.0gcm^(-3)xx(200"pm")^(3)xx10^(-30)cm^(3)xx6xx10(23)"atoms"))/(4)` `=12g "mol"^(-1)` Thus, 12 g `"mol"^(-1)` contains =`N_(A)` atoms `=6xx10^(23)` atoms `therefore 100g "contains"=(6xx10^(23))/(12)xx100=5xx10^(24)` atoms. |
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| 1453. |
Lithium has a bcc structure .Its density is `530 kg m^(-3)` and its atomic mass is `6.94 g mol^(-1)` .Calculate the edge length of a unit cell of lithium metal `(N_(A) = 6.02 xx 10^(23) mol^(-1))`A. 154pmB. 352pmC. 527pmD. 264pm |
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Answer» Correct Answer - B `rho=(ZxxM)/(N_(A)xxa^(3))` For bcc structure `Z=2.rho=530kg m^(-3)=0.530 g cm^(-1)` `0.530=(2xx6.94)/(6.02xx10^(23)xxa^(3))` `a^(3)=4.348xx10^(-23)cm^(3)` `a=3.52xx10^(-8)`cm a=352pm |
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| 1454. |
Sodium metal crystallizes in body centred cubic lattice with the cell edge a=4.28Å. What is the radius of the sodium atom ?A. `1.857xx10^(-8)cm`B. `2.371xx10^(-7)cn`C. `3.817xx10^(-8)cm`D. `9.312xx10^(-7)cm` |
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Answer» Correct Answer - A Radius of Na (in bcc lattice) `=(sqrt(3)a)/(4)=(sqrt(3)xx4.29)/(4)` `=1.8574overset(@)A=1.8572xx10^(-8)` cm |
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| 1455. |
For an ionic crystan of the type `AB` , the value of (limiting) earius ratio is `0.40` .The value suggest that the crystan struture should beA. OctahedralB. TerahedralC. Squre planerD. Plane triangle triangle |
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Answer» Correct Answer - b The crystals in which radius ratio value is found between `0.225 and 0.414` show tetrabudral crystal structure |
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| 1456. |
In Which of the following substances the carbon atom is arranged in a regular tetrahedral struture?A. DiamondB. BenzeneC. GraphiteD. Carbon black |
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Answer» Correct Answer - a In diamond , C atoms are arranged in a regular tetrabedral structure |
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| 1457. |
The number of atoms in `100 g` of an fcc crystal with density `= 10.0g cm^(-3)` and cell edge equal to `200 pm` is equal toA. `5 xx 10^(24`B. `5 xx 10^(25)`C. `6 xx 10^(23)`D. `2 xx 10^(25)` |
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Answer» Correct Answer - A `rho=(Z_(eff)xxMw)/(a^(3)xx10^(-30)xxN_(A))`, [For fcc, `Z_(eff) = 4//"unit cell"`] `:. Mw = (rho xx a^(3) xx 10^(-30) xx N_(A))/(Z_(eff))` `((10.0g cm^(-3) xx (200 "pm")^(3) xx 10^(-30)cm^(3)xx6xx10^(23)"atoms"))/(4)` `= 12 g mol^(-1)` Thus, `12 g mol^(-1)` contains `= N_(A)` atoms `= 6 xx 10^(23)` atoms `:. 100 g` contains `= (6 xx 10^(23))/(12) xx 100 = 5 xx 10^(24)` atoms |
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| 1458. |
The number of atoms in `100 g` of an fcc crystal with density `= 10.0g cm^(-3)` and cell edge equal to `200 pm` is equal toA. `3 xx 10^(25)`B. `5 xx 10^(24)`C. `1 xx 10^(25)`D. `5.96 xx 10^(-3)` |
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Answer» Correct Answer - b `M = (d+ a^(3) xx N_(0))/(n)` `= (10 xx (200 xx 10^(-3))^(3) xx (6.02 xx 10^(23)))/(4) = 12.04 g` No of atoms in `100 gm= (6.02 xx 10^(22))/(12.04) xx 100` `= 5 xx 10^(24)` |
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| 1459. |
In the closest packed struture of a metallic lattice , the number of mearest neighhours of a metallic atom isA. TwelveB. FourC. EightD. Six |
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Answer» Correct Answer - a Coordination number in `HPC` and `C CP` arrangement is `12` while in bcc arrrangement is `8` |
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| 1460. |
Potassium has a bcc structure with nearest neighour distance `4.52 Å`its atomic weight is `39` its density (in kg `m^(-3)`) will beA. 454B. 804C. 852D. 908 |
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Answer» Correct Answer - D For bcc, d`=sqrt(3)/(2)` a or `a=(2d)/sqrt(3)=(2xx4.52)/(1.732)=5.219overset(@)A`=522pm `rho=(ZxxM)/(a^(3)xxN_(0)xx10^(-30))=(2xx39)/((522)^(3)xx(56.023xx10^(23))xx10^(-30))` `=0.91g//cm^(3)=910kgm^(-3)` |
