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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.
| 51. |
A mixture of ideal gases is cooled up to liquid helium temperature `(4.22 K)` to form an ideal solution. Is this statement true or false? Justify your answer in not more than two lines. |
| Answer» An ideal gas cannot be liquefied because there are no intermolecular forces of attraction. So, this is false. | |
| 52. |
For the non-zero volume of the molecules, real gas equation for `n` mol of the gas will beA. `(p+(a)/(V^(2)))V=RT`B. `pV=nRT+nbp`C. `p(V-nb)=nRT`D. `(b) and (c) true |
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Answer» Correct Answer - D For non-zero volume, b cannot be neglected `:." "p(V-nb)=nRT` |
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| 53. |
The compressibility factor of a gas is greater than unity at `STP`. ThereforeA. `V_(m)gt22.4 L`B. `V_(m)lt22.4 L`C. `V_(m)=22.4 L`D. The gas will become less liquefiable |
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Answer» `Z_(real gas)gt1` `:. V_(m)gt22.4L` (Because `V_(ideal)=22.4 L`) Hence, when volume is higher, the gases are far apart and, therefore, difficult to liquefy. |
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| 54. |
The compressibility factor of gases is less than unity at `STP`. Therefore,A. `V_(m)(molar" volume")gt22.4" L"`B. `V_(m)lt22.4" L"`C. `V_(m)=22.4" L"`D. `V_(m)=44.8" L"` |
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Answer» Correct Answer - B `(pV)/(RT)=Z, Zlt1rArr(pV)/(RT)lt1` at STP,`(p)/(RT)=(1)/(0.0821xx273)=(1)/(22.4)` `V_(m)/(22.4)lt1` `:." "V_(m)lt22.4L` |
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| 55. |
The compressibility of a gas is less than unity at `STP`, therefore,A. `V_(m)gt22.4" L"`B. `V_(m)lt22.4" L"`C. `V_(m)=22.4" L"`D. `V_(m)=44.8" L"` |
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Answer» Correct Answer - B `Z=(pV)/(nRT)i.e.(pV)/(nRT)lt1orpVltnRT` (p = 1 atm at STP) `Vltxx0.0821xx273 orVlt22.4 " L"` |
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| 56. |
The compressibility factor of gases is less than unity at `STP`. Therefore,A. `V_(m)gt22.4 L`B. `V_(m)lt22.4 L`C. `V_(m)=22.4 L`D. `V_(m)=4.8 L` |
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Answer» `Z=(V_(real))/(V_(ideal))` If `Zlt1`, then `V_(real lt V_(ideal)` (i.e., `22.4 L` at `STP`) Hence, the correct choice is (`b`) |
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| 57. |
The compressibility of a gas is less than unity at N.T.P. Therefore.A. `V_(m) gt 22.4 L`B. `V_(m) lt 22.4 L`C. `V_(m) = 22.4 L`D. `V_(m) = 44.8 L` |
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Answer» Correct Answer - B `Z = (PV)/(nRT)` For ideal gas, `V_(m) = 22.4 L` at N.T.P. At a given temperature and pressure if `Z` is less than unity, then `V_(m) lt 22.4 L` |
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| 58. |
The r.m.s. velocity of hydrogen is `sqrt(7)` times the r.m.s. velocity `1.0 g cm^(-3)` and that of wate vapours is `0.0006 g cm^(-3)`, then the volume of water molecules in 1 L` of steam at this temperature isA. `T(H_(2)) = T(N_(2))`B. `T(H_(2)) gt T(N_(2))`C. `T(N_(2)) gt T(H_(2))`D. `T(H_(2)) = sqrt(7)T(N_(2))` |
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Answer» Correct Answer - C `sqrt((3RT_((H_(2))))/(2)) = sqrt(7)sqrt((3RT_((N_(2))))/(28))` or `(3RT_((H_(2))))/(2) = 7 xx (3RT_((N_(2))))/(28)` or `T_((N_(2))) = 2T_((H_(2)))` `:. T_((N_(2))) gt T_((H_(2)))` |
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| 59. |
In `CaF_2` crystal, `Ca^(2+)` ions are present in FCC arrangment . Calculate the number of `F^-` ions in the unit cell. |
| Answer» No. of `Ca^(2+)` ions per unit cell =`8xx1/8+6xx1/2=4`. Hence, no. of `F^-` ions per unit cell = 2 x 4 =8 | |
| 60. |
A compound `AB_2` possesses the `CaF_2` type crystal structure. Write the coordination numbers of `A^(2+)` and `B^-` ions in its crystals |
| Answer» Coordination no. of A =8, Coordination no. of B =4 | |
| 61. |
In fluorite `(CaF_2) , Ca^(2+)` ions forms the ______ structure whereas `F^-` ions are present in the _________ voids. |
| Answer» Correct Answer - fcc, tetrahedral | |
| 62. |
