Explore topic-wise InterviewSolutions in .

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.

301.

The composition of the equilibrium mixture for the equilibrium `Cl_(2) hArr 2 Cl` at 1400 K may be determined by the rate of diffusion of mixture through a pin hole. It is found that at 1400 K, the mixture diffuses 1.16 times as fast as krypton diffuses under the same conditions. Find the degree of dissociation of `Cl_(2)` equilibrium

Answer» Equilibrium of dissociation of `Cl_(2)` may be represented as :
`{:(,Cl_(2) (g),hArr,2Cl (g),),(t = 0,a,,0,),(t_(eq),a(1- alpha),,2 a alpha,):}`
Total moles `= a (1 - alpha) + 2a alpha = a (1 + alpha)`
`M_("min") = (a M_(Cl_(2)))/(a (1 + alpha)) = (M_(Cl_(2)))/((1 + alpha))`
`(R_("mix"))/(R_(Kr)) = sqrt((M_(Kr))/(M_("mix")))`
`1.16 = (84 (1 + alpha))/(M_(Cl_(2)))`
`((1.16)^(2) xx 71)/(84) - 1 = alpha, alpha = 0.1374`
302.

The composition of the equilibrium mixture `(Cl_(2)hArr2Cl)`, which is attained at `1200^(@)C` is determined by measuring the rate of effusion through a pin-hole. It is observed that at 1.80 mm Hg pressure, the mixture effuses 1.16 times as fast as Krypton effuses under the same conditions. Calculate the fraction of chlorine molecules dissociated into atoms (Atomic weight of Kr=84)

Answer» `(r_("mix"))/(r_(Kr))=sqrt((M_(Kr))/(M_("mix"))),` i.e., `1.16=sqrt((84)/(M_("mix")))" or " M_("mix")=62.43`
If x moles of `Cl_(2)` dissociate at equilibrium, then
`{:(,Cl_(2),hArr,2Cl),("At.eqm".,1-x,,2x):}`
Average , molecular mass of the mixture`=((1-x)xx71+2x xx35.5)/((1-x)+2x)=(71)/(1+x)`
`:.(71)/(1+x)=62.43` which gives x=0.137. Hence, fraction dissociated=0.137
303.

Assertion. Effusion rate of oxygen is smaller than nitrogen. Reason. Molecular size of nitrogen is smaller than oxygen.

Answer» Correct Answer - C
Correct R. Molecular size decreases from left to right along a period. Thus, molecular size of nitrogen is greater than that of oxygen.
304.

Assertion: Effusion rate of oxygen is smaller than nitrogen. Reason: Molecular size of nitrogen is smaller than oxygen.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.

Answer» Correct Answer - C
Assertion is true but reason is false because of effusion rate `prop (1)/(sqrt(M))` (molecular weight) but it does not depend on molecular size.
305.

If the pressure of a gas contained in a closed vessel is increased by 0.4% when heated by `1^(@)C`, then its initial temperature must be:A. `-23^(@)C`B. `+23^(@)C`C. `250^(@)C`D. `523^(@)C`

Answer» `LetT_(1)=T Rightarrow T_(2)=(T+1)`
`p_(1)=p Rightarrow p_(2)=p+(0.4p)/(100)=(100.4)/(100)p`
`(p_(1)V_(1))/(T_(1))=(p_(2)V_(2))/(T_(2)),(pV)/(T)=(100.4p)/(100)xx(V)/((T+1))`
`100T+100=100.4T`
`T=(100)/(0.4)=250K`
`(250-273)^(@)C=-23^(@)C`
306.

Assertion : The pressure of real gas is less than the pressure of ideal gas. Reason : Intermolecular forces of attraction in real gases are greater than in ideal gas.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.

Answer» Correct Answer - A
Reason is the correct explanation for assertion.
307.

KBr has fcc structure. The density of KBr is `2.75 "g cm"^(-3)`. Find the distance between `K^+` and `Br^-`. (At mass of Br = 80.0)

Answer» Correct Answer - 330 pm
308.