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| 1461. |
Potassium has a bcc structure with nearest neighour distance `4.52 Å`its atomic weight is `39` its density (in kg `m^(-3)`) will beA. `454`B. `804`C. `852`D. `908` |
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Answer» Correct Answer - d For fcc, `d= sqrt(3)/(2) xx or a = (2d)/(sqrt(3)) = (2 xx 4.52)/(1.732)` `= 5.129 Å = 522` pm `rho = (Z xx M)/(a^(3) xx N_(0) xx 10^(-30))` `= (2 xx 39)/((522)^(3) xx (0.23 xx 10^(23)) xx 10^(-30))` `= 0.91 g//cm^(3)` |
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| 1462. |
Aluminium crystallizes in a cubic close packed structure . Its metallic radius is 125 pm . What is the length of the side of the unit cell . |
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Answer» Given radius = 125 pm For a fcc lattice unit cell `r = (a)/(2sqrt2)` ` a = 2sqrt2 xx r` `a= 2sqrt2 xx r` `= 2 xx 1.414 xx 125` `= 353 . 5` pm `therefore` Length of the side of the unit cell = `353 . 5` pm . |
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| 1463. |
The number of atoms in `100 g an fcc` crystal with density `d = 10 g//cm^(3)` and the edge equal to 100 pm is equal toA. `4 xx 10^(25)`B. `3 xx 10^(25)`C. `2 xx 10^(25)`D. `1 xx 10^(25)` |
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Answer» Correct Answer - a `M= ( rho xx a^(3) xx N_(0) xx 10^(-30))/(z)` `= (10 xx (100)^(2) (6.02 xx 10^(23))xx 10^(-30))/(4) = 15.05` No of atoms in `100 g = (6.02 xx 10^(23))/(15.05) xx 100 = 4 xx 10^(23)` |
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| 1464. |
Ferrous oxide has cubes structure and each edge of the unit cell is `5.0 Å` .Assuming of the oxide as `4.0g//cm^(3)` then the number of `Fe^(2+) and O^(2)` inos present in each unit cell will beA. `Four Fe^(2+) and four O^(2-)`B. `Two Fe^(2+) and four O^(2-)`C. `Four Fe^(2+) and two O^(2-)`D. ` Three Fe^(2+) and three O^(2-)` |
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Answer» Correct Answer - a Let the units of ferrous oxide in a unit cell `= n`, molecular weight of ferrous `(FeO) = 56 + 16 = 72 g mol^(-1)` Weight of n unit`= (72 xx n)/(6.025 xxx 10^(23))` Volume of one unit `= ("length of corner")^(3)` `= (5 Å)^(3) = 125 xx 10^(-26) cm^(3)` Density `= ("wt, of cell" )/("volume")` `4.09 = (72 xx n)/(6.023 xx 10^(23) xx 125 xx 10^(-24))` `n = (3079.2 xx 10^(-1))/(72)` `= 42.7 xx 10^(-1) = 4.27 = 4` |
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| 1465. |
If the anions (A) form hexagonal closest packing and cations (C) occupy only 2/3 octahedral voids in it. Then the general formula of the compound isA. CAB. `CA_(2)`C. `C_(2)A_(3)`D. `C_(3)A_(2)` |
| Answer» Correct Answer - C | |
| 1466. |
Potassium has a bcc structure with nearest neighour distance `4.52 Å`its atomic weight is `39` its density (in kg `m^(-3)`) will beA. 454 kg `m^(-3)`B. 804 kg `m^(-3)`C. 852 kg `m^(-3)`D. 908 kg `m^(-3)` |
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Answer» Correct Answer - D For bcc, `d=(sqrt(3))/(2)a` or `a=(2d)/(sqrt(3))=(2xx4.52)/(1.732)=5.219 Å=522` pm `d=(ZxxM)/(a^(3)xxN_(0)xx10^(-30))=(2xx39)/((522)^(3)xx(6.02xx10^(23))xx10^(-30))` |
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| 1467. |
Potassium has a bcc structure with nearest neighour distance `4.52 Å`its atomic weight is `39` its density (in kg `m^(-3)`) will beA. `454`B. `804`C. `852`D. `910` |
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Answer» Correct Answer - d For fcc, `d= sqrt(3)/(2) a or a = (2d)/(sqrt(3)) =(2 xx 4.52)/(1.732)` `= 3.219 Å = 522` pm `rho = (Z xx M)/(a^(3) N_(0) xx 10^(-30))` `= (2 xx 39)/((522)^(2) xx (6.025 xx 10^(23)) xx 10^(-30))` `= 6.91 g//cm^(3) = 910 kg m^(-3)` |
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| 1468. |
Aluminium crystallizes in a cubic close-packed structre. Its metallic radius is `125 ppm` a. What is the length of the side of the unit cell? b. How many unit cell are there in `1.00 cm^(3)` of aluminium?A. `4.42xx10^(22)`B. `2.36xx10^(21`C. `2.26xx10^(22)`D. `3.6xx10^(18)` |
