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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.
| 501. |
Atmospheric pressure recorded in different citie are as follows `{:("Cities","Shimla","Bangalore","Delhi","Mumbai"),("p in "N//m^(2),1.01xx10^(5),1.2 xx 10^(5),1.02 xx 10^(5),1.21 xx 10^(5)):}` Consider the above data mark the place at which liquid will boil first.A. ShimlaB. BangaloreC. DelhiD. Mumbai |
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Answer» Correct Answer - A A liquid starts boiling when the pressure on its surface is equal to the atmopheric presssure. Thus liquid will boil first at Shimla because its atmospheric pressure is minimum. |
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| 502. |
Which of the following is true at the critical point?A. At the critical point three roots of van der Waals equation are equalB. Below the critical point two roots of the van der Waals equation are equal and imaginary but one root is realC. Above the critical point density of gas is greater than density of liquidD. Above the critical point three roots of van der Waals equation are real but unequal. |
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Answer» Correct Answer - A At the critical point three roots of the van der Waals equation are equal. |
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| 503. |
Unit of a would be......... |
| Answer» atm `L^(2) mol^(-2)` | |
| 504. |
Real gases behave idealy at ........and ....... |
| Answer» Low pressure, high temperature | |
| 505. |
In the microscopic model of the gas, all the moleculer are supposed to movek with the same velocities. |
| Answer» Correct Answer - True | |
| 506. |
What are the most favourble conditions to liquefy a gas ?A. High temperature and high pressureB. Law temperature and high pressureC. Low temperature and low pressureD. High temperature and low pressure |
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Answer» Correct Answer - B Gases can be liquefied at low temperature and high pressure. |
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| 507. |
Correct gas equation is :A. `(P_(1)V_(1))/(P_(2)V_(2)) = (T_(1))/(T_(2))`B. `(V_(1)T_(2))/(P_(1)) = (V_(2)T_(1))/(P_(2))`C. `(P_(1)T_(1))/(V_(1)) = (P_(2)T_(2))/(V_(2))`D. `(V_(1)V_(2))/(T_(1)T_(2)) = P_(1)P_(2)`. |
| Answer» Correct Answer - A | |
| 508. |
The unit cell of aluminium is a cube with an edge length of 405 pm. The density of aluminium is `2.70 g cm^(-3)` . What type of unti celll of alluminium is ? |
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Answer» Correct Answer - The unit cell is face-centred. Apply the formula , density `=(ZxxM)/(N_(0)xxV)` and find the value of Z. |
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| 509. |
An element cyrstallises in a structure having a fcc unit cell of and edge 200 pm. Calculate its density if 200 g of this element contains `24xx 10^(23)` atoms. |
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Answer» Correct Answer - `4.166 g cm^(-3)` Molar mass `=(200)/(24xx10^(23))xx6.023xx10^(23)=50.19 g "mol"^(-1)` for fcc, Z=4, `V=a^(3)=(200xx10^(-10))^(3)` Apply density `=(ZM)/(N_(0)xxV)` |
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| 510. |
Edge length of a cube is 400 pm , its body diagonal would beA. 500 pmB. 600 pmC. 566 pmD. 693 pm |
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Answer» Correct Answer - D Body diagonal =`sqrt3a` =1.732 x 400 pm =692.28 pm `approx` 693 pm |
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| 511. |
A gas having a molecular mass of `84.5 g mol^(-1)` enclosed in a flask at `27^(@)C` has a pressure of `1.5 atm`. Calculate the density of the gas. |