Distribution of molecules with velocity is represented by the curve Velocity corresponding to point `A` isA. `sqrt((3RT)/(M))`B. `sqrt((2RT)/(M))`C. `sqrt((8RT)/(piM))`D. `sqrt((RT)/(M))`

Answer» Point `A` represents the most probable distribution of molecules. Hence, the most probable velocity is `sqrt(2RT//M)`.
309.

2.9 g of a gas at `90^(@)C` occupie the same volume as 0.184 g of `H_(2)` at `17^(@)C` at the same pressure. What is the molar mass of the gas ?

Answer» Let the molar mass be M
`thereforen(g)=(2.9)/(M)`
`n(H_(2))=(0.184)/(2)=0.092`
volume of `H_(2)=(0.092xxRxx290)/(P)=(26.68R)/(P)`
volume of `(2.9)/(M)` mol of gas `=(2.9xxRxx368)/(MP)=(1067.2)/(MP)`
Now `(1067.2R)/(MP)=(26.68R)/(P)` or `M=40` g `mol^(-1)`
310.

At what temperature will hydrogen molecules have the same `KE` as nitrogen molecules at `280 K`?A. `280 K`B. `40 K`C. `400 K`D. `50 K`

Answer» `KE=(3)/(2)RT`
`KE prop T`
311.

Calculate the pressure of 4.0 mol of a gas occupying `5 dm^(3)` at 3.32 bar pressure. (R = 0083 bar `dm^(3) K^(-1) mol^(-1)`)

Answer» Correct Answer - 50 K or `-233^(@)C`
312.

There is a collection of crystalline substances in a hexagonal closed packing. If the density of matter is `2.6 g//cm^(3)`, what would be the average density of matter in collection? What fraction of space is actually unoccupied ?

Answer» In hexagonal closed packed, the packing efficiency is `74.05%`.
`therefore` Density of matter= packing fraction `xx` Total density `=(74.05)/(100)xx2.6=1.93 g cm^(-3)`
`%` empty space `=100-74.05=25.95 %`
313.

At `27^(@)C` a gas contains 10 molecules travelling with a speed of 4 `ms^(-1)`, 20 molecules travelling with a speed of 5 m `s^(-1)` and 40 molecules travelling with a speed of 8 m `s^(-1)`. Calculate the average speed, root mean square speed and most probable speed of the gas at `27^(@)C`.

Answer» Average speed `(overset(bar)(c )=(sumn_(i)v_(i))/(sumn_(i))=(10xx4+20xx5+40xx8)/(10+20+40)=(460)70)`
`=6.56 m s^(-1)`
Root mean square speed (c )`=sqrt((sumn_(i)v_(i))/(sumn_(i)))=sqrt((10xx4^(2)+20xx5^(2)+40xx82)/(10+20+40))`
`=sqrt((160+500+2560)/(70))=6.678 m s^(-1)`
Most probable speed `(c^(**))=0.816xxc=0.816xx6.78 m s^(-1)=5.53 m s^(-1)`.
314.

The speed of six different molecules in a gas are `25 m s^(-1), 20 ms^(-1), 30 m s^(-1), 15 m s^(-1), 10 m s^(-1)` and `25 m s^(-1)`. Calculate the average speed and also the root mean square of the gas.

Answer» Average speed `(U_(av)) = (25+20+30+15+10+25)/(6) = (125)/(6) = 20.8 ms^(-1)`
Root mean square speed `(U_(RMS)) = sqrt(((25^(2)) + (20)^(2) + (30)^(2) + (15)^(2) + (10)^(2) + (25)^(2))/(100)) = sqrt(950) = 21.88 ms^(-1)`
315.

A compound consisting of the monovalent ions `A^+ ,B^-` crystallizes in the body-centred cubic lattice. (i)What is the formula of the compound ? (ii)If one of `A^+` ions from the corner is replaced by a monovalent ions `C^+`, what would be the simplest formula of the resulting compound ?