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Answer» Correct Answer - C For an fcc unit cell, `r=(a)/(2sqrt(2))` [Given, r = 125 pm] `a=2sqrt(2)xxr=2xx1.414xx125` `=353.5` pm `~~354` pm Volume of unit cell `=a^(3)=(353.5xx10^(-10)cm)^(3)` `=442xx10^(-25)cm^(3)` Number of unit cell `=(1.00cm^(3))/(442xx10^(-25)cm^(3))` `=2.26xx10^(22)` unit cells |
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| 1469. |
In hexagonal close packing, the difference, in the number of respecitvely, and octahedral voids per unit cell is |
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Answer» Correct Answer - 6 In one hcp unit cell, atoms in the packing = 6. Hence, octahedral voids= 6, tetrahedral voids = 12 .Difference = 12 -6 = 6 |
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| 1470. |
`NH_(4)^(+) and Br^(-)` ions have ionic of 143 pm and 196 pm respecitvely. The coordination number of ` NH_(4)^(+)` ion in ` NH_(4)Br` is |
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Answer» Correct Answer - 4 ` r(NH_(4)^(+))//r(Br^(-)) ` = 143/196 = 0.730. Which lis in the range 0.414 - 0.732. Hence , coordination number = 6 |
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| 1471. |
A metal has bcc structure and the edge length of its unit cell is `3.04 Å`. The volume of the unit cell `cm^(3)` will beA. `1.6xx10^(23)cm^(3)`B. `1.6xx10^(-23)cm^(3)`C. `6.02xx10^(-23)cm^(3)`D. `6.6xx10^(-24)cm^(3)` |
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Answer» Correct Answer - B Volume of unit cell (V) `=a^(3)` `=(3.04xx10^(-8)cm)^(3)` `=2.81xx10^(-23)cm^(3)` |
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| 1472. |
The ratio of the volume of a tetragonal lattice unit cell to that of a hexagonal lattice unit cell is (both having same respective lengths)A. `(2)/sqrt(3)`B. `sqrt(3)/(2)abc`C. `(sqrt(3)a^(2))/(2bc)`D. `((2a^(2)c))/sqrt(3)b` |
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Answer» Volume of a lattice is given by: `V=abc(1-cos^(2)alpha-cos^(2)beta-cos^(2)gamma-2cos alpha cos beta cos gamma)^((1)/(2))` `V_("tetragonal")=a^(2)c` (because `a=bcancel=c,alpha=beta=gamma=90^(@)`) `V_("hexagonal")=a^(2)cxx(sqrt(2))/(2)`(because `a=bcancel=c,alpha=beta=90^(@),gamma=120^(@)`) `therefore(V_("tetragonal"))/(V_("hexagonal"))=(a^(2)cxx2)/(sqrt(3)xxa^(2)xxc)=(2)/sqrt(3)` |
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| 1473. |
An fcc lattice has a lattice parameter `a = 400` pm. Calculater the molar volume of the lattice including all the empty space.A. 7.6 mLB. 6.5 mLC. `10.8 ` mLD. 9.6 mL |
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Answer» Correct Answer - D Volume `=a^(3)=(400xx10^(-10)cm)^(3)=64xx10^(-24)cm^(3)` `V_("total")=VN_(A)=64xx10^(-24)xx6.02xx10^(23)=38.4` Molar volume `=(1)/(4)xxV_("total")=(38.4)/(4)=9.6ml`. |
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| 1474. |
An fcc lattice has a lattice parameter `a = 400` pm. Calculater the molar volume of the lattice including all the empty space.A. `19.8 mL`B. `96mL`C. `8.6mL`D. `9.6mL` |
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Answer» Correct Answer - 4 Volume of 4 atoms `=a^(3)=(4xx10^(-8))^(3)cm^(3)` volume of `N_(A)` atoms `=((4xx10^(-8)))/(4)xx6.023xx10^(23)=9.6ml`. |
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| 1475. |
An fcc lattice has a lattice parameter `a = 400` pm. Calculater the molar volume of the lattice including all the empty space.A. 10.8 mLB. 96 mLC. 8.6 mLD. 9.6 mL |
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Answer» Correct Answer - D Volume of 4 atoms `= a^(3) = (4 xx 10^(-8))^(3) cm^(3)` volume of `N_(A)` atoms `= ((4 xx 10^(-8)))/(4) xx 6.023 xx 10^(23) = 9.6 mL` |
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| 1476. |
An fcc lattice has a lattice parameter `a = 400` pm. Calculater the molar volume of the lattice including all the empty space.A. `10.8mL`B. `96mL`C. `8.6mL`D. `9.6mL` |
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Answer» Correct Answer - 4 Volume of 4 atoms `=a^(3)` `=(4xx10^(-8))^(3)cm^(3)` volume of `N_(A)` atoms`=((4xx10^(-8)))/(4)xx6.023xx10^(23)=9.6ml` |