In each of the compouds : NaCl , ZnS and `CaF_2` , write (i) ions occupying the voids (ii)types of voids occupied (iii) fraction of voids occupied. |
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Answer» (a)`NaCl=Na^+` ions in all the octahedral voids (b)`ZnS=Zn^(2+)` ions in alternate tetrahedral voids (c ) `CaF_2=F^-` ions in all the tetrahedral voids. |
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| 63. |
In Zinc blende , the percentage of tetrahedral voids occupied by `Zn^(2+)` ions is :A. `25%`B. `50 %`C. `75 %`D. `100 %` |
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Answer» Correct Answer - B In zinc blende , `S^(2-)` ions have ccp arrangement and `Zn^(2+)` ions occupy half , i.e., alternate tetrahedral voids. |
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| 64. |
In ZnS, `S^(2-)` ions from ________ structure while `Zn^(2+)` ions are present in _______ voids. |
| Answer» Correct Answer - fcc, alternate tetrahedral voids | |
| 65. |
Assertion. Frenkel defects are shown by silver halides. Reason. `Ag^+` ions have small size.A. If both assertion and reason are true, and reason is the true explanation of the assertion.B. If both assertion and reason are true, but reason is not the true explanation of the assertion.C. If assertion is true , but reason is falseD. If both assertion and reason are false. |
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Answer» Correct Answer - A R is correct explanation of A |
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| 66. |
Assertion : All the gases should be cooled below their critical temperature for liquification . Reason : Cooling slows down the movement of molecules therefore , intermolecular forces may hold the slowly moving molecules together and the gas liquifies .A. If both assertion and reason are true and reason is the correct explanation of assertion .B. If both assertion and reason are true but reason is not the correct explanation of assertion .C. If assertion is true but reason is false .D. If both assertion and reason are false . |
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Answer» Correct Answer - B Critical temperature of a gas is the highest temperature at which liquification of the gas first occurs. Above critical temperature, gas cannot be liquefied howsoever high pressure may be applied on the gas. |
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| 67. |
In which pair most efficient packing present ?A. hcp and bccB. hcp and ccpC. bcc and ccpD. bcc and simple cubic cell |
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Answer» Correct Answer - d In hcp and ccp, space occupied =74% while in bcc, it is 68% |
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| 68. |
Containers A, B and C of equal volume contain oxygen, neon and methane respectively at the same temperature and pressure. The correct increasing order of their masses isA. `A lt B lt C`B. `B lt C lt A`C. `C lt A lt B`D. `C lt B lt A` |
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Answer» Correct Answer - D Under similar conditions of temperature and pressure, equal volumes of different gases contain equal number of moles. `:.` Masses of `O_(2)`,Ne and `CH_(4)` will be in the ratio `32:20:16`. |
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| 69. |
If the unit cell of a mineral has a cubic close packed (ccp) array of oxygen atoms with m fraction of octahedral holes occupied by magnesium ions , m and n, respectively areA. `1/2,1/8`B. `1,1/4`C. `1/2,1/2`D. `1/4,1/8` |
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Answer» Correct Answer - A In ccp lattice, Z=4 . `therefore` No. of O-atoms per unit cell= `4(O^(2-))` . `therefore` No. of octahedral voids = 4 and No. of tetrahedral voids=8 As m fraction of octahedral voids is occupied by `Al^(3+)` ions, therefore , `Al^(3+)` ions present =4 m . Similarly, `Mg^(2+)` ions =8 n. Hence, formula of the mineral is `Al_(4m) Mg_(8n)O_4`. As total charge on the compound is zero, hence 4 m (+3)+8 n(+2)+4 (-2)=0 or 12 m +16 n -8 =0 Substituting the given values of m and n , equation is satisfied only when `m=1/2` and `n=1/8` (as `12 xx1/2+16xx1/8-8=0`) |
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| 70. |