Answer» (i)Contribution of `A^+` ions present at the corners towards unit cell =`8xx1/8=1`
Contribution of `B^-` ion present at the body-centre=1
`therefore` Ratio of `A^+ :B^-` =1:1 .Hence , the formula is AB.
(ii)`A^+` ions at the corners =7. Hence, their contribution towards unit cell=`7/8`
`C^+` ion at one corner contributes =`1/8`
`B^-` ion at the body centre has contribution =1
Hence, ratio of A:B:C=`7/8:1/8:1=7:1:8 therefore ` Formula will be `A_7BC_8`
316.

Assertion:The packing efficiency is maximum for the fcc structure. Reason: The coordination number is 12 in fcc structures.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.

Answer» Correct Answer - b
Correct explanation. Fcc is a close packed structure.
317.

Why does zinc oxide exhibit enhanced electrical conductivity on heating ?

Answer» On heating , zinc oxide losses oxygen as : `ZnO oversetDeltaZn^(2+)+ 1/2 O_2 + 2e^(-)`
The `Zn^(2+)` ions thus formed are entrapped into the interstitial sites while the electrons are also entrapped in the neighbouring interstitial sites . These entrapped electrons enhance the electrical conductivity of ZnO.
318.

The arrangement of `X^(ɵ)` ions around `A^(o+)` ion in solid `AX` is given in the figure (not drawn to scale). If the radius of `X^(ɵ)` is `250 pm`, the radius of `A^(o+)` is A. 104 pmB. 125 pmC. 183 pmD. 57 pm

Answer» Correct Answer - A
For octahedral voids : `(r^(+))/(r_(-))=0.414-0.732`
`r^(+)=0.414xxr^(-)=0.414xx250=10.3.5` pm
or `r^(+)=0.732xx250=183` pm
No, suitable information is given therefore 104 pm will be most suitable answer by considering the extact fit ion in the void.
319.

If two mole of an ideal gas at `546 K` occupies a volume of `44.8 litres`, the pressure must be :A. `2 atm`B. `3 atm`C. `4 atm`D. `1 atm`

Answer» Correct Answer - A
`PV= nRT implies P=(2xx0.082xx546)/(44.8)xx10^(-1)`
320.

Which of the following gases would have the highest `rms` speed at `0^(@)C`?A. `O_(2)`B. `CO_(2)`C. `SO_(3)`D. `CO`

Answer» Correct Answer - D
`V_(rms) prop (1)/(sqrt(M))`
321.

Which of the following differentiate between diffusion and effusion?A. Diffusion is the intermixing of the gas molecules at any direction and effusion is the reverse of diffusionB. Diffusion is the property of the gas molecules and effusion is the property of the gas container onlyC. Diffusion occurs at any direction, whereas effusion occurs under the potential differenceD. Diffusion is the intermixing gas molecules, whereas effusion is the passage of gas molecules through the pores in one direction

Answer» Correct Answer - D
Diffusion is the ability of gas to mix spontaneously and to form a homogenous mixture while effusion is a process in which a gas under pressure escape out from out from a fine hole in one direction.
322.

Which one of the following gases has the highest critical temperature ?A. NitrogenB. AmmoniaC. Water vapoursD. Carbon dioxide

Answer» Correct Answer - C
The gas which can be liquefied most easily has the highest critical temperature. Water vapours i.e., `H_(2)O(g)` molecules can be liquefied most easily due to presence of intermolecular hydrogen bonding. Therefore, they have maximum critical temperature .
323.

The compressibility factor for an ideal gas isA. `prop`B. `1.5`C. `1.0`D. `2.0`

Answer» Correct Answer - C
Compressibility factor `(Z)` is given as
`Z = (pV)/(nRT)`
For an ideal gas , `Z = 1` at all temperatures and pressures because `pV = nRt` . The graph of `Z` versus `p` will be a straight line parallel to the value of `Z` deviates from unity.
324.

A metallic element crystallises into a lattice containing a sequence of layers of ABABAB….. . Any packing of spheres leaves out voids in the lattice. What percentage by volume of this lattice is empty space ?

Answer» Correct Answer - 0.26
325.

