1085 + Interview Questions in STATES-OF-MATTER IN GENERAL AWARENESS Page 16 InterviewSolution

751.	The ratio of van der Waals constant (a/b) has the dimensions :A. atm `L^(-1)`B. L atm `L^(-1)`C. atm mol `L^(-1)`D. atm L `mol^(-2)`
Answer» Correct Answer - A::B

Discussion

752.	Dimensions of surface tension areA. `kg s^(-1)`B. `kg s`C. `kg s^(-2)`D. `kg s^(2)`
Answer» Correct Answer - C Surface tension is defined as the force acting per unit length perpendicular to the line drawn on the surface of liquid. It is denoted by the Greek letter `gamma` (gamma) . It has dimensions of force//length , i.e., `(kg m s^(-2))/(m) rArr kgs^(-2)`. In `SI` unit , it is expressed as `N m^(-1)`.

Discussion

753.	The effect of increase of temperature on surface tension and viscosity is that ………………… .
Answer» Correct Answer - both decrease with increase of temperature

Discussion

754.	Relative humidity of air is `60^(@)C` and the saturation vapour pressure of water vapour in air is `3.6kPa`. The amount of water vapours present in `2 L` air at `300 K` isA. `52 g`B. `31.2 g`C. `26 g`D. `5.2 g`
Answer» `PV=(w)/(m)RT` (for vapours of `H_(2)O`) `P=3.6xx10^(3)Pa`, `V=2xx10^(-3)m^(3)`, `T=300 K` `:. W_(H_(2)O)=(3.6xx10^(3)xx18xx2xx10^(-3))/(8.314xx300=0.052)` `w_(H_(2)O)=52 g` Since, realtive humidity `=60%`, therefore, amount of `H_(2)O=52xx0.6=31.2 g`

Discussion

755.	A neon-dioxygen mixture contains 70.6 g dioxygen and 167.5 g neon. If pressure of the mixture of gases in the cylinder is 25 bar. What is the partial pressure of dioxygen and neon in the mixture ?
Answer» Number of moles of dioxygen `=(70.6g)/(32g mol^(-1))` =2.21 mol Number of moles of neon `=(167.5g)/(20g mol^(-1))` =8.375 mol Mole fraction of dioxygen `=(2.21)/(2.21+8.3745)` `=(2.21)/(10.585)` =0.21 Mole fraction of neon `=(8.375)/(2.21+8.375)` `" " =0.79` Alternatively, mole fraction of neon =1-0.21=0.79 Partial pressure mole fraction `xx` of a gas total pressure `rArr"Partial pressure" =0.21xx(25"bar")` of oxygen =5.25 bar Partial pressure =0.79 `xx` (25 bar) of neon =19.75 bar

Discussion

756.	Value of gas constant R in the ideal gas equation PV = nRT depends uponA. temperature of the gasB. pressure of the gasC. units in which P , V and T are measuredD. nature of the gas .
Answer» Correct Answer - C

Discussion

757.	Assertion: The value of van der Waals constant a is larger for ammonia than for nitrogen. Reason: Hydrogen bonding is present in ammonia.A. Statement I is true: Statement II is true, Statement II is the correct explanation of Statement I.B. Statement I is true, Statement II is true, Statement II is not the correct explanation of Statement I.C. Statement I is true, Statement II is false.D. Statement I is false, Statement II is true.
Answer» Correct Answer - A

Discussion

758.	Statement-1 : Boiling point of `H_(2)O` is more than HF . and Statement-2 : Intermolecular hydrogen bonding in HF is stronger than `H_(2)O`.A. H - F has highest van der Waals forces and dipole moment.B. H - F has highest London forces.C. H - F has highest dipole moment hence has dipole - dipole , London forces and hydrogen bonding.D. H - F has strong intermolecular interactions like dipole - induced dipole.
Answer» Correct Answer - C H-F has dipole-dipole interaction, London forces and hydrogen bonding due to highest electronegativity of F. Hence, boiling point of H-F is highest.