KF has ccp structure. Calculate the radius of unit cell if the side of the cube or edge length is 400 pm. How many `F^-` ions and octahedral voids are there in this unit cell ? |
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Answer» ccp is same as fcc, Hence, `r=a/(2sqrt2)`=0.3535 a=0.3535 x 400 nm=141.4 pm In fcc structure , there are `4F^-` ions in the packing and hence 4 octahedral voids. |
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| 71. |
In hexagonal close packing , the difference in the number of tetrahedral 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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| 72. |
A metal crystallizes into two cubic phases , face-centred cubic (fcc) and body -centred cubic (bcc) whose unit cell lengths are 3.5 and 3.0 Å respectively. Calculate the ratio of the densities of fcc and bcc. |
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Answer» Density `(rho)=(ZxxM)/(a^3xxN_0)` For fcc, Z=4,a=3.5 Å =`3.5xx10^(-8)` cm `therefore rho_"fcc"=(4xxM)/((3.5xx10^(-8))xxN_0)` For bcc, Z=2, a=3.0 Å =`3.0xx10^(-8)`cm `therefore rho_"bcc"=(2xxM)/((3.0xx10^(-8))^3xxN_0) therefore rho_"fcc"/rho_"bcc"=(4xx(3.0xx10^(-8))^3)/(2xx(3.5xx10^(-8))^3)=1.259` |
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| 73. |
Assertion: Total number of octahedral voids present in unit cell of cubic close packing including the one that is present at the body centre, is four . Reason: Besides the body centre, there is one octahedral void present at the centre of each of the six faces of the unit cells and each of which is shared between two adjacent unit cells.A. Assertion and reason both are correct statements and reason is correct explanation for assertion.B. Assertion and reason both are correct statements but reason is not correct explanation for assertionC. Assertion is correct statement but reason is wrong statementD. Assertion is wrong statement but reason is correct statement. |
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Answer» Correct Answer - c Correct Reason.Besides the body centre, there are 12 octahedral voids present at the centres of the 12 edges |
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| 74. |
In the cubic close packing , the unit cell has ______A. 4 tetrahedral voids each of which is shared by four adjacent unit cells.B. 4 tetrahedral voids within the unit cellC. 8 tetrahedral voids each of the which is shared by four adjacent unit cells.D. 8 tetrahedral voids within the unit cells. |
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Answer» Correct Answer - d ccp=fcc |
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| 75. |
The edge lengths of the unit cells in terms of the radius of spheres constituting fcc, bcc and simple cubic unit cell are respectively _____A. `2sqrt2r, (4r)/sqrt3, 2r`B. `(4r)/sqrt3, 2sqrt2r , 2r`C. `2r, 2sqrt(2r), (4r)/sqrt3`D. `2r, (4r)/sqrt3 , 2sqrt2r` |
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Answer» Correct Answer - a For fcc, `r=a/(2sqrt2)` or `a=2sqrt2` for bcc, `r=sqrt3/4a` or `a=4r//sqrt3` For simple, r=a/2 or a=2r |
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| 76. |
The percentage of empty space in a body centred cubic arrangement is _____A. 74B. 68C. 32D. 26 |
| Answer» Correct Answer - c | |
| 77. |
At `1425^(@)C` , Fe crystallises in a body-centred cubic lattice whose edge length is `2.93 Ã…` . Assuming the atoms to be packed spheres, calculate : (a) the radius of the spheres, (b) the distance between centres of neighbouring spheres, (c) the number of atoms of Fe per unit lattice and (d) the total volume occupied by an atom of Fe. |
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Answer» (a) `asqrt(3)=4r` where, a=edge length `r=(asqrt(3))/(4)=(2.93xxsqrt(3))/(4)=1.268 Ã…` (b) Distance between the centres of neighouring spheres `=2r = 2 xx 1.1268=2.537 Ã…` (c) No. of atoms per unit cell `=8xx(1)/(8)+1=2` (d) Volume occupied by an atom of iron `=(4)/(3)pir^(3)`. |
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| 78. |
A liquid is transferred from a smaller vessel to a bigger vessel at the same temperature. What will be the effect on the vapour pressure? |
| Answer» No effect is depends only on the nature of the liquid and temperature. | |
| 79. |