The `SI` unit of viscosity coefficient isA. `N s m^(-2)`B. `Pa s`C. `kg m^(-1) s^(-1)`D. All of these

Answer» Correct Answer - D
The `SI` unit of viscosity coefficient is 1 newton second per square meter `(N s m^(-2)) = "pascal second" (Pa s = 1 kg m^(-1) s^(-1))`.
326.

At what temperature, the root-mean-square velocity of `SO_(2)` will be the same as that of `CH_(4)` at `27^(@)C`?A. `3000 K`B. `1345 K`C. `1200 K`D. `1700 K`

Answer» Correct Answer - C
Root-mean-square velocity `=sqrt((3RT)/(M))`
The value of `CH_(4)` at `27^(@)C` or `300 K= sqrt((3Rxx300)/(16))`
Now for `SO_(2)`
Root-mean-square velocity `=sqrt((3RT)/(64))`
Or, `sqrt((3Rxx300)/(16))=sqrt((3RxxT)/(64))` or `T= 1200K`
327.

The volume occupied by 4.4 g of `CO_(2)` at STP isA. 22.4 LB. 44.8 LC. 12.2 LD. 2.24 L

Answer» Correct Answer - D
44g of `CO_(2)` occupies 22.4 L
4.4g of `CO_(2)` occupies `(22.4)/(44)xx4.4=2.24 L`
328.

The volume occupied by 4.4 g of `CO_(2)` at STP isA. `2.4 L`B. `2.24 L`C. `44 L`D. `22.4 L`

Answer» Correct Answer - B
`n_(CO_(2)) = ("Mass"_(CO_(2)))/("Molar mass"_(CO_(2))) = (4.4 g)/(44 g "mol"^(-1)) = 0.1 "mol"`
Under `STP` conditions ,
`n_(CO_(2)) = (V_(CO_(2)) "inliters at STP")/(22.4 L" mol"^(-1))`
`0.1 "mol" = (V_(CO_(2)) "inliters at STP")/(22.4 L mol^(-1))`
or `V_(CO_(2)) = (0.1 "mol") (22.4 L "mol"^(-1))`
`= 2.24 L`
329.

Which of the following is the correct value of `R` in `SI` units ?A. `0.083 bar dm^(3) K^(-) mol^(-1)`B. `1.98 cal K^(-1) mol^(-1)`C. `8.314 xx 10^(7) erg K^(-1) mol^(-1)`D. `8.314 JK^(-1) mol^(-1)`

Answer» Correct Answer - D
The volume of one mole of an ideal gas under `STP` conditions (`273.15 K` and `1 bar` pressure) is `22.710981 L mol^(-1)`. The value of `R` of an ideal gas can be calculated under these conditions as follows:
`R = (10^(5) Nm^(-2)) ((22.71 xx 10^(-3) m^(3)))/((1 mol)(273.15 K) `
`= 8.314 Pa m^(3) K^(-1) mol^(-1) (or 8.314 JK^(-1) mol^(-1))`
330.

The volume occupied by `7.40 g of CO_(2)` (at `STP`) isA. `3.8 L`B. `4.5 L`C. `5.6 L`D. `2.9 L`

Answer» Correct Answer - A
Since `1 mol` of an ideal gas occupies `22.4 L at STP`, we have
`V_(CO_(2)) = (n_(CO_(2))) (22.4 (L)/(mol))`
`= ((7.40 g)/(44 g mol^(-1))) (22.4 L mol^(-1))`
`= 3.76 L`
331.

If `C_(1), C_(2), C_(3) …..` represent the speeds on `n_(1), n_(2) , n_(3)…..` molecules, then the root mean square speed isA. `((n_(1)C_(1)^(2)+n_(2)C_(2)^(2)+n_(3)C_(3)^(2)+……….)/(n_(1)+n_(2)+n_(3)+.....))^(1//2)`B. `((n_(1)C_(1)^(2)+n_(2)C_(2)^(2)+n_(3)C_(3)^(2)+…...)^(1//2))/(n_(1)+n_(2)+n_(3)+....)`C. `(n_(1)C_(1)^(2))^(1//2)/(n_(1))+(n_(2)C_(2)^(2))^(1//2)/(n_(2))+((n_(3)C_(3)^(2))^(1//2))/(n_(3))`D. `[((n_(1)C_(1)+n_(2)C_(2)+n_(3)C_(3)+.....)^(2))/((n_(1)+n_(2)+n_(3).....))]^(1//2)`

Answer» Correct Answer - A
Root mean sqaure speed
`[(n_(1)C_(1)^(2)+n_(2)C_(2)^(2)+n_(3)C_(3)^(2)+)/(n_(1)+n_(2)+n_(3)+.....)]^(1//2)`
332.