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| 1477. |
A metal of density `7.5 xx 10^(3) kg m^(-3)` has an fcc crystal structure with lattice parameter `a = 400 pm`. Calculater the number of unit cells present in `0.015 kg` of metal.A. `6.250 xx 10^(22)`B. `3.125 xx 10^(23)`C. `3.125 xx 10^(22)`D. `1.563 xx 10^(22)` |
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Answer» Correct Answer - C The volume availagble `= (0.015)/(7.5 xx 10^(3))` `("Number of unit cells") xx (400 xx 10^(-12))^(3) = (0.015)/(7.5 xx 10^(3))` `rArr` Number of unit cell `= 3.125 xx 10^(22)` |
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| 1478. |
The atomic fraction `(d)` of tin in bronze (fcc) with a density of `7717 kg m^(-3)` and a lattice parameter of `3.903 Å` is `(Aw Cu = 63.54, Sn = 118.7, 1 amu = 1.66 xx 10^(-27 kg))`A. `0.01`B. `0.05`C. `0.10`D. `3.8` |
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Answer» Correct Answer - B (Refer Section `1.13` alternative method) `rho=[{:[(sum("Number of atom of each kind")),(xx("Mw of each kind") xx 1.66 xx 10^(-27)kg)]:}]/a^(3)` `7717kgm^(-3) = [[{:(("Number of Sn atoms")xx),((118.7xx1.66xx10^(-27))+),(("Number of Cu atoms")xx),((63.54 xx 1.66 xx 10^(-27))):}]]/((3.903 xx 10^(-10))^(2)m^(3))` `276.4 = n_(Sn)(118.7) + n_(Cu)(63.54)` `4.35 = 1.86n_(Sn) + n_(Cu)` `n_(Cu) = 4 rArr n_(Sn) = 0.188` Atomic fraction `= (n_(Sn))/(n_(Sn) + n_(Cu)) = 0.05` |
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| 1479. |
If `NaCl` is doped with `10^(-4)mol%`of `SrCl_(2)` the concentration of cation vacancies will be `(N_(A)=6.02xx10^(23)mol^(-1))`A. `6.02xx10^(15)"mol"^(-1)`B. `6.02xx10^(16)"mol"^(-1)`C. `6.02xx10^(17)"mol"^(-1)`D. `6.02xx10^(14)"mol"^(-1)` |
| Answer» Correct Answer - C | |
| 1480. |
Classify the following solids as ionic , metallic , molecular , covalent network or amorphous . i) Si ii) `I_(2)` iii) `P_(4)` iv) Rb v) SiC vi) LiBr vii) Ammonium Phosphate `(NH_(4))_(3) PO_(4)` viii) Plastic ix) graphite x) Tetra phosphorous decoxide |
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Answer» i) Si - Covalent network solid ii) `I_(2) ` - Molecular solid with covalent bonds iii) `P_(4)` - Molecular solid with covalent bonds iv) Rb - Metallic solid v) SiC - Giant molecular Network Solid with covalent bonds vi) LiBr - Ionic solids vii) Ammonium Phsophate `(NH_(4))_(3) PO_(4) - ` Ionic Solid viii) Plastic - Amorphous solid ix ) Graphite - Hexagonal network solid with covalent bonds (giant molecule) x) Tetra phosphorous decoxide `(P_(4)O_(6))` - Molecular solid with covalent bonds . xi) Brass - Metallic solid |
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| 1481. |
Assertion : Graphite is a good conductor of electricity, however diamond belongs to the category of insulators. Reason : Graphite is soft in nature, on the other hand diamond is very hard and brittleA. Assertion and reason both are corrct statements and reason is correct explanation for assertion.B. Assertion and reson both are correct statements but reason is not correct explanation for asseration .C. Assertion is correct statement but reason is wrong statement .D. Assertion is wrong statement but reason is correct statement . |
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Answer» Correct Answer - c Correct explanation . In graphite, each carbon atoms is linked to three othe carbon atoms in layers and the fourth electron is free to move. Diamond is a covalent network solid in which each carbon atom is linked to four other carbon atoms. |
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| 1482. |
Give the nature of bonding in the following types of solids. (i)Copper (ii) water (iii) sodium chloride (iv) graphite. |
| Answer» (i) metallic (ii) hydrogen bonding (iii) ionic bonding (iv) covalent bonding | |
| 1483. |
The radius of the `Na^(+)` is 95 pm and that of CI ion is 181 pm Predict the coordination number of `Na^(+)` ?A. `4`B. `6`C. `8`D. unpredictable |
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Answer» Correct Answer - b Between ` 0.414` and `0.732` `rArr` coordination no = 6 |
