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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.
| 1001. |
If the ration of the masses of `SO_(3)` and `O_(2)` gases confined in a vessel is `1 : 1` , then the ratio of their partial pressure would beA. `5 : 2`B. `2 : 5`C. `2 : 1`D. `1 : 2` |
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Answer» Correct Answer - B Let m be the mass of `SO_(3)` and `O_(2)` enclosed in the vessel. Number of moles of `SO_(3)=(m)/(80)` Number of moles of `O_(2)=(m)/(32)` Partial pressure of `SO_(3), P_(A)=(m)/(80)xx(R xx T)/(V)` Partial pressure of `O_(2), P_(B)=(m)/(32)xx(R xx T)/(V)` Now, `(P_(A))/(P_(B))=(m)/(80)xx(32)/(m)=(2)/(5)` Hence, ratio of partial pressure of m g of `SO_(3)` and `O_(2)` is 2 : 5. |
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| 1002. |
What is the molecular weight of a gas whose density `40^(@)C and 785 mm` of `Hg` pressure is `1.3 g L^(-1)`?A. `32`B. `40`C. `15`D. `98` |
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Answer» Correct Answer - A We know that `PV=nRT` `PV=(omega)/(M)RT or P=(omega)/(V)xx(1)/(M)RT` Or `P=(d)/(M)RT` Now, `(785)/(760)=(1.3)/(M)xx0.0821xx313` or `M= 32.4 or 32` |
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| 1003. |
The pressure `p` of a gas is plotted against its absolute temperature `T` for two different constant volumes, `V_(1)` and `V_(2)` when `V_(1) gt V_(2)`, theA. curves have the same slope and do not intersectB. curves must intersect at some point other than `T=0`C. curves for `V_(2)` has a greater slope than that for `V_(1)`D. curve for `V_(1)` has a greater slope than that for `V_(2)`` |
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Answer» Correct Answer - C At constant volumes `P prop T` `P= "constant" T, PV= nRT :. P=(nR)/(V)T` Slope `=m=(nR)/(V) :. V_(2) lt V_(1)` `(m_(1))/(m_(2))= (V_(2))/(V_(1))` is curve for `V_(2)` has a greater slope than for `V_(1)` |
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| 1004. |
Virial equation is: `PV_(M)=RT[A+(B)/(V_(M))+(C )/(V_(M^(2)))+…]`, where `A`, `B`, `C`, …. are first second,third, … virial coefficent, respectively, For an ideal gasA. `A=` unity and `B,C` are zero.B. `A,B,C` are all equal to unityC. `A` is dependent of temperatureD. All `A,B,C` depend on temperature. |
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Answer» Correct Answer - A `PV= RT` for ideal gases. |
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| 1005. |
Assertion: Pressure exerted by a mixture of gases is equal to the sum of their partial pressure. Reason: Reacting gases react to form a new gas having pressure equal to the sum of both.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 - C New gas may have pressure equal to sum of both reacting gases or less more depending on the reacting gases and product Formed. |
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| 1006. |
Calculate the total pressure in a mixture of `4 g` of oxygen and `2 g` of hydrogen confined in a total volume of `1 L` at `0^(@)C`. |
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Answer» Correct Answer - `25.49 "bar"` No. of mass of oxygen `= ("Mass")/("Molar mass ") = 4/32 = 0.125 ` mol No. of moles of hydrogen `= ("Mass")/("Molar mass") = (2)/(2) = 1.0` mol Total no. of moles `= 0.125 + 1= 1.125` mol. According to ideal gas equation , `PV = nRT` `n = 1.125 mol, V = 1L, T = 0 + 273 = 273 K, R = 0.083 L "bar" K^(-1) mol^(-1)` `:. P = (nRT)/(V) = ((1.125 mol) xx (0.083 L "bar" K^(-1) mol^(-1)) (273K))/((1 L)) = 25.49 "bar"` |
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| 1007. |
What is the presser exerted by a mixture of `3.5 g` of nitrogen and 2 g of helium when confined in a vessel of volume `40` lires at `0^(@)C`? |