Discussion

759.	Assertion: The value of van der Waals constant a is larger for ammonia than for nitrogen. Reason: Hydrogen bonding is present in ammonia.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.

Discussion

760.	Temperature at which `r.m.s` speed of `O_(2)` is equal to that of neon at `300 K` is:A. `280 K`B. `480 K`C. `680 K`D. `180 K`
Answer» Correct Answer - B `U_(rms)=sqrt((T)/(M_(w))) T propM_(w)` `(T_(N_(e )))/(T_(O_(2)))=(20)/(32) " " T_(O_(2))= 480 K`

Discussion

761.	For the given ideal gas equation `PV=nRT`, answer the following questions: Which of the following does not represent ideal gas equation?A. `PV=(1)/(3)mNv`B. `PV=nRT`C. `P=rho(RT)/(M)`D. `PV=RT`
Answer» The ideal gas equation is `PV=nRT` or `P=(rho RT)/(M)`, or `PV=RT(n=1)`

Discussion

762.	Which of the following species are capable of hydrogen bonding among themselves?A. `H_(2)S`B. `C_(6)H_(6)`C. `CH_(3)OH`D. `CH_(4)`
Answer» Correct Answer - C Hydrogen bonding among the molecules of a substances is possible only if `H` directly bonded to a highly electronegative element such as `O,F,or N`.

Discussion

763.	Average kinetic energy per molecule of a gas is related to its temperature as `bar KE`=………………… .
Answer» Correct Answer - `(3)/(2)kT`

Discussion

764.	The rms velocity of hydrogen is `sqrt(7)` times the rms velocity of nitrogen. If `T` is the temperature of the gas, thenA. `T_(H_(2))=T_(N_(2))`B. `T_(H_(2))gtT_(N_(2))`C. `T_(H_(2)) lt T_(N_(2))`D. `T_(H_(2))=sqrt(7)T_(N_(2))`
Answer» `sqrt((3RT_(H_(2)))/(2))=sqrt(7)sqrt((3RT_(N_(2)))/(28))` `(T_(H_(2)))/(2)=7(T_(N_(2)))/(28)` `T_(N_(2))=2T_(H_(2))`

Discussion

765.	The rms velocity of hydrogen is `sqrt(7)` times the rms velocity of nitrogen. If `T` is the temperature of the gas, thenA. `T(H_(2))=T(N_(2))`B. `T(H_(2)) gt T(N_(2))`C. `T(H_(2)) lt T(N_(2))`D. `T(H_(2))=sqrt(7)T(N_(2))`
Answer» Correct Answer - C `U_(rms)=sqrt((3RT)/(M))` `sqrt((3RT_(H2))/(2))=sqrt(7)sqrt((3RT_(N_(2)))/(28))` `T_(N_(2))=2T_(H_(2)) or T_(N_(2))gt T_(H_(2))`

Discussion

766.	At what temperature will the total `KE` of `0.3 mol` of `He` be the same as the total KE of `0.40 mol` of `Ar` at `400 K`A. `533 K`B. `400 K`C. `346 K`D. `300 K`
Answer» Correct Answer - A `KE=(3)/(2)nRT` `(3)/(2)xx0.30xxT=(3)/(2)xx0.4xx400` `T= 533K`

Discussion

767.	Calculate the average kinetic energy (in joule) per molecule in `8.0 g` of methane at `27^(@)C`.A. `6.21 xx10^(-20) J// "molecule"`B. `6.21 xx10^(-21) J// "molecule"`C. `6.21 xx10^(-22) J// "molecule"`D. `3.1 xx10^(-22) J// "molecule"`
Answer» Correct Answer - B Average `KE=(3)/(2)xx(8.314)/(6.023xx10)^(23)` `6.21xx10^(-21)J//"molecule"`