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| 1484. |
Gold (atoic radius = 0.144 nm) crystallizes in a facelcentred unit cell. What is the length of a side of the cell? |
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Answer» For a face centred cubic unit cell (fcc) Edge length `(a)=2sqrt(2r)=2xx1.4142xx0.144mm=0.407nm` |
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| 1485. |
For a gaseous reaction `A+3BhArr 2C,DeltaH^(@)=-90.0kJ, DeltaS^(@)=-200.0jk^(-1)` AT `400k.` The `DeltaG^(@)` for the reaction `(1)/(2)A+(3)/(2)B rarr C` at `400K` is `:` |
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Answer» Correct Answer - 5 For, `A+3BhArr 2C` `DeltaG^(@)=DeltaH^(@)-TDeltaS^(@)` `=-90-400xx(-0.200)` `=-90+80=-10kJ` For, `(1)/(2)A+(3)/(2)B hArrC` `DeltaG^(@)=-5kJ` |
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| 1486. |
How do you distinguish between hexagonal close - packing and cubic close - packing structures ? |
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Answer» Hexagonal close packing : The spheres of the `3^(rd)` layer are exactly aligned with those of first layer . This pattern of spheres is repeated in alternate layers . Tetrahedral voids of the `2^(nd)` layer may be convered by the spheres of `3^(rd)` layer . This structure is called hexagonal close packed (hcp) structure . Cubic close packing : The spheres of `3^(rd)` layer cover the octehedral voids of `2^(nd)` layer . But the spheres of `4^(th)` layers are aligned with those of first layer . This structure is called cubic close packing . |
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| 1487. |
How are the intermolecular forces among the molecules affect the melting point ? |
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Answer» `to` As the intermolecular forces between constituent particles of solid increases , stability of the compound increases . `to` As the stability of crystal increases melting point of solid also increases (high) . |
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| 1488. |
Which of the following is not true about crystalline solidsA. They are rigid and hardB. They possess plane surfacesC. They are obtained by rapid cooling of molten substancesD. They have definite geometric configuration. |
| Answer» Correct Answer - 3 | |
| 1489. |
a. "Stability of a crystal is reflected is reflected in the magnitude of its melting points" Comment. b. Melting points of some compounds are given below water `= 273 K`, ethyl alcohol `= 153.7 K`, diethyl ether `= 156.8 K`, methane `= 90.5 K`. What can you say about the intermolecular forces between the molecules of these compounds? |
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Answer» Stability of a crystal is reflected in the magnitude of its melting point . Explanation . `to ` The stability of a crystal mainly depends upon the magnitude of forces of attraction between the constituent particles . `to` As the attractive forces between the constituent particles increases stability of the crystal also increases . `to` As the stability of crystal increases melting point of solid will be higher . |
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| 1490. |
The forces operating between non-polar molecules like He, ` H_(2) , CH_(2)` etc.When present as crystalline solids are called ________ (a type of van der waals forces |
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Answer» Correct Answer - Rhombus with an angle of `60^(@)` |
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| 1491. |
Which of the following are not the characteristics of crystalline solids ?A. They are isotropicB. They exhibit polymorphismC. After melting, they become non - crystallineD. They donot have thermodynamic defects |
| Answer» Correct Answer - A::D | |
| 1492. |
Which of the following are not the characteristics of crystalline solids ?A. They are exhibit polymorphismB. The are isotropicC. They do not have thermodynamic defectsD. After melting, they become crystalline |
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Answer» Correct Answer - b,c b,c |