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Answer» Correct Answer - `0.354 "bar"` Moles of `N_(2) (.^(n)N_(2)) = ("Mass of" N_(2))/("Molar mass") = ((3.5 g))/((28 g mol^(-1))) = 0.125` mol Moles of `He (.^(n)He) = ("Mass of" He)/("Molar mass") = ((2.0 g))/((4 g mol^(-1))) = 0.5` mol. Partial pressure of `N_(2) (.^(n)N_(2)) = (.^(n)N_(32)RT)/(V) = ((0.125 mol) xx (0.083 L "bar" K^(-1) mol^(-1)) xx (273 K))/((40 L)) = 0.071 "bar"`. Partial pressure of `He (.^(p)He) = (.^(n)HeRT)/(V) = ((0.5 mol) xx (0.083 L "bar"K^(-1) mol^(-1)) xx (273 K))/((40 L)) = 0.283 "bar"`. Total pressure of gaseous mixture ` = 0.071 + 0.283 = 0.354 "bar"` |
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| 1008. |
Why at one time gases like helium, hydrogen and nitrogen were called permanent gases bt not now ? |
| Answer» These gases were called permanent gases because they could not be liquefied at room temperature by applying any amount of pressure on them. Now, they are not called so because they can be liquefied by first cooling them to a temperature below a particular temperature called critical temperature of the gas then applying suitable pressure on the gas. | |
| 1009. |
Equal masses of `H_(2)O_(2)` and methane have been taken in a container of volume V at temperature `27^(@)C` in identical conditions. The ratio of the volumes of gases `H_(2) : O_(2)` : methane would beA. `8:16:1`B. `16:8:1`C. `16:1:2`D. `8:1:2` |
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Answer» Correct Answer - C Suppose mass of each of `H_(2), O_(2)` and `CH_(4)` taken =w g `w g H_(2)=(w)/(2)"mole", w g O_(2)=(w)/(32)"mole"`, `w g CH_(4)=(w)/(16)"mole"` As under identical conditions, volume are in the ratio of moles. Hence, volume of `H_(2) : O_(2): CH_(4)=(w)/(2) : (w)/(32) : (w)/(16)` `=(1)/(2) : (1)/(32) : (1)/(16)=16 : 1 : 2` |
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| 1010. |
At `100^(@)C` and 1 atm, if the density of the liquid water is `1.0 g cm^(-3)` and that of water vapour is `0.0006 g cm^(-3)` , then the volume occupied by water molecules in `1 L` steam at this temperature isA. `6 c c`B. `60 c c`C. `0.6 c c`D. `0.06 c c` |
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Answer» Correct Answer - C Mass of `1L` water water vapour `=Vxxd= 1000xx0.0006= 0.6g` `:.` volume of liquid water `=(0.6)/(1)= 0.6 c c` |
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| 1011. |
In the temperature of 1 mole of a gas is increased by `50^(@)C`. Calculate chagne in kinetic energy:A. `62.32 J`B. `6.235 J`C. `623.5 J`D. `6235.0 J` |
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Answer» Correct Answer - C `E = (3)/(2) RT` (Kinetic energy of 1 mole gas) `Delta E = (3)/(2) R (T + 50) - (3)/(2) RT` `= (3)/(2) R xx 50 = (3)/(2) xx 8.314 xx 50 = 623.55 J` |
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| 1012. |
Assertion : Compressibility factor for ideal gas is one. Reason : For an ideal gas `PV = nRT` equation is obeyed.A. If both assertion and reason are correct and reason is correct explanation for assertionB. If both assertion and reason are correct and reason is not correct explanation for assertionC. If assertion is correct but reason is incorrectD. If assertion and reason are both incorreect. |
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Answer» Correct Answer - A Reason is the correct explanation for assertion. |
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| 1013. |
What would have happened to the pressure of a gas if the collisions of its molecules had not been elastic? |