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Answer» Here, `P=1.5 atm`, `T=300 K`, `M=84.5 g mol^(-1)`, `R=0.0821 L atm K^(-1)mol^(-1)` The realtionship between pressure and density is `P=(rho RT)/(M)` or `rho=(PM)/(RT)=(1.5xx84.5)/(0.0821xx300)=5.416 g L^(-1)` |
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| 512. |
When `2 g` of a gas `A` is introduced into an evacuated flask kept at `25^(@)C`, the pressure is found to be `1 atm`. If `3 g` of another gas `B` is then heated in the same flask, the total pressure becomes `1.5 atm`. Assuming ideal gas behaviour, calculate the ratio of the molecular weights `M_(A)` and `M_(B)`.A. `1:1`B. `1:2`C. `2:3`D. `1:4` |
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Answer» Correct Answer - C After the introduction of `3 g` of `B` the pressure became half its number of mole of `B` `PV=nRT` `n=(w)/(M)implies (w_(A))/(M_(A))=(w_(B))/(M_(B))` or `(2)/(M_(A))=(3)/(M_(B)) or (M_(A))/(M_(B))=(2)/(3)` |
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| 513. |
The compressibility factor `(Z=PV//nRT)` for `N_(2)` at `223 K` and `81.06 MPa` is `1.95`, and at `373 K` and `20.265 MPa`, it is `1.10`. A certain mass of `N_(2)` occupies a volume of `1.0 dm^(3)` at `223 K` and `81.06 MPa`. Calculate the volume occupied by the same quantity of `N_(2)` at `373 K` and `20.265 MPa`.A. `3.774 dm^(3)`B. `2.77 dm^(3)`C. `5.07 dm^(3)`D. `9.30 dm^(3)` |
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Answer» Correct Answer - A For T = 223 K, P = 81.06 Mpa, Z = 1.95, and `V=1.0 dm^(3)=10^(3)cm^(3)`, we have `n=(PV)/(ZRT)=(81.06xx10^(3))/(1.95xx8.314xx223)=22.42` mol Now, at T = 373 K, P = 20,265 Mpa, Z = 1.10, the volume occupied will be `V=(ZnRT)/(P)=(1.10xx22.42xx8.314xx373)/(20.265)=3774.0 cm^(3)` `therefore V=3.774 dm^(3)` |
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| 514. |
As regards magnetic behaviour, `TiO_2` is _______ |
| Answer» Correct Answer - diamagnetic | |
| 515. |
At `25^(@)C` and `730 mm` pressure, `730 mL` of dry oxygen was collected. If the temperature is kept constant what volume will oxygen gas occupy at `760 mm` pressure?A. `701 mL`B. `449 mL`C. `569 mL`D. `621 mL` |
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Answer» `P_(1)=730 mm` `V_(1)=730 mL` `V_(2)=?` `P_(2)=760 mm` `:. P_(1)V_(1)=P_(2)V_(2)` `V_(2)=(P_(1)V_(1))/(P_(2))=(730xx730)/(760)=701 mL` |
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| 516. |
What is the meaning of pressure of the gas? |
| Answer» Rate of diffusion of `H_(2)` is faster than that of air. | |
| 517. |
The ratio between the three speeds, `u_(mp) : u_(av) : u_(rms)` is given asA. `1 : 1.224 : 1.128`B. `1 : 1.128 : 1.224`C. `1.224 : 1.128 : 1`D. `1.128 : 1.224 : 1` |
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Answer» Correct Answer - B `u_(mp) : u_(av) : u_(rms) = sqrt((2 RT)/(M)) : sqrt((8 RT)/(pi M)): sqrt((3 RT)/(M))` `= sqrt(2) : sqrt((56)/(22)) : sqrt(3)` `= 1.414 : 1.595 : 1.732` Dividing by `1.414` gives `u_(mp): u_(av) : u_(rms) = 1 : 1.128 : 1.224` |
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| 518. |
Which of the following is used in deriving the kinetic gas equation ?A. most probable speedB. Average speedC. Root mean square speedD. Mean square speed |
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Answer» Correct Answer - C Root mean square speed is used because it is the most accurate form in which speed can be used to describe the behaviour of an ideal gas. |
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| 519. |
