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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.
| 851. |
A gas such as carbon monoxide would be most likely to obey the ideal gas law atA. low temperature and high pressureB. high temperature and high pressure.C. low temperature and low pressureD. high temperature and low pressure |
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Answer» Correct Answer - D The deviation of real gases form ideal gas behaviour is minimum when the temperature is high and pressure is low. |
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| 852. |
Dominance of strong repulsive forces among the molecules of the gas (`Z =` compressibility factor)A. depends on `Z` and is indicated by `Z = 1`B. depends on `Z` and is indicated by `Z gt 1`C. depends on `Z` and is indicated by `Z lt 1`D. is independent on `Z` |
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Answer» Correct Answer - B `Z gt 1` implies that the gas is less compressible than the ideal gas , i.e., when we compress the gas , the decrease in volume is less than expected. This is because of the dominance of strong repulsive forces between the molecules preventing them to come closer. |
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| 853. |
`50 mL` of each gas `A` and of gas `B` takes `150 and 200` seconds respectively for effusing through a pin hole under the similar conditon. If molecular mass of gas `B` is `36`, then the molecular mass of gas `A` will beA. `96`B. `128`C. `32`D. `64` |
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Answer» Correct Answer - C `(V_(A))/(t_(B))//(VB)/(t_(B))=sqrt((M_(B))/(M_(A)))` `implies (200)/(150)=sqrt((36)/(M_(A)))implies(4)/(3)=sqrt((36)/(M_(A)))` `implies(16)/(9)=(36)/(M_(A))implies M_(A)=(81)/(4)=20.25` So, most approximate answer is `32`. |
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| 854. |
Dominance of strong repulsive forces among the molecules of the gas (`Z =` compressibility factor)A. depends on Z and indicates that `Z = 1`B. depends on Z and indicates that `Z gt 1`C. depends on Z and indicates that `Z lt 1`D. is independent of Z. |
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Answer» Correct Answer - B Because of strong repulsinve forces, molecules of a gas can not be compressed and `Z gt 1`. |
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| 855. |
A gas volume `100 cc` is kept in a vessel at pressure `10.4 Pa` maintained at temperature `24^(@)C`. Now, if the pressure is increased to `105Pa`, keeping the temperature constant, then the volume of the gas becomesA. `10cc`B. `100cc`C. `1 cc`D. `1000 cc` |
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Answer» Given `V_(1)=100 mL` `P_(1)=10.4 Pa` `V_(2)=?` `P_(2)=105 Pa` ` P_(1)V_(1)=P_(2)V_(2)` `:. V_(2)=(P_(1)V_(1))/(P_(2))=(105xx100)/(10.4)=10 mL` |
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| 856. |
It is desired to increase of the volume of a gas by 20% without changing the pressure. To what temperature, the gas must be heated if the initial temperature of the gas is `27^(@)` ? |
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Answer» Correct Answer - C Suppose volume of gas at `27^(@)C=V cm^(3)` Increase in volume desired =20% of `V=(20)/(100)xxV=0.2 V :.` Final volume =V+0.2 V=1.2 V Now, `V_(1)=V cm^(3),T_(1)=300 K, V_(2)=1.2 V,T_(2)`=? At constant P, `(V_(1))/(T_(1))=(V_(2))/(T_(2)):.(V)/(300)=(1.2 V)/(T_(2))" or T_(2)=360 K=360-273^(@)C=87^(@)C`. |
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| 857. |
A spherical ballon of `21 cm` diameter is to be filled with hydrogen at `STP` from a cylinder containing the gas at `20 atm` and `27^(@)C` . If the cylinder can hold `2.82 L` of water, calculate the number of balloons that can be filled up . |