Discussion

768.	What will be the temperature when the rms velocity is four times of that at 300 K ?A. 300 KB. 900 KC. 4800 KD. 1200 K
Answer» (c ) `U_(rms)("velocity")=sqrt((3RT)/(M))` Hence, `upropsqrttrArr(u_(1))/(u_(2))=sqrt((T_(1))/(T_(2)))` From Eq. (i), `(1)/(4)=sqrt((300)/(T_(2)))` `(1)/(16)=(300)/(T_(2))` `T^(2)=16xx300` 4800 K

Discussion

769.	The volume of a gas is directly proportional to the temperature (constant `n,p`), if the temperature is expressed on the (i) thermodynamic scale ,(ii) Kelvin scale (iii) absolute scale , (iv) Celsius scaleA. `(i),(ii),(iii),(iv)`B. `(ii),(ii),(iii)`C. `(ii),(iii)`D. `(i),(ii)`
Answer» Correct Answer - B A scale of temperature based on the absolute zero of temperature is known as the absolute scale of temperature. Other mames are Kelvin scale (as Lord Kelvin first suggested this scale ) and thermodynamic scale (as this scale can be justified by thermodynamics). This scale is used in each scientific work because many equations assume simple form on this scale.

Discussion

770.	According to Avogadro,s law, `V = k_(4)T`. The value of `k_(4)` (proportionality constant) depends upon (i) temperature (ii) pressure (iii) volume (iv) nature of gasA. `(i),(ii),(iii)`B. `(ii),(iii),(iv)`C. `(i),(ii)`D. `(i),(ii),(iv)`
Answer» Correct Answer - C The value of `k_(4)` depends on the temperature and pressure of the gas. The units of `k_(4)` are determined by the units of volume. In fact, the value of `k_(4)` increases with the increase of temperature but decreases with the increase of pressure.

Discussion

771.	The ratio, `("rms velocity of" SO_(2))/("rms velocity of He")`, of sulpur dioxide and helium gases at `30^(@)C` is equal to:A. `4`B. `0.25`C. `0.10`D. `8`
Answer» Correct Answer - B `((U_(rms))SO_(2))/((U_(rms))_(He))=sqrt((M_(He))/(M_(SO_(2))))` `sqrt((4)/(64))=(1)/(4)= 0.25`

Discussion

772.	A temperature at which `rms` speed of `SO_(2)` molecule is half of that of helium molecules at `300 K`A. `1200 K`B. `600 K`C. `800 K`D. `900 K`
Answer» Correct Answer - A `r_(SO_(2))=sqrt((3RT)/(64)), r_(He)=sqrt((3Rxx300)/(4))` `:. sqrt((3Rxx300)/(4))=2sqrt((3RT)/(64)) or (3Rxx300)/(4)` `=4xx(3RT)/(64)` or `T=(64)/(14)xx300= 1200K`

Discussion

773.	The average velocity of gas molecules is `400 m s^(-1)`. Calculate their `rms` velocity at the same temperature.
Answer» Correct Answer - `435ms^(-1)`

Discussion

774.	At what temperature will the `rms` velocity of `SO_(2)` be the same as that of `O_(2) at 303 K`?A. `606 K`B. `403 K`C. `273 K`D. `303 K`
Answer» Correct Answer - A According to the kinetic molecular theory of gases , we have `u_(rms) = sqrt(3RT)/(M)` We are given `(u_(rms))_(SO_(2) = (u_(rms))_(O_(2))` `sqrt((T_(SO_(2)))/(M_(SO_(2)))) = sqrt((T_(O_(2)))/(M_(O_(2))))` Squaring both sides , we get `(T_(SO_(2)))/(M_(SO_(2))) = (T_(O_(2)))/(M_(O_(2)))` `T_(SO_(2)) = (T_(O_(2)M_(SO_(2))))/(M_(O_(2)))` `= ((303 K)(64 g mol^(-1)))/((32 g mol^(-1)))` `= 606 K`