Liquefied natural gas `(LNG)` is mainly methane. A `10 m^(3)` tank is constructed to store `LNG` at - `164^(@)C` and `1 atm` pressure,under this condition density of LNG is `415 kg//m^(3)`. The volume of strorage tank capable of holding …. Mass of `LNG` as a gas at `20^(@)C` and `1 atm` pressure will beA. `1250 m^(3)`B. `5280 m^(3)`C. `6230 m^(3)`D. `9870 m^(3)` |
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Answer» Correct Answer - C `n=(10xx415xx1000)/(16)= 2.59xx10^(5)` `V=(nRT)/(P)=(2.59xx10^(5)xx0.0821xx193)/(1)` `= 6.23xx10^(6)L= 6230m^(3)` |
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| 520. |
Calculate the ioninc radius of a `Cs^(+)` ion, assuming that the cell edge length for CsCl is 0.4123 nm and the ionic radius of a `Cl^(-)` ion is 0.181nm.A. 0.176nmB. 0.231nmC. 0.357nmD. 0.116nm |
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Answer» (a) `r_(Cs^(-))+r_(Cl)=(d_(body))/(2)=(0.711)/(2)nm=0.3571nm` `r_(Cl^(+))=0.357-0.181=0.176nm` |
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| 521. |
What is the relation between mass and number of moles of the gas ? |
| Answer» No. of mole of the gas `(n) = ("Mass of the gas (m)")/("Molar mass of the gas (M)")` | |
| 522. |
The critical temperature of water is higher than that of `O_(2)` because the `H_(2)O` molecule hasA. fewer electrons than `O_(2)`B. two covalent bondsC. `V`-shapeD. dipole moment |
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Answer» Correct Answer - D |
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| 523. |
A gas can be liqufiedA. above its critical temperatureB. at its critical temperatureC. below its critical temperatureD. at any temperature |
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Answer» Correct Answer - C gas can be liquefied below its critical temperature. |
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| 524. |
The incorrect statement among the following isA. The boiling point of a liquid at one bar is called standard boiling oint of the liquidB. The vapour pressure of a liquid is constant temperatureC. The SI unit of coefficient of viscosity of a pascal secondD. The boiling point of liquid is same external pressures. |
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Answer» Correct Answer - D |
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| 525. |
The vapour density of a gas is 11.2. The volume occupied by 11.2 g of the gas is NTP isA. 22.4 LB. 11.2 LC. 1 LD. 2.24 L |
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Answer» (b) Molecular weight of gas = 2`xxVD=2xx11.2=22.4` Hence, number of moles of gas taken=`(11.2)/(22.4)=(1)/(2)` Hence, volume of gas at NTP=`(1)/(2)xx22.4L=11.2 L` |
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| 526. |
Equal molecules of `N_(2)` and `O_(2)` are kept in a closed container at pressure `P`. If `N_(2)` is removed from the system, then what will be the pressure of the container?A. `P`B. `2P`C. `P//2`D. `P^(2)` |
| Answer» `P//2` because `P_(total)=P_(N_(2))+P_(O_(2))` | |
| 527. |
Molecular mass of a gas is 78 . Its density at `98 .^(@)C` and 1 atm will beA. `200 g L^(-1)`B. `2 . 56 g L^(-1)`C. `256 g L^(-1)`D. `78 g L^(-1)` |
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Answer» Correct Answer - B P = 1 atm, `T = 98^(@)C+273 = 371 K, M = 78g mol^(-1)` `R = 0.0821" L atm K"^(-1)mol^(-1)` PM=dRT or `d=(PM)/(RT)=(1xx78)/(0.0821xx371)=2.56 g L^(-1)` |
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| 528. |
The drain cleaner Drainex contains small bits of aluminimum which react with caustic soda to form hydrogen. What volume of hydrogen at `20^(@)C` and `1 "bar"` will be released when `0.15 g` of aluminimum reacts? |