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Answer» Correct Answer - 10 |
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| 858. |
It is desired to fill a cylinder of `1 L` capacity at `82 atm` and `27^(@)C` with hydrogen. What will be the density of the hydrogen in the cylinder? What will be the volume of hydrogen under standard conditions of temperature and pressure? |
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Answer» Given `V_(1)=1 L`, `V_(2)=?` `P_(1)=82 atm`, `P_(2)=1 atm` `T_(1)=27^(@)C` or `300.15 K`, `T_(2)=0^(@)C` or `273.15 K` `density=?` `:. V_(2)=(P_(1)V_(1))/(T_(1))xx(T_(2))/(P_(2))=(82xx1)/(300.15)xx(273.15)/(1)=74.6 L` |
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| 859. |
The rate of diffusion of methane is twice that of `X`. The molecular mass of `X` is divided by `32`. What is value of `x` is ?A. `1`B. `2`C. `3`D. `4` |
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Answer» `r_(CH_(4))=2r_(X)`, we know `r=sqrt((1)/("Molecular mass"))` `:. (r_(CH_(4)))/(r_(X))=sqrt((M_(X))/(M_(CH_(4))implies (2r_(X))/(r_(X))=sqrt((M_(X))/(16))` or `M_(X)=64` Given Molecular mass is divided by `32` therefore, `X=(M_(X))/(32)=(64)/(32)=2` |
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| 860. |
A quantity of gas is collected in a graduated tube over the mercury. The volume of gas at `18^(@)C` is 50 mL and the level of mercuty in the tube is 100mm above the outside mercuty level. The barometer reads 750 torr. Hence, volume of gas 1 atm and `0^(@)C` is approximately:A. 22 mlB. 40 mlC. 20 mlD. 44 ml |
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Answer» Correct Answer - B Under given condition, pressure=750-100=650 Torr `(p_(1)V_(1))/T_(1)=(p_(2)V_(2))/T_(2)rArr(650xx50)/291=(760xxV_(2))/273` `:.V_(2)=40.1" mL"` |
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| 861. |
Assertion: A mixture of `He and O_(2)` is used for respiration for deep sea divers. Reason: `He` is soluble in blood.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 The assertion that a mixture of helium and oxygen is used for deep sea divers is correct. He is not soluble in blood. Hence, this mixture is used. |
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| 862. |
Assertion: Gases do not settle at the bottom of container. Reason: Gases have high kinetic energy.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 It is correct that gases do not settle at the bottom of container and it is because due to higher kinetic enrgy of gaseous molecules they diffuse. |
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| 863. |
A certain mass of dry gas at `27^(@)C` and `760 mm` pressure has density `28`. What will be its density at `7^(@)C` and `740 mm`? |
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Answer» Given `d_(1)=28` `P_(1)=760 mm` `T_(1)=27^(@)C=300.15 K` `d_(2)=?` `P_(2)=740 mm` `T_(2)=280.15K` `:. (d_(1))/(d_(2))=(P_(1)xxT_(2))/(T_(1)xxP_(2))` `:. D_(2)=(d_(1)xxT_(1)xxP_(2))/(P_(1)xxT_(2))=(28xx300.15xx740)/(760xx280.15)=29.21` |
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| 864. |
A 1 L flask contains 32 g `O_(2)` gas at `27^(@)C`. What mass of `O_(2)` must be released to reduce the pressure in the flask to 12.315 atm?A. 8 gB. 16 gC. 24 gD. 0 g |
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Answer» Correct Answer - B `pV=nRT=w/mRT` `:.w=(pVm)/(RT)=(12.315xx1xx32)/(0.0821xx300)=16g` `:. 0_(2)` to be released = 32 -16=16 g |
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| 865. |
The excluded volume (b) is = 4 (volume of one gas molecule) |
| Answer» Correct Answer - True | |
| 866. |
A `10 L` box contains `41.4 g` of a mixture of gases `C_(x)H_(8)` and `C_(x)H_(12)`. The total pressure at `44^(@)C` in flask is `1.56 atm`. Analysis revelated that the gas mixture has `87%` total `C` and `13%` total `H`. Find out the value of `x`A. `1`B. `3`C. `5`D. `2` |