Discussion

775.	At what temperature, the rms velocity of `SO_(2)` be same as that of `O_(2)` at 303K ?A. 273 KB. 606 KC. 303 KD. 403 K
Answer» (b) `(U_(rms)(SO_(2)))/(U_(rms)(O_(2)))=sqrt((T(SO_(2)))/(M(SO_(2)))xx(M(O_(2)))/(T(O_(2))))` i.e. `1=sqrt((T(SO_(2)))/(64)xx(32)/(303))or T(SO_(2))=606 K`

Discussion

776.	Sodium metal crystallizes in a body -centred cubic lattice with a unit cell edge of 4.29 Å. The radius of sodium metal is approximatelyA. 5.72 ÅB. 0.93 ÅC. 1.86 ÅD. 3.22 Å
Answer» Correct Answer - C For BCC, `r=sqrt3/4 a =1.732/4xx4.29 Å=1.86 Å`

Discussion

777.	A crystalline solid of a pure substance has a face-centred cubic structure with a cell edge of 400 pm. If the density of the substance in the crystal is `8 g cm^(-3)`, then the number of atoms present in 256g of the crystal is `N xx 10^(24)`. The value of `N` is
Answer» Correct Answer - 2 We know, `Z=(a^(3)xxd xx N_(A))/(m)` Molar mass of solid can be calculated as, `m=(a^(3)xxd xx N_(A))/(Z)` `=((400xx10^(-10))^(3)xx8xxN_(A))/(4)` `=(64xx2xx10^(-24))N_(A)` Number of atoms `=("Mass")/("Molar mass")xxN_(A)` `=(256)/((64xx2xx10^(-24)N_(A)))xxN_(A)` `=2xx10^(24)` `therefore " "N=2`

Discussion

778.	A 500 mL sample of a gas weighs 0.326 g at `100^(@)C` and 0.500 atm. What is the molecular mass of the gas
Answer» Correct Answer - A::C

Discussion

779.	Cu metal crystallies in face centred cubic lattic with cell edge, a =361.6pm. What is the density of Cu crystal ? (Atomic mass of copper =63.5 amu, `N_(A)=6.023xx10^(23)`)
Answer» We know `Z=(a^(3)xxrhoxxN_(A))/(M)` `:. Rho=(ZM)/(a^(3)N_(A))` Z=4 for fcc unit cell , `M =63.5 g mol^(-1)` `a= 361.6xx10^(-10) cm, N_(A)=6.023xx10^(23)` Putting these values in (i), we get `rho=(4xx63.5)/((36.6xx10^(-10))^(3)xx6.023xx10^(23))=8.94 g cm ^(-3)`

Discussion

780.	If a metal crystallises in a face-centred cubic structure with metallic radius of 25 pm, the number of unit cells in `1 cm^(3)` of lattice is :A. `2.828 xx 10^(28)`B. `1.414xx10^(28)`C. `1.414xx10^(24)`D. `2.828xx 10^(24)`
Answer» Correct Answer - D `asqrt(2)=4r` `axx 1.414=4xx25` `a=70.72` pm `=70.72xx10^(-10)` cm Volume of one unit cell `=a^(3)=(70.72xx10^(-10))^(3)` `=3.536xx10^(-25) cm^(3)` Number of unit cells in `1 cm^(3)` lattice `=(1)/(3.536xx10^(-25))` `=2.828xx10^(24)`

Discussion

781.	Two gases A and B having the same volume diffuse through a porous partition in 20 and 10 seconds respectively. The molecular mass of A is 49 u. Molecular mass of B will beA. 25.00 uB. 50.00 uC. 12.25 uD. 6.50 u
Answer» Correct Answer - A::B::C

Discussion

782.	The density of a gas is found to be 3.43 g/litre at 300K and 1.00 atm pressure. Calculate the molar mass of the gas.
Answer» Correct Answer - A::D