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Answer» The chemical reaction taking place is `underset(2xx27 g)underset(2 mol)(2AI)+2NaOH+2H_(2)O rarr 2NaAIO_(2)+underset(3mol)(3H_(2)` `54 g` of `Al` produces hydrogen `=3 mol` `0.15 g` of `Al` produces hydrogen `=(3xx0.15)/(54) mol=8.33xx10^(-3)mol` Calculation of volume of `8.33xx10^(-3)mol` of hydrogen at `20^(@)C` and `1` bar: `V=(nRT)/(P)` `=(8.33xx10^(-3)(mol)xx0.083("bar" L K^(-1) mol^(-1))xx293 (K))/(1 "bar")` `=0.20282 L=202.82 mL` |
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| 529. |
The drain cleaner Drainex contains small bits of aluminium which react with caustic soda to produce hydrogen What volume of hydrogen at `20^(@)C` aand one bar will be released when `0.15 g` of aluminium reacts ? .A. 204 mLB. 200 mLC. 203 mLD. 400 mL |
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Answer» Correct Answer - C The reaction between aluminium and caustic soda is `underset(2xx27=54g)(2Al)+2NaOH+2H_(2)O rarr 2NaAlO_(2)+underset(3xx22.4" L at STP")(3H_(2))` `therefore 54 g` of Al produces `H_(2)` at `STP = 3xx22.4 L` 0.15 g of Al will produce `H_(2)` at STP `=(3xx22.4)/(54)xx0.15=0.186 L` `{:("At STP","Given conditions"),(P_(1)="1 atm",P_(2)="1 bar "=0.987 " atm"),(V_(1)=0.186 L,V_(2)=?),(T_(1)=273 K,T_(2)=273+20=293 K):}` Applying ideal gas rquation, `(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))` `(1xx0.1867)/(273)=(0.987xx V_(2))/(293)` `V_(2)=(293)/(0.987)xx(1xx0.1867)/(273)=0.2030 L = 203 mL` . |
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| 530. |
What is the kinetic energy of `1 g` of `O_(2)` at `47^(@) C`?A. `3.24 xx 10^(2) J`B. `2.24 xx 10^(2) J`C. `1.24 xx 10^(2) J`D. `1.24 xx 10^(3) J` |
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Answer» Correct Answer - C According to the kinetic molecular theory of gases , we have Kinetic energy `(KE) "of n moles" = (3)/(2) nRT` `n_(O_(2)) = ("Mass of" O_(2))/("Molar mass of" O_(2)) = (1g)/(32 g "mol"^(-1))` `T = 47 + 273 = 230 K` Thus , `KE = (3)/(2)((1)/(32) "mol") (8.314 J K^(-1) "mol"^(-1)) (320 K)` `= 124.71 J = 1.24 xx 102 J` |
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| 531. |
The ratio of most probable velocity to that of average velocity isA. `pi//2`B. `sqrt(pi)//2`C. `2//sqrt(pi)`D. `2//pi` |
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Answer» Correct Answer - B According to the kinetic molecular theory of gases , we have `u_(mp) : u_(aV) = sqrt((2 RT)/(M)) : sqrt((8 RT)/(pi M))` ` = (sqrt(2))/(sqrt(8//pi))` ` = (sqrt(2pi))/(sqrt(8)) = (sqrt(2 pi))/(sqrt(4.2))` `= (sqrt(2 pi))/(2 sqrt(2)) = (sqrt(pi))/(2)` |
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| 532. |
What is the coordination number of an octahedral void ? |
| Answer» Correct Answer - Six | |
| 533. |
In a crystal of zinc sulphide , zinc occupies tetrahedral voids . What is the coordination number of zinc ? |
| Answer» Correct Answer - 4 | |
| 534. |
Statement-1: The octahedral voids have double the size of the tetrahedral voids in a crystal. Statement-2: The number of tetrahedral voids is double the number of octahedral voids in a crystalA. Statement 1 is True , Statement-2 is True , Statement-2 is a correct explanation for statement-1B. Statement 1 is True , Statement-2 is True , Statement-2 is NOT a correct explanation of Statement-1C. Statement-1 is True , Statement-2 is FalseD. Statement -1 is False , Statement -2 is True |
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Answer» Correct Answer - D Correct statement-1. Octahedral voids are larger in size than tetrahedral voids but not double in size. Statement-2 is correct. |