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Answer» Given `P=1.56 atm`, `V=10 L` `T=317 K`, `R=0.082` Total moles `(m)=(PV)/(RT)=(1.56xx10)/(0.082xx317)=0.6 mol` Let `CxH_(8)` be a mol, therefore moles of `C_(x)H_(12)=(0.6-a)mol`, mas of `C` in a mol of `C_(x)H_(12)-12ax g`, mass of `C` in `(0.6-a) mol` of `C_(x)H_(12)=12xx(0.6-a)g` `:. Total mass of C in mixture=12ax+12x(0.6-a)g` `=41.4 g` `%` of `C` in mixture `=(7.2x)/(41.4)xx100` Given `%` of `C=87%` or `(720 x)/(41.4)=87` or `x=5` |
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| 867. |
What is the ratio of rate of diffusion of gas `A` and `B`. The molecular mass of `A` is `11` and molecular mass of `B` is `44`.A. `1`B. `2`C. `3`D. `4` |
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Answer» Rate of diffustion `prop sqrt((1)/("Molecular mass"))` `:. (r_(A))/(r_(B))=sqrt((44)/(11))=sqrt(4)=2` |
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| 868. |
The ratio of excluded volume (`b`) to molar volume of a gas molecule isA. `1`B. `2`C. `3`D. `4` |
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Answer» `b=4` (volume of gas molecule) `:. (b)/("Volume of gas molecule")=4` |
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| 869. |
Calculate the moles of an ideal gas at pressure `2 atm` and volume `1 L` at a temperature of `97.5 K`A. `1`B. `2`C. `3`D. `4` |
| Answer» `PV=nRT`, `n=(RT)/(PV)=(0.082xx97.5)/(2xx1)=(8)/(2)=4` | |
| 870. |
`5 L` of nitrogen measured at `750 mm` have to be compressed into an iron cylinder of `1 L` capacity. If temperature is kept constant, calculate the pressure in atmospheres required to do so. |
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Answer» Given `V_(1)= 5L` `P_(1)=750 mm` or `(750)/(760)atm` `V_(2)=1 L` `P_(2)=?` ` :. P_(1)V_(1)=P_(2)V_(2)` or `P_(2)=(P_(1)V_(1))/(V_(2))=((750)/(760)xx5)/(1)` `:. P_(2)=4.93 atm` |
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| 871. |
Calculate kinetic energy of `4g N_(2)` at `-13^(@)C`. |
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Answer» We know, `KE=(3)/(2)xxnRT` Number of moles of `4 g N_(2)=(4)/(28)g mol^(-1)` `T=-13^(@)C=260.15K` `R=8.314 J K^(-1) mol^(-1)` `:. K.E.=(3)/(2)xx(4)/(28)xx8.314xx260.15` `=463.47 J mol^(-1)` |
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| 872. |
`152 mL` of a gas at `STP` was taken to `20^(@)C` and `729 mm` pressure. What was the change in volume of the gas? |
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Answer» Given `V_(1)=152 mL` `P_(1)=1 atm=760 mm` `T_(1)=273.15 K` `V_(2)=?` `P_(2)=729 m` `T_(2)=20^(@)C` or `293.15 K` `:. V_(2)=(P_(1)V_(1))/(T_(1))xx(T_(2))/(P_(2))=(760xx152xx293.15)/(273.15xx729)=170 mL` `:. Change in volume=V_(2)-V_(1)=170-152=18 mL` |
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| 873. |
The value of compressibility factor (`Z`) for an ideal gas isA. `2`B. `1`C. `3`D. `4` |
| Answer» `Z_(ideal gas)=(V_(m))/(V_(ideal))=1` | |
| 874. |
What is the average speed of a molecule, having a molecular mass of `529.5 g mol^(-1)`. At temperature `100 K`A. `1`B. `2`C. `3`D. `4` |
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Answer» `u_(av)=sqrt((8RT)/(piM))`, `T=100 K`, `M=529.5 g mol^(-1)` `=sqrt((8xx8.314xx100)/(3.14xx529.5))=2` `:.` Average speed of molecule is `2` or (`b`) |
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| 875. |
Initial volume of a gas is `1 L` at temperature `100 K`. What is the volume of a gas at `300 K`.A. `1`B. `2`C. `3`D. `4` |
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Answer» Given `V_(1)=1 L`, `T_(1)=100 K`, `T_(2)=300 K`, `V_(2)=?` We know, `(V_(1))/(T_(1))=(V_(2))/(T_(2)) implies V_(2)=(V_(1)T_(2))/(T_(1))` `:. V_(2)=(1xx300)/(100)=3` |