Discussion

783.	Assertion (A) Gases do not liquefy above their critical temperature, even on applying high pressure. Reason (R) Above critical temperature, the molecular speed is high and intermolecular attractions cannot hold the molecules together because they escape because of high speed.A. Both A and R are true and R is the correct explanation of A.B. Both A and R are true and R is not the correct explanation of A.C. A is true but R is false.D. A is false and R is true.
Answer» Correct Answer - A Reason is the correct explanation for assertion

Discussion

784.	Copper crystallises in face-centred cubic lattice with a unit cell length of 361 pm. What is the radius of copper atom in pm?A. 108B. 128C. 157D. 181
Answer» Correct Answer - B `asqrt(2)=4r` `361xxsqrt(2)=4r` r=128 pm

Discussion

785.	Assertion (A) : At critical temperature, liquid passes into gaseous state imperceptibly and continuously. Reason (R ) : The density of liquid and gaseous phase is equal at critical temperature.A. Both A and R are true and R is the correct explanation of A.B. Both A and R are true R is not the correct explanation of A.C. A is true but R is false.D. A is false but R is true.
Answer» Correct Answer - A

Discussion

786.	In the equation PV=RT, the value of R will not depend on (one or more)A. the nature of the gasB. the temperature of the gasC. the pressure of the gasD. units of measurement.
Answer» Correct Answer - A,B,C,D

Discussion

787.	Statement-1. The pressure of a fixed amount of an ideal gas is proportional to its temperature. Statement-2. Frequency of collisions and their impact both increase in proportion to the square root of the temperature.A. Statement-1 is correct, Statement-2 is correct , Statement-2 is the correct explanation for Statement-1.B. Statement-1 is correct, Statement-2 is correct , Statement-2 is not a correct explanation of Statement-1.C. Statement-1 is correct, Statement-2 is incorrect.D. Statement-1 is incorrect, Statement-2 is correct.
Answer» Correct Answer - A

Discussion

788.	Which the following represents the molar volume of the gas correctlyA. `22.4" L at "0^(@)C` and 1 atm pressureB. `22.7" L at "0^(@)C` and 1 bar pressureC. `24.8" L at "` SATP conditionsD. `22.5" L at "25^(@)C` and 1 bar pressure .
Answer» Correct Answer - A,B,C

Discussion

789.	The density of lead is `11.35 "g cm"^(-3)` and the metal crystallizes with fcc unit cell. Estimate the radius of lead atom. (At. Mass of lead = `207 "g mol"^(-1)` and `N_A=6.02xx10^23 "mol"^(-1)`).
Answer» Correct Answer - 174.7 pm `rho=(ZxxM)/(a^3xxN_A)` or `a^3=(ZxxM)/(N_Axxrho)=(4xx207 g mol^(-1))/(6.02xx10xx23 mol^(-1)xx11.35 g cm^(-3))=12.118 xx10^(-23) =121.18xx10^(-24)` or `a=(121.18)^(1//3)xx10^(-8)` Let `x=(121.18)^(1//3) therefore log x =1/3 log 121.18=1/3(2.083)=0.694` or x =antilog 0.694=4.943 `therefore a=4.943xx10^(-8) cm=494.3xx10^(-10) `cm=494.3 pm For fcc, `r=a/(2sqrt2)`=0.3535a=0.3535 x 494.3 pm =174.7 pm

Discussion

790.	Assertion (A) : Gases do not liquefy above their critical temperature even on applying high pressure. Reason (R ) : Above critical temperature, the molecular speed is high and intermolecular attractions cannot hold the molecules together because they escape because of high speed.A. Both A and R are true and R is the correct explanation of A.B. Both A and R are true R is not the correct explanation of A.C. A is true but R is false.D. A is false but R is true.
Answer» Correct Answer - A

Discussion

791.	Graphite is a good conductor of electricity due to the presence of _____A. lone pair of electronsB. free valence electronsC. cationsD. anions
Answer» Correct Answer - b