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| 535. |
The density of neon gas will be highest atA. `STP`B. `0^(@)C`, `2 atm`C. `273^(@)C`, `1 atm`D. `273^(@)C`, `2 atm` |
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Answer» `P=(n)/(V)RT=(w)/(mV)=RT` `P=(dRT)/(m)` `d prop (P)/(T)` |
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| 536. |
The density of neon will be highest atA. `STP`B. `0^(@)C`, `2 atm`C. `273^(@)C`, `1 atm`D. `273^(@)C`, `2 atm` |
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Answer» Density `rho=(PM)/(RT)` For maximum density, `P//T` will be maximum which is for option (`b`). |
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| 537. |
The density of a gas is found to be 1.56 g/litre at 745 mm pressure and `65^(@)C`. Calculate the molecular mass of the gas. |
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Answer» Here, we have `" " d=1.56" g litre"^(-1), P=745 mm=(745)/(760)" atm"` `T=65^(@)C=65+273=338" K", R=0.0821" litre atm" K^(-1) mol^(-1)` `M=(dRT)/(P)=(1.56xx0.0821xx338)/(745//760)=44.2 u` |
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| 538. |
The density of a gas is found to be `1.56 g dm^(-3)` at `0.98` bar pressure and `65^(@)C`. Calculate the molar mass of the gas. |
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Answer» From the available data : `rho = 1.56 g dm^(-3) , P = 0.98 "bar"` `T = (65+273) = 338 K, R = 0.083 dm^(3) "bar" K^(-1) mol^(-1)`. We know that, `M = (rhoRT)/(P) = ((1.56 g dm^(-3)) xx (0.083 dm^(3) "bar" K^(-1) mol^(-1)) xx (338 K))/((0.98 "bar"))` `= 44.66 g mol^(-1)` |
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| 539. |
Molar volume of a gas at `0^(@)C` and 1 bar pressure is ………………………………. |
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Answer» Correct Answer - 22700 mL |
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| 540. |
Which of the following assumptions is incorrect according to kinetic theory of gases ?A. Particles of a gas move in all possible directions in straight lines .B. All the particles , at any particular time , have same speed and same kinetic energy .C. There is no force of attraction between the particles of a gas at ordinary temperature and pressure .D. The actual volume of the gas is negligible in comparison to the empty space between them . |
| Answer» Correct Answer - B | |
| 541. |
What is the temperaure at which the kinetic energy of 0.3 mole of helium is equal tp the kinetic energy of 0.4 mole of argon at 400KA. 400KB. 873 KC. 533KD. 300K |
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Answer» (c ) We know that, the kinetic energy=Nrt Kinetic energy of helium =0.3xxRxxT …(I) Kinetic energy of argon=0.3xxRxx400…(ii) According to question Kinetic energy of helium=kinetic energy of argon `0.3xxRxxT=0.4xxRxx400` T=533K |
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| 542. |
A compund is made of two elements `P` and `Q` are in `ccp` arrangement while atoms `P` occupy all the tetrahedral voids. What is the formula of the compound?A. PQB. `PO_(2)`C. `P_(2)Q`D. `P_(3)Q` |
| Answer» (c ) Suppose number of atoms Q in the ccp arrangement=100 Then, number of tetrahedral sites=200 As all the tetrahedral sites are occupied by atoms P, therefore their number=200 Hence, ratio P:Q=2:1 i.e. the formula is `P_(2)Q`. | |
| 543. |
A compound `M_(p)X_(q)` has cubic close packing `(ccp)` arrangement of `X`. Its unit cell structure is shown below. The empirical formula of the compound is A. MXB. `MX_(2)`C. `M_(2)X`D. `M_(5)X_(14)` |