| Answer» There might have been decrease in the kinetic energy during the collisions because energy is likely to decrease in each collision under the circumstance. Since the collisions are very frequent, the energy of the molecules might have considerably decreased and ultimately the gas molecules might have come to rest. | |
| 1014. |
The kinetic theory of gases presumes the collisions between the molecules to be perfectly elastic becauseA. occur in a zig - zag pathB. occur in a straight lineC. change velocity and energyD. result in settling down of molecules. |
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Answer» Correct Answer - B Particles of a gas collide in a straight line in all possible directions. |
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| 1015. |
The compressibility factor for an ideal gas isA. `1.5`B. `1.0`C. `2.0`D. `infty` |
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Answer» Compressibility factor, `Z=(PV)/(nRT)` For ideal gases, `PV=nRT`, so `Z=1`. |
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| 1016. |
The kinetic theory of gases presumes the collisions between the molecules to be perfectly elastic becauseA. collisions will not split the moleculesB. the molecules are tinyC. the molecules are rigidD. the temperature remains constant irrespective of collisions |
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Answer» Correct Answer - B If the molecules are tiny , the energy and momenta of the molecules are conserved during the collisions. Consequently , the temperature remains constant irrespective of collisions. |
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| 1017. |
Two flasks `A` and `B` have equal volume. `A` is maintained at `300 K` and `B` at `600 K`. While `A` contains `H_(2)` gas, `B` has an equal mass of `CH_(4)` gas. Assuming ideal behaviour for both the gases, which of the following statement is true about the velocities of molecules?A. The molecules in flasks `A` and `B` are moving with the same veocity.B. The molecules in flask `A` are moving two times faster than the molecules in flask `B`.C. The molecules in flask `B` are moving two times faster than the molexules in flask `A`.D. The molecules in flask `A` are moving four times faster than the molecules in flask `B`. |
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Answer» `KE` per mole `=(3)/(2)RT` ` :. KE prop T` `(KE_(1))/(KE_(2))=(T_(A))/(T_(B))=(100)/(200)=(1)/(2)` |
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| 1018. |
Two flasks `A` and `B` have equal volume. `A` is maintained at `300 K` and `B` at `600 K`. While `A` contains `H_(2)` gas, `B` has an equal mass of `CH_(4)` gas. Assuming ideal behaviours for both the gases, answer the following: Flask in which molecules are moving fasterA. `A`B. `B`C. Both `A` and `B`D. None |
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Answer» Since `(mu_(av))_(A)=sqrt((8RT_(A))/(pi M_(A)))` `(mu_(av))_(B)=sqrt((8RT_(B))/(pi M_(B)))` Hence `((mu_(av))_(A))/((mu_(av))_(B))=sqrt((T_(A))/(M_(A)))xxsqrt((M_(B))/(T_(B)))=sqrt((300)/(2))xx(16)/(600)=2` `:. (mu_(av))_(A)gt(mu_(av))_(B)` |
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| 1019. |
Two flasks `A` and `B` have equal volume. `A` is maintained at `300 K` and `B` at `600 K`. While `A` contains `H_(2)` gas, `B` has an equal mass of `CH_(4)` gas. Assuming ideal behaviours for both the gases, answer the following: Flask with greater molar kinetic energyA. `A`B. `B`C. Both `A` and `B`D. None |
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Answer» Since `KE per mole=(3)/(2)RT`, `KE per mole of A=(3)/(2)RT_(A)` `KE per mole of B=(3)/(2)RT_(B)` Hence `(KE per mole of A)/(KE per mole of B)=(T_(A))/(T_(B))=(300)/(600)=(1)/(2)` `:. KE per mole of Alt KE per mole of B` |
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| 1020. |