Discussion

792.	`2.9 g` of a gas at `95^(@) C` occupied the same volume as `0.184 g` of hydrogen at `17^(@)C` at same pressure What is the molar mass of the gas ? .A. `40 g mol^(-1)`B. `50 g mol^(-1)`C. `60 g mol^(-1)`D. `30 g mol^(-1)`
Answer» Correct Answer - A According to the ideal gas law `(pV = nRT)`, we have `p_(gas)V_(gas) = n_(gas)RT_(gas)` `p_(H_(2)) = n_(H_(2))RT_(H_(2))` Since `p_(gas) = pH_(2)` and `V_(gas) = V_(H_(2))` `n_(gas)RT_(gas) = n_(H_(2))RT_(H_(2))` or `n_(gas) T_(gas) = n_(H_(2))T_(H_(2))` Since `n = (Mass(m))/(Molar mass(M))` we can write `(m_(gas))/(M_(gas)) T_(gas) = (m_(H_(2)))/(M_(H_(2)))T_(H_(2))` or `M_(gas) = (M_(H_(2))m_(gas)T_(gas))/(m_(H_(2))T_(H_(2)))` `= ((2 g mol^(-1))(2.9g) (95 + 273 K))/((0.184 g)(17 + 273 K))` `= 40 g mol^(-1)`

Discussion

793.	Calculate the density of silver which crystallizes in a face-centred cubic structure. The distance between the nearest silver atoms in this structure is 287 pm. (Molar mass of Ag =`107.87 " g mol"^(-1), N_A=6.02xx10^23 "mol"^(-1)`)
Answer» Correct Answer - `10.71 "g cm"^(-3)` For FCC, `d=a/sqrt2` or a=`sqrt2`d=1414 x 287 pm = 406 pm , Z=4, M=`107. 87 "g mol"^(-1)` `rho=(ZxxM)/(a^3xxN_0)=(4xx107.87 "g mol"^(-1))/((406xx10^(-10) cm)^3xx(6.02xx10^23 mol^(-1)))=10.71 g cm^(-3)`

Discussion

794.	Which kind of defect is found in KCl crystal?A. FrenkelB. SchottkyC. LinearD. Impurity
Answer» Correct Answer - b

Discussion

795.	Write the coordination numbers of cations and anions in the following ionic compounds : (a)Zinc blende (b)Fluorite
Answer» Zinc blende=ZnS, `Zn^(2+)=4 , S^(2-)`=4 Fluorite=`CaF_2, Ce^(2+)`=8 , `F^-` = 4

Discussion

796.	`3.7 gm` of gas at `25^(@)C` occupied the same volume as `0.184 gm` of hydrogen at `17^(@)C ` and at the same pressure. What is the molecular mass of the gas ?
Answer» For hydrogen, `w = 0.184 g , T = 17 + 273 = 290 K, M = 2` We know that, `PV = (w)/(M) RT` `= (0.184)/(2) xx R xx 290`.....(i) For unknown gas, `w = 3.7 g, T = 25 + 273 = 298 K, M = ?` `PV = (3.7)/(M) xx R xx 298`....(ii) Equating both the equations, `(3.7)/(M) xx R xx 298 = (0.184)/(2) xx R xx 290` or `M = (3.7 xx 298 xx 2)/(0.184 xx 290) = 41.33`

Discussion

797.	The lattice site in a pure crystal cannot be occupied by _____A. moleculeB. ionC. electronD. atom
Answer» Correct Answer - c

Discussion

798.	Niobium crystallizes in a body centred cubic structure. If density is `8.55 "g cm"^(-3)`, calculate atomic radius of niobium, given that its atomic mass is 93 u.
Answer» `a^3=(MxxZ)/(rhoxxN_0xx10^(-30))=(93 "g mol"^(-1)xx2)/("8.55 g cm"^(-3)xx6.02xx10^23 "mol"^(-1)xx10^(-30))=3.61xx10^7=36.1xx10^6` `therefore a=(36.1)^(1//3)xx10^2 "pm"=3.304xx10^2 "pm=330.4 pm"` `[x=(36.1)^(1//3), log x=1/3 "log"36.1 =1/3 xx"1.5575=0.519 or x =antilog 0.519=3.304"]` For body-centred cubic, `r=sqrt3/4 a` =0.433 a =0.433 x 330.4 pm = 143.1 pm