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| 1493. |
Which of the following characteristics is/are possessed by non- polar molecular solids?A. They have low melting point and are usually in liquid or gaseous state at room temperature and presureB. Atoms or molecules are held by weak dispersion foeces or Londom forcesC. They are soft and non-conductors of electricityD. All of the above |
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Answer» Correct Answer - d Monoatomic gasses such and He, Ar and diatomic molecules like `CI_(2),H_(2) and I_(1)` formed by none- polar covalent honds are non- polar molecular solid and shown the following characteratics (i) Being covalent honded by weak forces , they have low melting point and are liquid or gases at mom temperature and presure (ii) Atoms or molecules are held by weak dispersion force (iii) Being non-polar they are non-conductor. |
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| 1494. |
Which is not a property of solids?A. Solids are always crystalline in natureB. solid have hight density and low compressinilityC. The diffusion of solid is very slowD. Solids have definete volume |
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Answer» Correct Answer - a Solid are also non-crystalline in nature. |
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| 1495. |
Which solid will have the weakest intermolecular forces?A. IceB. PhosphorusC. NaphthaleneD. Sodium fluoride |
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Answer» Correct Answer - A Ice has the lowest melting point out of the given solids. Hence it has the weakest intermolecular forces. |
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| 1496. |
Which one of the following is the most correct statement?A. Brass is an interstitial alloy, while steel is a substitutionaly alloyB. Brass and steel are both substitutional alloysC. Brass is a substtitutional alloy, while steel is an interstitial alloyD. Brass and steel are both interstial alloys |
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Answer» Correct Answer - b Brass `Cu = 80 %, Za = 20%` subtantional alloy. |
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| 1497. |
`Na` and `Mg` crystallize in bcc- and fcc-type crystals, respectively, then the number of atoms of `Na` and `Mg` present in the unit cell of their respective crystal isA. `4` and `2`B. `9` and `14`C. `14` and `9`D. `2` and `4` |
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Answer» Correct Answer - d number of unit cell in bcc is `2` and in fcc is `4` . |
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| 1498. |
`Na` and `Mg` crystallize in bcc- and fcc-type crystals, the ratio of number of atoms present in the unit cell of their respective crystal isA. `1`B. `0.5`C. `3`D. `4` |
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Answer» Correct Answer - B For bcc, `Z_("eff") = 2//"unit cell"` For fcc, `Z_(eff) = 4//"unit cell"` `:.` Ratio `= 2/4 = 0.5` |
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| 1499. |
`Na` and `Mg` crystallize in bcc- and fcc-type crystals, respectively, then the number of atoms of `Na` and `Mg` present in the unit cell of their respective crystal isA. 4 and 2B. 9 and 14C. 14 and 9D. 2 and 4 |
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Answer» Correct Answer - D bcc has atoms at corner and body centre . Hence , total number of atoms = 2 and in fcc, atoms are present at corner and faces hence, total number of atoms = 4 |
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| 1500. |
`Na` and `Mg` crystallize in bcc- and fcc-type crystals, the ratio of number of atoms present in the unit cell of their respective crystal isA. 4 and 2B. 9 and 14C. 14 and 9D. 2 and 4 |
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Answer» Correct Answer - D In BCC (Na) No. of atoms per unit cell `=8xx1//8+1=2` In FCC. (Mg) No. of atoms per unit cell `=8xx1//8+6xx1//2=4` The number of atoms in unit cells are respectively 2 and 4 |
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