Two flasks `A` and `B` have equal volume. `A` is maintained at `300 K` and `B` at `600 K`. While `A` contains `H_(2)` gas, `B` has an equal mass of `CH_(4)` gas. Assuming ideal behaviours for both the gases, answer the following: Flask in which pressure is higherA. `A`B. `B`C. Both `A` and `B`D. None |
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Answer» Since `P_(A)V=n_(A)RT_(A)` and `P_(B)V=n_(B)RT_(B)` `:. (P_(A))/(P_(B))=(n_(A)T_(A))/(n_(B)T_(B))=8xx(300)/(600)=4` `:. P_(A)gtP_(B)` or `P_(A)=4P_(B)` |
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| 1021. |
The ratio between the root mean square speed of `H_(2)` at `50 K` and that of `O_(2)` at `800 K` isA. `4`B. `2`C. `1`D. `1//4` |
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Answer» `v_(rms)=sqrt((3RT)/(M))` For `H_(2)` at `50 K`, `v_(1)=sqrt((3Rxx50)/(2))` For `O_(2)` at `800 K`, `v_(2)=sqrt((3Rxx800)/(32))` `:. (v_(1))/(v_(2))=(sqrt((3Rxx50)/(2)))/(sqrt((3Rxx800)/(32)))=1` |
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| 1022. |
At constant volume, for a fixed number of moles of a gas, the pressure of the gas increases with the rise in temperature due toA. Increase in average molecular speedB. Increase in the rate of collisions among the moleculesC. Increase in the molecular attractionD. Decrease in the mean free path |
| Answer» Pressure on the walls of the container is equal to the change of momentum per unit time per unit area. At constant volume, for a fixed number of moles of a gas, the pressure increases with rise in temperature due to increase in average molecular speed. This increases the change in momentum during collisions. | |
| 1023. |
Equal weights of methane and hydrogen are mixed in an empty container at `25^(@)C`. The fraction of the total pressure exerted by hydrogen isA. `1//2`B. `8//9`C. `16//19`D. `1//9` |
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Answer» Correct Answer - B `P_(n_(2)) prop n_(H_(2)), P_(t otal) prop(n_(H_(2))+n_(CH_(4)))` `(P_(H_(2)))/(P_(t otal))=(w//2)/(w//2+w//16)implies (8)/(9)` |
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| 1024. |
Equal weights of methane and hydrogen are mixed in an empty container at `25^(@)C`. The fraction of the total pressure exerted by hydrogen isA. `8//9`B. `1//9`C. `16//17`D. `1//2` |
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Answer» Correct Answer - A The fraction of the total pressure exerted by `H_(2)` gas is gieven by `p_(f) = (p_(H_(2)))/(p_("total"))` which is equal to the mole freaction of `H_(2)` gas . Let us assume that `"Mass"_(H_(2)) = "Mass"_(CH_(4)) = mg` Thus , `n_(H_(2)) = ("Mass"_(H_(2)))/("Molar mass" _(H_(2))) = (mg)/(2g "mol"^(-1))` `n_(CH_(4)) = ("Mass"_(CH_(4)))/("Molar mass" _(CH_(4))) = (mg)/(16g "mol"^(-1))` `chi_(H_(2)) = (n_(H_(2)))/(n_(H_(2)) + n_(CH_(4))) = (m//2)/((m)/(2) + (m)/(16))` `= (m//2)/(9m//16) = (m)/(2) xx (16)/(9 m) = (8)/(9)` |
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| 1025. |
Two flasks `A` and `B` have equal volume. `A` is maintained at `300 K` and `B` at `600 K`. While `A` contains `H_(2)` gas, `B` has an equal mass of `CH_(4)` gas. Assuming ideal behaviours for both the gases, answer the following: Flask in which the compressibility factor is greaterA. `A`B. `B`C. Both `A` and `B`D. None |
| Answer» Both the gases will have same cpmpressibility factor, i.e., `1`, since both of them are ideal gases. | |
| 1026. |
Two flasks `A` and `B` have equal volume. `A` is maintained at `300 K` and `B` at `600 K`. While `A` contains `H_(2)` gas, `B` has an equal mass of `CH_(4)` gas. Assuming ideal behaviours for both the gases, answer the following: Flask in which the total kinetic energy is greaterA. `A`B. `B`C. Both `A` and `B`D. None |