If two gases of moleuclar weight `M_(1) and M_(2)` at temperature `T_(1) and T_(2), T_(1)M_(2)=T_(2)M_(1)`, then which property has the same magnitude of both the gases?A. TemperatureB. PressureC. `KE` per moleD. Root mean square velocity

Answer» Correct Answer - D
`u_(1)= u_(2)`
`u_(1)^(2)=u_(2)^(2)`
`(3RT_(i))/(M_(1))=(3RT_(2))/(M_(2))`
`T_(1)M_(2)=T_(2)M_(2)`
333.

If two gases of moleuclar weight `M_(1) and M_(2)` at temperature `T_(1) and T_(2), T_(1)M_(2)=T_(2)M_(1)`, then which property has the same magnitude of both the gases?A. densityB. pressureC. `KE` per moleD. `V_(rms)`

Answer» Correct Answer - D
(i) density of a gas `(rho) =(PM)/(RT)`
Since `(M_(B))/(T_(B))=(M_(A))/(T_(A))`
`:.` at the same pressure `rho_(A)=rho_(B)`
But if pressure is different, then `rho_(A)!=rho_(B)`
(ii) Pressure of the gas would be equal if their densities are equal otherwise not
(iii) `KE` per mol `=(3)/(2)RT`
`:.` It will be different for the two gases
(iv) `V_(rms)=sqrt((3RT)/(M))`, since `(T_(A))/(M_(A))=(T_(B))/(M_(B))`
`V_(rms)` of `A=V` of`B`
334.

Calculate the total number of electrons present 1.4 g of dinitrogen gas.

Answer» Correct Answer - `4.2154xx10^(23)` electrons
Molar mass of dinirogen `(N_(2))=28g mol^(-1)`
Thus, 1.4 g of `N_(2)=(1.4)/(28)=0.05 mol`
`=0.05xx6.02xx10^(23)` number of molecules
`=3.01xx10^(23)` number of molecules
Now, 1 molecule of `N_(2)` contains 14 electrons.
Therefore, `3.01xx10^(23)` molecules of `N_(2)` contains `=14xx3.01xx1023=4.214xx10^(23)` electrons
335.

At `0^(@)C` , the density of a gaseous oxide at `2` bar is the same as that of dinitrogen at `5` bar. The molar mass of the oxide isA. `60 g mol^(-1)`B. `50 g mol^(-1)`C. `70 g mol^(-1)`D. `30 g mol^(-1)`

Answer» Correct Answer - C
According to the ideal gas law,
`pM = dRT`
At the same temperature `(T)` and for the same density `(d) ,pM` becomes a constant. Thus,
`p_(o x i de)M_(o x i de) = p_(N_(2))M_(N_(2))`
or `M_(Ox i de) = p_(N_(2)) M_(N_(2))//p_(o x ide)`
`= ((5 ba r) (28 g mol^(-1)))/((2 ba r))`
`= 70 g mol^(-1)`
336.

A gas- filled freely collapsible balloon is pushed from the surface level of a lake to a depth of `100 m`. Approximately what percentage of its original volume will the balloon finally have? Assume that the gas behaves ideally

Answer» Let the volume of balloon at the surface of the lake be `V`.
`P=1 atm=76xx13.6xx981 dyn cm^(-1)`
`=981xx1033.6`
Pressure at the depth of `100 m`
`=76xx13.6xx981+100xx100xx1xx981`
`=981(76xx13.6xx10000)`
`=981xx11033.6`
`P_(1)V_(1)=P_(2)V_(2)`
`981xx1033.6xxV=981xx11033.6xxV_(2)`
`V_(2)=(1033.6)/(11033.6)xxV`
`=(1033.6xxV)/(11033.6xxV)xx100%`
`=9.67%`
337.