Which of the following changes decrease the vapour pressure of water kept in a sealed vessel ?A. Decreasing the quantity of waterB. Adding salt to waterC. Decreasing the volume of the vessel to one halfD. Decreasing the temperature of water |
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Answer» Correct Answer - B::D Vapour pressure does not depend upon the quantity of water or size of the vessel. It decreases on adding salt to water or decreasing the temperature of water. |
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| 80. |
A thin tube of uniform cross-section is sealed at both ends. It lies horizontally, the middle 5 cm containing mercury and the two equal end containing air at the same pressure P. When the tube is held at an angle of `60^@` with the vetical direction, the length of the air column above and below the mercury column are 46 cm and 44.5 cm respectively. Calculate the pressure P in centimeters of mercury. (The temperature of the system is kept at `30^@C`). |
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Answer» At horizontal position, let the length of air column in tube be L cm. `:. 2L + 5 = 46 + 5 + 44.5 cm` `L = 45.25 cm` When the tube is held at `60^(@)` with the vertical, the mercury column will slip down. `P_(Y) + 5 cos 60^(@) = P_(X)` `P_(X) - P_(Y) = (5)/(2) = 2.5 cm Hg`.....(i) From end X, `P_(0) xx 45.25 = P_(X) xx 44.5` `P_(X) = (45.25)/(44.5) P_(0)`.....(ii) From end Y, `P_(0) xx 45.25 = P_(Y) xx 46` `P_(Y) = (45.25)/(46) P_(0)`.....(iii) Substituting the values of `P_(X) " and " P_(Y)` in equation (i) we get `P_(0) = 75.4` |
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| 81. |
Which of the following changes decrease the vapour pressure of water kept in a sealed vessel ?A. Decreasing the quantity of waterB. Adding salt to waterC. Decreasing the volume of the vessel to one-halfD. Decreasing the temperature of water. |
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Answer» Correct Answer - B::D are the correct answer. |
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| 82. |
Under which of the following conditions applied together, a gas deviates most from the ideal behaviour ?A. Low pressureB. High pressureC. Low temperatureD. High temperature |
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Answer» Correct Answer - B::C Deviation is maximum under high pressure and low temperature. |
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| 83. |
The absolute temperature of an ideal gas is….. to/than the average kinetic energy of the gas molecules. |
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Answer» Correct Answer - less |
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| 84. |
Point `A` in the given curve shifts to higher value of velocity if A. `T` is increasedB. `P` is decreasedC. `V` is decreasedD. Molecular weight `M` is decreased |
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Answer» Hint: Average speed `mu_(av)=sqrt((8RT)/(pi M))` `:. Mu_(av) prop (T)/(M)` |
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| 85. |
Which of the following processes would lead to an increase in the average speed of the molecules of an ideal gas system?A. Decreasing the temperature of the systemB. Compressing the gas with a pistonC. Expanding the gas into a vacuumD. Heating the system keeping `V` and `P` constant. |
| Answer» Hint: Average speed `mu_(av) prop T` | |
| 86. |
At room temperature, the following reaction proceeds nearly to completion: `2NO+O_(2)to2NO_(2)toN_(2)O_(4)` The dimer, `N_(2)O_(4)`, solidfies at `262 K`. A `250 mL` flask and a `100 mL` flask are separated by a stopcock. At `300 K`, the nitric oxide in the larger flask exerts a pressure of `1.053 atm` and the smaller one contains oxygen at `0.789 atm`. The gase are mixed by opening the stopcock and after the end of the reaction the flasks are cooled to `220 K`. Neglecting the vapour pressure of the dimer, find out the pressure and composition of the gas remaining at `220 K`. (Assume the gases to behave ideally) |
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Answer» Correct Answer - 0.221 atm |
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| 87. |