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Answer» Since total `KE=KExx` Number of molecule `:. Total(KE)_(A)=(3)/(2)kT_(A)N_(A)` Total `(KE)_(B)=(3)/(2)kT_(B)N_(B)` Hence `(Total (KE)_(A))/(Total(KE)_(B))=(T_(A)N_(A))/(T_(B)N_(A))=(300)/(600)xx8=4` `:. Total (KE)_(A)gtTotal(KE)_(B)` |
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| 1027. |
A gas occupies 100.0 mL at `50^(@)C` and I atm pressure. The gas is cooled at constant pressure so that its volume is reduced to 50.0 mL. what is the final temperautre? |
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Answer» Given `V_(1)=100mL` `T_(1)=50^(@)C` `P_(1)=1atm` `V_(2)=50mL` `T_(2)=?` `T_(2)=V_(2)T_(1)//V_(1)=-111.5^(@)C |
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| 1028. |
500 mL of nitrogen at `27^(@)C` is cooled to `-5^(@)C` at the same pressure. The new volume becomesA. 326.32 mLB. 446.66 mLC. 546.32 mLD. 771.56 mL |
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Answer» Correct Answer - B Initial volume, `V_(1)=500" mL"` Initial temperature, `T_(1)=27^(@)C=27+273=300K` Final temperature, `T_(2)=-5+273=268K` `V_(2)=?` `(V_(1))/(T_(1))=(V_(2))/(T_(2))` `V_(2)=(V_(1)T_(2))/(T_(1))=(500xx268)/(300)` `=446.66" mL"` |
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| 1029. |
`300 ml` of a gas at `27^(@)C` is cooled to `-3^(@)C` at constant pressure, the final volume isA. `540 ml`B. `135 ml`C. `270 ml`D. `350 ml` |
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Answer» Correct Answer - C `V_(2)=(T_(2))/(T_(1))implies V_(1)=(270K)/(300K)xx300ml= 270ml` |
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| 1030. |
A mixture of gases contains `H_(2)` and `O_(2)` gases in the ratio of `1:4 (w//w)`. What is the molar ratio of the two gases in the mixture?A. `1:4`B. `4:1`C. `16:1`D. `2:1` |
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Answer» Correct Answer - B Let the mass of `H_(2)` gas be xg and mass of `O_(2)` gas 4xg `{:("Molar",H_(2),:,O_(2)),("mass",2,:,32),("i.e.",1,:,16):}` `therefore" Molar ratio"=(n_(H_(2)))/(n_(O_(2)))=(x//2)/(4x//32)=(x xx32)/(2xx4x)=(4)/(1)=4:1` |
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| 1031. |
The vacant space in bcc lattice unit cell isA. `26%`B. `48%`C. `23%`D. `32%` |
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Answer» Correct Answer - D `because" Packing efficiency in bcc lattice"=68%` `therefore" Vacant space in bcc lattice"` `=100-68=32%` |
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| 1032. |
Sodium metal crystllizes in a body centred cubic lattice with a unit cell edge of `4.29 Å`. The radius of sodium atom is approximatelyA. `0.93 Ã…`B. `1.86 Ã…`C. `3.22 Ã…`D. `5.72Ã…` |
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Answer» Correct Answer - B For body centred cubic unit cell : `asqrt(3)=4r` `r=(4.29xx1.732)/(4)=1.857=1.86 Ã…` |
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| 1033. |
A given metal crystalline out with a cubic structure having edge length of `361` pm .if there are four metal atoms in one unit cell, what is the radius of metal atom?A. 40 pmB. 127 pmC. 80 pmD. 108 pm |
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Answer» Correct Answer - B Given, edge length = 361 pm Four metal atoms in one unit cell i.e. effective number in unit cell (z) = 4 (given) `therefore" It is a FCC structure"` `therefore"Face diagonal"=4r` `sqrt2a=4r` `r=(sqrt2xx361)/(4)=127" pm"` |
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| 1034. |
Two flasks A and B have equal volume. Flask A contains hydrogen at 300 K while flask B has an equal mass of `CH_(4)` at 600 K. which flask contains larger number of moleculars? b. In which flask is the pressure greater and by how many times? |