If the pressure of a gas is doubled and the temperature is tripled, by how much will the mean free path of a gas molecule in a vessel change?A. Increase `3` timesB. Decrease `3` timesC. Increase `1.5` timesD. Decrease `1.5` times

Answer» `lambda_(1)=(T_(1))/(P_(1))`, `lambda_(2)=(T_(2))/(P_(2))`
`:. (lambda_(2))/(lambda_(1))=(T_(2)xxP_(1))/(T_(2)xxT_(1))=(3T_(1)xxP_(1))/(2P_(1)xxT_(1))=(3)/(2)`
`lambda_(2)=(3)/(2)lambda_(1)`
338.

By now much will the mean free path of a gas molecule in a vessel at constant `T` change if the pressure is reduced by `10%`?A. `10%` increaseB. `10%` decreaseC. `11.1%` increaseD. `11.1%` decrease

Answer» `lambda prop (T)/(P)`
`lambda(mean free path)=(overline (c ))/(sqrt(2)pi sigma^(2)overline(c )N)=(1)/(sqrt(2) pi sigma^(2)N^(*))`
` :. Lambda prop (1)/(sigma^(2))` (`sigma=` molecular diameter, `overline(c )=` average velocity)
`lambda=(KT)/(sqrt(2)pi sigma^(2)P)(K= Boltzmann constant =(R)/(N)=1.38xx10^(-23)JK^(-1)mol^(-1))`
`lambda prop (1)/(P)` (at constant `T`)
`lambda prop T` (at constant `P`)
`lambda_(1)P_(1)=lambda_(2)P_(2), (lambda_(1))/(T_(1))=(lambda_(2))/(T_(2))`
`:. lambda_(1)P_(1)=lambda_(2)xx0.9 P_(1)`
` implies lambda_(1)=0.9lambda_(2) implies lambda_(2)=1.11 lambda_(1)`
`:. lambda_(2)=111.1 %`
Increases in `lambda_(2)=111.1-100=11.1%`
This shows `lambda` will increase by `1//9`, if `P` is reduced by `10%`.
339.

By how much will the mean free path of a gas molecule is a vessel at constant `P` change if the temperature is reduced by `20%`?A. `12.5%` decreaseB. `12.5%` increaseC. `80%` decreaseD. `80%` increase

Answer» `lambda prop (T)/(P)`, i.e., `(lambda_(1))/(T_(1))=(lambda_(2))/(T_(2))` (at constant pressure)
If temperature is reduced, `lambda` will decrease.
`(lambda_(2))/(lambda_(1))=(T_(2))/(T_(1))=(0.8T_(1))/(T_(1))=0.8` i.e., `80%` decrease
340.

`100 mL` of hydrogen was confined in a diffusion tube and exposed to air, and at equilibrium, a volume of `26.1 mL` of air was measured in the tube. Again, when `100 mL` of `CO_(2)` was placed in the same tube and exposed to air, `123 mL` of air was measured in the tube at the equilibrium. Find the molecular weight of `CO_(2)`.

Answer» In the first case, when `100 mL` of `H_(2)` is diffused, the volume of air diffused was `26.1 mL`. Thus
`(Rate of diffusion of H_(2))/(Rate of diffusion of air)=(100)/(26.1)` …(`i`)
`(Rate of diffusion of CO_(2))/(Rate of diffusion of air)=(100)/(123)` ...(`ii`)
From equation (`i`) and (`ii`), we get
`(r_(H_(2)))/(r_(CO_(2)))=(123)/(261)=sqrt((M_(CO_(2)))/(M_(H_(2))))`
`Molecular of weight of CO_(2)=(123xx123)/(26.1xx26.1)=2=44.42`
341.

What is the binding force between molecules if a subsatance is a gas under ordinary conditions of temperature and pressure?

Answer» van der Waals forces
342.

What is the equation of state of an ideal gas for n moles ?

Answer» Correct Answer - PV=nRT.
343.

What is the molar volume of a gas at SATP condition ?