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| 876. |
What volume will a sample of gas occupy at `87^(@)C` and `720 mm` pressure if its volume at `27^(@)C` and `750 mm` pressure is `250 mL` ? |
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Answer» Given `V_(1)=250 mL` `T_(1)=27^(@)C` or `300.15 K` `P_(1)=750 mm` `V_(2)=?` `T_(2)=87^(@)C` or `360.15 K` `P_(2)=720 mm` `:. (P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))` or `V_(2)=(P_(1)V_(1))/(T_(1))xx(T_(2))/(P_(2))=(750xx250)/(300.15)xx(360.15)/(720)=312.5 mL` `:. Change in volume =V_(2)-V_(1)=170-152=18 mL` |
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| 877. |
`100 mL` of gas is collected at `750 mm` pressure. What volume will it occupy at `74.5 mm` pressure? |
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Answer» Given `V_(1)=100 mL` `P_(1)=750 mm` `V_(2)=?` `P_(2)=745 mm` ` :. P_(1)V_(1)=P_(2)V_(2)` or `V_(2)=(P_(1)V_(1))/(P_(2))=(750xx1000)/(745)` ` :. V_(2)=100.67 mL` |
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| 878. |
A certain gas is at a temperature of `350 K`. If the temperature is raised to `700 K`, the average translational kinetic energy of the gas will increase byA. `2`B. `3`C. `4`D. `5` |
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Answer» `T_(i)=350 K`, `T_(f)=700 K` `KE_(avg)(Translational)=(3)/(2)KT` `:. KE` increase by a factor of `(700)/(350)=2` |
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| 879. |
Three gases `CO_(2) ,O_(2)` , and `Cl_(2)` are at the same temperature. Which of the following relations for the average translational kinetic energy per mole `(E_(A))` is true?A. `E_(K) (C O_(2)) gt E_(K) (O_(2)) lt E_(K) (C l_(2))`B. `E_(K) (CO_(2)) gt E_(K) (O_(2)) gt E_(K) (C l_(2))`C. `E_(K) (CO_(2)) lt E_(K) (O_(2)) lt E_(K) (Cl_(2))`D. `E_(K) (CO_(2)) = E_(K) (O_(2)) = E_(K) (Cl_(2))` |
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Answer» Correct Answer - D Average translational to the absolute temperature: `E_(K) = (3)/(2)RT` It is independent of the nature of the gas. |
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| 880. |
The ratio of most probable velocity to the average velocity isA. `(pi//2)`B. `2//pi`C. `sqrt(pi)//2`D. `2//sqrt(pi)` |
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Answer» Correct Answer - B::C |
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| 881. |
Which gas has the highest partical pressure in atmosphere?A. `CO_(2)`B. `H_(2)O`C. `O_(2)`D. `N_(2)` |
| Answer» (d) `N_(2)` has the highest partial paressure in atmosphere | |
| 882. |
Which one of the following statements is NOT true about the effect of an increase in temperature on the distribution of molecular speeds in a gas ?A. The most probable speed increases.B. The fraction of the molecules with the most probable speed increases.C. The distribution becomes broader.D. The area under the distribution curve remains the same as under the lower temperature. |
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Answer» Correct Answer - B |
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| 883. |
The distribution of speeds of molecules of a gas depends on (i) temperature , (ii) volume (iii) pressure , (iv) molecular massA. `(i),(ii),(iii)`B. `(i),(iv)`C. `(i),(ii),(iii),(iv)`D. `(i),(iii)` |
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Answer» Correct Answer - B Maxwell and Boltzmann have shown that the actual distribution of molecular speeds depends on the temperature and the molecular mass of the gas. On increasing the temperature, the most probable speed (corresponding to the maximum in the curve) increases. Also , the speed distribution curve broadens at higher temperature. Broadening of the curve shows that the number of molecules moving at higher speed increases. Speed distribution also depends on the molecular mass of the molecules. At the same temperature, the gas molecules with higher molecular mass have slower speed than the molecules with lower moleculat mass. |