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Answer» a. A has larger number of molecules b. pressure is higher in B c. molecules move faster in B d. B. |
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| 1035. |
Two flasks `A` and `B` have equal volumes. Flask `A` containing `H_(2)` gas is maintained at `27^(@)C` while `B` containing an equal mass of `C_(2)H_(6)` gas is maintained at `627^(@)C`. In which flask and by how many times are molecules moving faster, assuming ideal behaviour for both the gases? |
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Answer» Calculate the relative velocities of molecules Average velocity of a gas `mu=sqrt((8RT)/(pi M))` `mu_(H_(2))=sqrt((8RT_(1))/(pi M_(1)))`, `mu_(C_(2)H_(6))=sqrt((8RT_(2))/(pi M_(2)))` `(mu_(H_(2)))/(mu_(C_(2)H_(6)))=sqrt((T_(1)M_(2))/(T_(2)M_(1)))` [`T_(1)=300K`, `T_(2)=900 K`] `M_(1)=2g mol^(-1)` `M_(2)=30 g mol^(-1)` `(mu_(H_(2)))/(mu_(C_(2)H_(6)))=sqrt((300xx30)/(900xx2))=sqrt((5))/(1)=2.237:1` Thus, `H_(2)` molecules in flask `A` will be moving `2.237` times faster than `C_(2)H_(6)` molecules in `B`. |
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| 1036. |
Two flasks `A` and `B` of equal volume containing equal masses of `H_(2)` and `CH_(4)` gases are at `100 K` and `200 K` temperature, respectively. Which of the following is true about the total `KE` (Kinetic energy)?A. Total `KE` of `H_(2)` is four times that of `CH_(4)`.B. Total `KE` of `CH_(4)` is four times that of `H_(2)`.C. Total `KE` of `H_(2)` is two times that of `CH_(4)`.D. Total `KE` of `CH_(4)` is two times that of `H_(2)`. |
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Answer» Total `KE=Koverline(E)(per mole)xxNumber of molecules` Total `KE_(A)=(3)/(2)KT_(A)N_(A)` `n_(A)=(m)/(2)`, `n_(B)=(m)/(16) :. (n_(A))/(n_(B))=8` `:. (Total KE_(A))/(Total KE_(B))=(T_(A)N_(A))/(T_(B)N_(B))=(1)/(2)xx8=4` |
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| 1037. |
Which of the following FCC structure contains cations in alternate tetrahedral voids ?A. NaClB. ZnSC. `Na_2O`D. `CaF_2` |
| Answer» Correct Answer - b | |
| 1038. |
Which of the following solids is not an electrical conductor ? (A) Mg(s), (B)TiO(s) , (C )`I_2`(s) , (D)`H_2O` (s)A. (A) onlyB. (B) onlyC. (C ) and (D)D. (B) , (C ) and (D) |
| Answer» Correct Answer - c | |
| 1039. |
The sharp melting point of crystalline solids is due to ______A. a regular arrangement of constituent particles observed over a short distance in the crystal lattice.B. a regular arrangement of constituent particles observed over a long distance in the crystal lattice.C. same arrangement of constituent particles in different directionsD. different arrangement of constituent particles in different directions |
| Answer» Correct Answer - b | |
| 1040. |
Which of the following statement is not true about amorphous solids ?A. On heating they may become crystalline at certain temperatureB. They may become crystalline on keeping for long timeC. Amorphous solids can be moulded by heatingD. They are anisotropic in nature |
| Answer» Correct Answer - d | |
| 1041. |
Which of the following is not the charcteristic of ionic solids ?A. Very low value of electrical conductivity in the molten state.B. Brittle nature .C. Very strong forces of interactionsD. Anisotropic nature |
| Answer» Correct Answer - a | |
| 1042. |
In an ionic compound `A^+ B^-`, radius of `A^+` 88 pm while that of `B^-` is 200 pm. The coordination number `A^+` will be ______ |
| Answer» 6(`r_+ //r_-`=0.44 which lies in the range 0.414-0.732) | |
| 1043. |
Which of the following represents correct order of conductivity in solids ?A. `K_"metals" gt gt K_"insulators" lt K_"semiconductors"`B. `K_"metals" lt lt K_"insulators" lt K_"semiconductors"`C. `K_"metals" , K_"semiconductors" gt K_"insulators" = zero`D. `K_"metals" lt K_"semiconductors" gt K_"insulators" ne zero` |