Which of the following statements is incorrect?A. Joul-Thomson coefficient is zero at inversion temperature of a real gasB. Ideal gas do not show, Joule-Thomson effectC. Inversion temperature of `H_(2)` and `He` is very -very lowD. Joule-Thomson coefficient `mu=((deltaT)/(deltaP))_(H)` |
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Answer» Correct Answer - D Joule-Thomoson coefficient `mu=((deltap)/(deltaT))_(H), mu=0` at inversion temperature. Also ideal gas does not show Joule-Thomson effect. `H_(2)` and `He` have low value of `Ti` and thus show heating effect if subjected for Joule-Thomson effect. |
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| 88. |
It is eaiser to liquefy oxygen than hydrogen because.A. oxygen has a higher `T_(c )` and lower inversion temperature `(T_(i))`B. oxygen has lower `T_(c )` and higher `T_(i)` than hydrogenC. oxygen has a high `T_(c )` and high `T_(i)` than hydrogenD. oxygen has lower `T_(c )` and low `T_(i)` than hydrogen |
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Answer» Correct Answer - C `(c )` is the correct answer because `O_(2)` has higher molecular weight than `H_(2)`, more surface area, move van der Waals forces of attraction, hence has higher `T_(C )` and higher `T_(i)` (inversion temperature). |
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| 89. |
The equation of `(P_(c )V_(c ))/(RT_(c ))` = .......... |
| Answer» `(P_(c )V_(c ))/(RT_(C )) = (3)/(8)` | |
| 90. |
For which of the following gas/gases, `(P_(C )V_(C ))/(RT_(C ))` close to `0.22`?A. `Cl_(2)`B. `CH_(3)OH`C. `C_(2)H_(4)`D. `CH_(4)` |
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Answer» Correct Answer - B Molecule having hydrogen bonding values of `(P_(C )V_(C ))/(T)` close to `0.22` |
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| 91. |
Calculate the volume occupied by `2.0 mol` of `N_(2)` at `200 K` and `10.1325MPa` pressure if `(P_(c)V_(c))/(RT_(c))=(3)/(8)` and `(P_(r)V_(r))/(T_(r))=2.21`. |
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Answer» Since `(P_(c)V_(c))/(RT_(c)) =(3)/(8)` and `(P_(r)V_(r))/(T_(r))=2.21` `(P_(c)V_(c))/(RT_(c)) xx (P_(r)V_(r))/(T_(r))=(3)/(8)xx2.21` or `(PV_(m))/(RT)=(3)/(8)xx2.21` `:. V_(m)=((3)/(8)xx2.21)(RT)/(P)` `=((3)/(8)xx2.21)((8.314MPa cm^(3)K^(-1)mol^(-1))(200 K))/((10.1325MPa))` `=136.0cm^(3)mol^(-1)` Hence, the volume of `2 mol` of `N_(2)` is `(2.0 mol)(136.0 cm^(3)mol^(-1))=272.0 cm^(3)` |
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| 92. |
Amorphous solids may be considered as:A. supercooled liquidsB. supercooled solidsC. molecular solidsD. covalent network solids |
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Answer» Correct Answer - A Amorphous solids like glass and plastic are considered as supercooled liquids. |
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| 93. |
At `27^(@)C` and one bar pressure a has volume `VL`. What will be its volume at `177 ^(2)C` and under pressure of `1.5"bar"`?A.B.C.D. |
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Answer» Correct Answer - VL `V_(1) , = VL , V_(2) , = ?` `P_(1) = 1 "bar" , P_(2) = 1.5 "bar"` According to ideal gas equation, `(P_(1)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2))` or `V_(2) = (P_(1)V_(1)T_(2))/(T_(1)P_(2)) = ((1 "bar") xx (VL) xx (450 K))/((300 K) xx (1.5 K)) = VL` |
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| 94. |
A gas cylinder containing cooking gas can withsand a pressure of 14.9 atmosphere. The pressure gauge of the cylinder indicates 12 atmosphere at `27^(@)C`. Due to a sudden fire in the building the temperature starts rising. At what temperature will the cylinder explode ? |
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Answer» Since, the gas is confined in a cylinder, its volume will remain constant `{:("Initial conditions","Final conditions",),(P_(1) = 12 atm,P_(2) = 14.9 atm,),(T_(1) = 27 + 273 = 300 K,T_(2) = ?,):}` Applying pressure - temperature law, `(P_(1))/(T_(1)) = (P_(2))/(T_(2))` So, `T_(2) = (P_(2) xx T_(1))/(P_(1)) = (14.9 xx 300)/(12) = 372.5 K` Temperature in `.^(@)C = (372.5 - 273) = 99.5^(@)C` |
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| 95. |
A `10 L` flask at `298 K` contains a gaseous mixture of `CO` and `CO_(2)` at a total pressure of `2.0 "bar"` if `0.20 ` mole of `CO` is present, find its partial pressure and also that of `CO_(2)`. |