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| 884. |
The molecular speeds of gaseous molecules are analogous to those to rifle bullets , why do then odour of the gaseous molecular not detected so fast ? |
| Answer» No doubt molecules in the gaseous state move alomost with the same speed of rifle bulles, but these molecules donot follow a straight path. Since they undergo collsions with one another a very fast rate, the path followed is zig-zag. Thus, the odour cannot be detected at the same speed at which the molecules move. | |
| 885. |
What is the relation between three types of molecular speeds at a given temperature? |
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Answer» a. volume of a gas molecule is negligible in comparison to the total volume of container. b. there is no force of attraction between the gaseous molecules. |
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| 886. |
Which of the following expressions represents the value and unit of van der Waals constant a?A. `a = (V)/(n) ," L mol"^(-1)`B. `a = (PV)/(n) , "atm L"^(2) mol^(-1)`C. `a = (PV^(2))/(n^(2)) , "atm L"^(2) mol^(-1)`D. `a = (P)/(n) ," atm mol"^(-1)` |
| Answer» Correct Answer - C | |
| 887. |
In the van der Waals equationA. `b` is the volume occupied by the gas moleculesB. `b` is four times the volume occupied by the gas moleculesC. `b` is the correction factor for intermolecular attractionD. None of these |
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Answer» `b`(co-volume)`=4((4)/(3)pi r^(3))` `=4` (Volume occupied by gaseous molecule) |
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| 888. |
Which of the following is true about gaseous state?A. Thermal energy `=` Molecular attractionB. Thermal energy `gtgt` Molecular attractionC. Thermal energy `ltlt` Molecular attractionD. Molecular forces `gtgt` Those in liquids |
| Answer» Thermal energy `gtgt` Molecular attraction | |
| 889. |
Which of the following is not a correct postulate of the kinetic molecular theory of gases ?A. Molecules of gases remain in continuous motion.B. while moving, they collide with each other and with the walls of the container.C. Collisions of gas molecules are inelastic.D. Speeds and energies of the molecules of the gas at any instant are not the same. |
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Answer» Correct Answer - C Collisions of gas molecules are perfectly elastic, i.e., the total energy of molecules before and after the collision remains the same. |
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| 890. |
A closed vessel contains equal number of nitrogen and oxygen molecules at pressure of `P mm`. If nitrogen is removed from the system, then the pressure will be:A. PB. `2 P`C. `P//2`D. `P^(2)` |
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Answer» Correct Answer - C By removing the nitrogen molecules (half in number) the partial pressure will be also reduced to half i.e, it will becomes `p//2` |
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| 891. |
Temperature at which most probable speed of `O_(2)` becomes equal to root mean square speed of `N_(2)` is [Given : `N_(2)` at `427^(@)C`]A. `732 K`B. `1200 K`C. `927 K`D. `800 K` |
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Answer» Correct Answer - B `1200 K` |
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| 892. |
Match the statements of ColumnI with values of Column II |
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Answer» Correct Answer - `Atoq,r;Btop,q,r;Ctop,q,r;Dtoq,s` |