| Answer» Correct Answer - a | |
| 1044. |
Which of the following is true about the charge acquired by p-type semiconductors ?A. positiveB. neutralC. negativeD. depends on concentration of p impurity |
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Answer» Correct Answer - b All p-type semiconductors are neutral.p only indicates that holes move like positive charge , i.e., towards cathode. |
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| 1045. |
The electrical conductivity of semiconductors ___________ with increase of temperature |
| Answer» Correct Answer - increases | |
| 1046. |
Assertion: Semiconductors are solids with conductivities in the intermediate range from `10^(-6) -10^4 "ohm"^(-1) m^(-1)` Reason : Intermediate conductivity in semiconductor is due to partially filled valence band .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. Intermediate conductivity of semiconductors is due to small energy gap between filled valence band and empty conduction band. |
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| 1047. |
Which of the following defects decrease the density ?A. Interstitial defectB. Vacancy defectC. Frenkel defectD. Schottky defect |
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Answer» Correct Answer - b,d In vacancy defect and Schottky defect, some lattice sites are vacant. Thus, mass decreases but volume remains the same . Hence , density decreases. |
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| 1048. |
No crystal is found to be prefect at room temperature . The defects present in the crystals can be stoichiometric or non-stoichiometric. Due to non-stoichiometric defects, the formula of the ionic compound is different from the ideal formula. For example , the ideal formula of ferrous oxide should be FeO but actually in one sample , it was found to be `Fe_0.93 O`. This is because the crystal may have some ferric ions in place of ferrous ions. These defects change the properties of the crystals. In some cases , defects are introduced to have crystals of desired properties as required in the field of electronics . Doping of elements of Group 14 with those of Group 13 or 15 is most common. In ionic compounds , usually impurities are introduced in which the cation has higher valency than the cation of the parent crystal , e.g. of `SrCl_2` into NaCl Which one of the following doping will produce p-type semiconductor ?A. Silicon doped with arsenicB. Germanium doped with phosphorusC. Germanium doped with aluminiumD. Silicon doped with phosphorus |
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Answer» Correct Answer - C Germanium belongs to Group 14 while aluminium belongs to Group 13. Hence, germanium doped with aluminium will produce p-type semiconductor . |
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| 1049. |
In a face centred cubic lattice unit cell is shared equally by how many unit cells?A. 8B. 4C. 2D. 6 |
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Answer» Correct Answer - D In a face centred cubic (fcc) lattice , a unit cell is shared equally by six unit cells. |
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| 1050. |
Surface tension vanishes atA. critical pointB. triple pointC. boiling pointD. condensation point |
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Answer» Correct Answer - A Surface tension of a liquid is the amount of enerf=gy required to stretch or increase the surface by unit area . It becomes zero at the critical temperature because at this temperature , the meniscus (surface) between the liquid and its vapor disappears. |
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