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Answer» Correct Answer - B Number of atoms of (X) `=8xx(1)/(8)+6xx(1)/(2)=4` Number of atoms of (M)`=4xx(1)/(4)+1=2` `{:(M:X),(2:4),(1:2):}` Empirical formula `=MX_(2)` |
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| 544. |
A compound `M_p X_q` has cubic close packing (ccp) arrangement of X. Its unit cell structure . The empirical formula of the compound isA. MXB. `MX_2`C. `M_2X`D. `M_5X_14` |
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Answer» Correct Answer - B No. of M atoms `{:(=4xx1/4," " 1,=2),("(edge centres)","(body-centred)",):}` No. of X atoms `{:(=8xx1/8,6xx1/2,=4),("(corners)","(face-centres)",):}` `therefore` Unit formula =`M_2X_4` and empirical formula =`MX_2` |
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| 545. |
A 100 mL flask contained `H_(2)` at 200 Torr, and a 200 mL flask contained He at 100 Torr. The two flask were then connected so that each gas filled their combined volume. Assuming no change in temperature,total pressure isA. 300 TorrB. 66.66 TorrC. 150 TorrD. 133.33 Torr |
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Answer» Correct Answer - D `n_(H2)=(pV)/(RT)=(20000)/(RT)rArrn_(He)=(pV)/(RT)=(20000)/(RT)` `:." "n_(H2)+n_(He)=(40000)/(RT)` Total volume = 300 mL `p=(n)/(V)RT=(40000RT)/(RTxx300)=(400)/(3)=133.33" Torr"` |
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| 546. |
Which assumption of kinetic molecular theory is not followed when a real gas shows non-ideal beahaviour?A. Gas molecules move at random with no attractive forces between themB. The velocity of gas molecules is dependent on temperatureC. The amount of space occupied by a gas is much greater than the space occupied by the actual gas moleculesD. In collisions with the walls of the container or with other molecules, energy is conserved |
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Answer» Correct Answer - A Gas molecules move at random motion with no atractive forces between them. |
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| 547. |
Assertion: Molar specific heat at constant volume of an ideal diatomic gas is `[(3)/(2)R+R]`. Reason: On heating `1` mole an ideal diatomic gas at constant pressure of `1^(@)C` rise in temperature, the increase in internal energy of gas is `(7)/(2)R`.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - A Average energy of diatomic molecule is `(5)/(2)RT` at constant `V`. `:.` Average energy of diatomic molecule `=(7)/(2)RT` `:.` Increase in internal energy `=(7)/(2)R(T+1)` `=(7)/(2)RT=(7)/(2)R` |
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| 548. |
If silver iodide crystallizes in a zinc blende structure with `I^-` ions forming the lattice, then calculate fraction of the tetrahedral voids occupied by `Ag^+` ions. |
| Answer» In AgI, if there are n `I^-` ions, there will be n `Ag^+` ions. As `I^-` ions from the lattice, number of tetrahedral voids=2n. As there are n `Ag^+` ions to occupy these voids, therefore, fraction of tetrahedral voids occupied by `Ag^+` ions =n/2n=1/2=50% | |
| 549. |
Which of the following oxides behaves as conductor or insulator depending upon temperature ?A. TiOB. `SiO_2`C. `TIO_3`D. MgO |
| Answer» Correct Answer - c | |
| 550. |
The packing efficiency of the two dimensional square unit cell shown below is: A. `39.27%`B. `68.02 %`C. `74.05%`D. `78.54%` |
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Answer» Correct Answer - D `asqrt(2)=4r` , `a=2sqrt(2)r` packing fraction `=("Occupied area")/("Total area")xx100` `=(2pir^(2))/((2sqrt(2)r)^(2))xx100=78.5%` |
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