Answer» 24.8 L (at `25^(@)C`, 1 bar pressure, called standard ambient temperature and pressure)
344.

A real gas acts as an ideal gas under which condition ?A. High temperature, low pressureB. Low temperature, high pressureC. High temperature, high pressureD. Low temperature, low pressure

Answer» Correct Answer - A
345.

Calculate the root mean square speed (rms) of ozone kept in a closed vessel at `50^(@)C` and pressure of 82 cm of Hg.

Answer» Volume occupied by 1 mole of `O_(3)` at `20^(@)C` and 82 cm pressure can be calculated by applying gas equation as follows :
`(76xx22400)/(273)=(82xxV)/(293)" or V=22282 cm^(3)`
`P=82xx13.6xx981" dynes"//cm^(2)`
`u=sqrt((3 PV)/(M))=sqrt((3xx82xx13.6xx981xx22282)/(48))=3.90xx10^(4)" cm"//sec`.
346.

Which of the following is true about pressures in flasks `A` and `B`?A. The pressure in flask `A` is four times that in flask `B`.B. The pressure in flask `B` is four times that in flask `A`.C. Both flasks have some pressure.D. The pressure in flask `A` is eight times that in flask `B`.

Answer» `P_(A)V=n_(A)RT_(A)`, `P_(B)V=n_(B)RT_(B)`
`(P_(A))/(P_(B))=(n_(A)T_(A))/(n_(A)T_(B))=8xx(300)/(600)=4`
`n_(A)=(m)/(2)`, `n_(B)=(m)/(16)`, `(n_(A))/(n_(B))=8`
347.

The relationship between the coefficient of viscosity of a liquid and temperature can be expressed asA. `eta = Ae^(RT//E)`B. `eta = Ae^(E//RT)`C. `eta = Ae^(ERT)`D. `eta = ET//R`

Answer» Correct Answer - B
The variation of viscosity with temperature is given approximately by
`eta = A exp ((E ) /(RT))`
348.

Calculate the distance between (111) planes in a crystal of calcuim . Repeat the calculation for (222) planes . Which planes are closer ? (a=0.556 nm)

Answer» `d=a/(sqrt(h^2+k^2+l^2))`
`d_111=0.556/(sqrt(1^2+1^2+1^2))=0.556/sqrt3`=0.321 nm
`d_222=0.556/(sqrt(2^2+2^2+2^2))=0.556/sqrt12`=0.161 nm
Thus (222) planes are closer and their distance of separation is half of that of (111) planes
349.

The density of sodium chloride at `25^@C` is `2.163xx10^3 "kg m"^(-3)`. When X-rays from a palladium target having wavelength of 58.1 pm are used , the (200) reflection of sodium chloride occurs at an angles of `5.90^@` . How many `Na^+` and `Cl^-` ions are present in the unit cell ? (Molar mass of NaCl =`58.5 "g mol"^(-1)`, sin `5.9^@=0.1028`)

Answer» `2d sin theta=n lambda`
For n=1, `d_200=lambda/(2 sin theta)="58.1 pm"/(2 sin 5.9^@) =58.1 /(2xx0.1028)`=282 pm
`d_200=a/(sqrt(h^2+k^2+l^2))`
`therefore "282 pm"=a/(sqrt(2^2+0^2+0^2))=a/2` or a=564 pm
`rho=(ZxxM)/(a^3xxN_0)`
or `Z=(rhoxxa^3xxN_0)/M=((2.163xx10^3 "kg m"^(3))(564xx10^(-12)m)^3 (6.023xx10^23 "mol"^(-1)))/(58.5xx10^(-3) "kg mol"^(-1))`
=3.999 `approx` 4
Thus, a unit cell contain 4 NaCl units or 4 `Na^+` ions and 4 `Cl^-` ions.
350.

A gaseous system has a volume of `580 cm^(3)` at a certain pressure. If its pressure in increased by 0.96 atm, its volume becomes `100 cm^(3)`. Determine the pressure of the system

Answer» Correct Answer - `0.2 atm`
`P xx 580 = (P + 0.96) 100`