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| 893. |
At what temperature most probable speed of `O_(2)` molecules have the same value of root mean square speed of `O_(2)` molecules at `300 K`?A. `150 K`B. `600 K`C. `750 K`D. `450 K` |
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Answer» Correct Answer - D `(U_(MP))/(U_(rms))=sqrt(((2RT_(1))/(M))/((3RT_(2))/(M)))=sqrt((2T_(1))/(3T_(2)))implies (2)/(3)(T_(1))/(T_(2))=1` `T_(1)=(3)/(2)xx300= 450K` |
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| 894. |
Assertion At same temperature if equal number of molecules is considered, `O_(2)` has greater fractions of molecules moving with most probable than `CO_(2)` Reason Most probable speed is inversely related to square root of molar mass. |
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Answer» Correct Answer - d |
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| 895. |
A mixture of `CO " and " CO_(2)` is found to have a density of `1.5 g L^(-1) " at " 30^(@)C` and 730 torr. What is the composition of the mixture ? |
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Answer» Let the average molecular mass be M. `M = (dRT)/(P) = 1.5 xx 0.0821 xx 303 xx (760)/(730) = 38.85` Let `x` mole of `CO " and " (1 - x)` mole of `CO_(2)` be present `x xx 28 + (1 - x) xx 44 = 38.85` `x = 0.3218` Mole % of `CO = 32.18` and mole % of `CO_(2) = 67.82` |
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| 896. |
A gasesous mixture contains 56 g`N_(2)`,44 g `CO_(2)` and 16 g `CH_(4)`. The total pressure of the mixture is 720 mm Hg. What is the partial pressure of `CH_(4)` ? |
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Answer» Correct Answer - A Similar to Solved Problem 4. |
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| 897. |
A mixture of CO and `CO_(2)` is found to have a density of `1.50" g "L^(-1)` at `20^(@)C` and 740 mm pressure. Calculate the composition of the mixture. |
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Answer» Calculation of average molecular mass of the mixture : `M=(dRT)/(P)=(1.50" g "L^(-1)xx0.0821" L atm "mol^(-1)xx293" K")/((740//760)" atm")=37.06` Calculation of percentage composition : Suppose mol % of CO in the mixture=x Then mol% of `CO_(2)` in the mixture=(100-x) Average molecular mass `=(x xx28+(100-x)xx44)/(100)` `:. " " (28x+4400-44x)/(100)=37.06+" or " 16x=4400-3706=694" or " x=694//16=43.38` `:.` Mol % CO=43.38 and Mol % of `CO_(2)=100-43.8=56.62`. |
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| 898. |
The edge length of a face centered cubic cell of an ionic substance is 508 pm. If the radius of the cation is 110 pm, the radius of the anion isA. 618 pmB. 144 pmC. 288 pmD. 398 pm |
| Answer» Correct Answer - b | |
| 899. |
The edge length of face centred cubic unit is 508 pm. If the radius of the cation is 110 pm, the radius of the anion isA. 144 pmB. 288 pmC. 618 pmD. 398 pm |
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Answer» Correct Answer - A For face-centred cubic unit cell (e.g., that of NaCl), edge length = 2 x Distance between cation and anion =`2(r_c +r_a)`=508 pm (given ), Putting `r_c`=110 pm, we get `r_a`=144 pm |
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| 900. |
The edge length of a face-centred cubic unit cell is `508 pm`. If the radius of the cation is `110 pm` the radius of the anion isA. 288 pmB. 398 pmC. 154 pmD. 618 pm |
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Answer» Correct Answer - C For centred unit cell, `2(r^(+)+r^(-))=a` `r^(+)-"radii of cation"` `r^(-)="radii of anion"` `2(1000+r^(-))=508` `100+r^(-)=(508)/(2)` `r^(-)=(508)/(2)-100` `=254-100=154" pm"` |
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