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Answer» Correct Answer - `p_(CO) = 0.495 "bar", p_(CO_(2)) = 1.505 "bar"` Partial pressure of `CO (pCO) = ((.^(n)CO)RT)/(V) = ((0.2 mol) xx (0.083 L "bar" K^(-1) mol^(-1)) xx (298 K))/((10 L)) = 0.495 "bar"` Partial pressure of `CO_(2) (pCO) = P - pCO = 2 - 0.495 = 1.505 "bar"`. |
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| 96. |
A gas jar of 10 litre volume filled with `O_(2)` at 300 K is connected to glycerine manometer. The maonometer shows 5 m difference in the level as shown in figure. What will be the number of moles of `O_(2)` in the gas jar ? (Given `d_("glycerine") = 2/72 g // mL , d_("mercury") = 13.6 g//mL`)A. `0.64 mol`B. `0.4 mol`C. `0.94 mol`D. `0.36 mol` |
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Answer» Correct Answer - C `(h xx d xx g)_(Hg) = (h xx d xx g)_("glycerine")` `(h xx 13.6)_(Hg) = (5 xx 2.72)_("glycerine")` `h_(Hg) = 1m` `P_("gas") = (1 + 0.76) m` `= 1760 mm Hg` `PV = nRT` `(1760)/(760) xx 10 n xx 0.0821 xx 300` `n = 0.94 mol` |
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| 97. |
Ashutosh was getting late for the office. He tried to sip boiling hot coffee from a cup. He felt very uncomfertable. His wife immediately brought a plate and asked him to sip the coffee from the plate. Ashutosh followed the advice and did not face any problem. (a) Why was Ashutosh feeling uncomfortable(b) How did sipping coffee from a plate was more comfertable. (c) how did wife help Ashutosh ? (d) what is the value associated with it? |
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Answer» (a) Since coffee was very hot, it was not possible for Ashutosh to sip it hurriedly. It could burn its tongue. (b) A plate has more surface area than the cup. Evaporation becomes faster in this case. Since cooling is always caused during evaporation, the temperature got lowered. It became comfortable to sip coffee from a plate. (c ) With the timely help by his wife, Ashutosh could sip coffee quite comfortably and could reach office in time. (d) We shold never drink milk, tea or coffee when these are boiling hot,. These are injurious to our throat. It is always better to allow them to cool. In case allow them to cool. In case we are in a hurry, we place them in a container with large surface area. |
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| 98. |
Mohan was asked by his teacher to determine the boiling point of a liquid. He set up the appartus as shown in the figure. He rocorded the boiling point of the liquid as `140^(@)C` and reported the reading to this teacher. The teacher asked him to repeat the experiment after dipping the bulb of the thermometer in the liquid and also placing the beaker on a tripod stand covered with a wire gauze. After reading this narration, answer the following question : (i) Was the reading taken by the student correct ? (ii) What was the necessity of wire gauze? (iii) How did teacher help the student? (iv) What is teh value based information associated with this? |
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Answer» (i) No, it was wrong, The bulb of the thermometer must dip in the liquid. (ii) If the beaker is in direct contact with the flame, it might crack. (iii) Teacher guided the student properly. By direct heating, an accident could occur and could cause serious injuries to the student. (iv) In the laboratory or in the kitchan, we should never heat glass beakers or containers directly on the flame. These are likely to crack. However, Borocil beakers can be heated on direct flame. |
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| 99. |
Why does the boiling point of liquid rise on increasing pressure ? |
| Answer» Boiling point of a liquid is the temperat,ure when its vapour pressure becomes equal to the atmospheric pressure. When pressure is increased, the vapour pressure of the liquid has to be increased to mathc the external pressure. Therefore, the liquid has to be heated or its boiling point increases. | |
| 100. |
Water boils at lower temperature on high altitude because:A. atomspheric pressure is low thereB. atomospheric pressure is high thereC. water is weakly hydrogen bonded thereD. water in pure form is found there |
| Answer» Correct Answer - A | |