Discussion

799.	Schottky defect is observed in crystal when ______A. some cations move from their lattice sites to interstitial sitesB. equal number of cations and anions are missing from the lattice.C. some lattice sites are occupied by electronsD. some impurity is present in the lattice.
Answer» Correct Answer - b

Discussion

800.	In the fluorite structure the coordination number of `Ca^(2+)` ion isA. 4B. 6C. 8D. 3
Answer» Correct Answer - C In fluorine, `Ca^(2+)` ions are in ccp and `F^(-)` ions occupy all the tetrahedral voids. Coordination number will be : (8:4).

Discussion

Explore topic-wise InterviewSolutions in .

The ratio of van der Waals constant (a/b) has the dimensions :A. atm `L^(-1)`B. L atm `L^(-1)`C. atm mol `L^(-1)`D. atm L `mol^(-2)`

Dimensions of surface tension areA. `kg s^(-1)`B. `kg s`C. `kg s^(-2)`D. `kg s^(2)`

The effect of increase of temperature on surface tension and viscosity is that ………………… .

Relative humidity of air is `60^(@)C` and the saturation vapour pressure of water vapour in air is `3.6kPa`. The amount of water vapours present in `2 L` air at `300 K` isA. `52 g`B. `31.2 g`C. `26 g`D. `5.2 g`

A neon-dioxygen mixture contains 70.6 g dioxygen and 167.5 g neon. If pressure of the mixture of gases in the cylinder is 25 bar. What is the partial pressure of dioxygen and neon in the mixture ?

Value of gas constant R in the ideal gas equation PV = nRT depends uponA. temperature of the gasB. pressure of the gasC. units in which P , V and T are measuredD. nature of the gas .

Temperature at which `r.m.s` speed of `O_(2)` is equal to that of neon at `300 K` is:A. `280 K`B. `480 K`C. `680 K`D. `180 K`

For the given ideal gas equation `PV=nRT`, answer the following questions: Which of the following does not represent ideal gas equation?A. `PV=(1)/(3)mNv`B. `PV=nRT`C. `P=rho(RT)/(M)`D. `PV=RT`

Which of the following species are capable of hydrogen bonding among themselves?A. `H_(2)S`B. `C_(6)H_(6)`C. `CH_(3)OH`D. `CH_(4)`

Average kinetic energy per molecule of a gas is related to its temperature as `bar KE`=………………… .

The rms velocity of hydrogen is `sqrt(7)` times the rms velocity of nitrogen. If `T` is the temperature of the gas, thenA. `T_(H_(2))=T_(N_(2))`B. `T_(H_(2))gtT_(N_(2))`C. `T_(H_(2)) lt T_(N_(2))`D. `T_(H_(2))=sqrt(7)T_(N_(2))`

The rms velocity of hydrogen is `sqrt(7)` times the rms velocity of nitrogen. If `T` is the temperature of the gas, thenA. `T(H_(2))=T(N_(2))`B. `T(H_(2)) gt T(N_(2))`C. `T(H_(2)) lt T(N_(2))`D. `T(H_(2))=sqrt(7)T(N_(2))`

At what temperature will the total `KE` of `0.3 mol` of `He` be the same as the total KE of `0.40 mol` of `Ar` at `400 K`A. `533 K`B. `400 K`C. `346 K`D. `300 K`

Calculate the average kinetic energy (in joule) per molecule in `8.0 g` of methane at `27^(@)C`.A. `6.21 xx10^(-20) J// "molecule"`B. `6.21 xx10^(-21) J// "molecule"`C. `6.21 xx10^(-22) J// "molecule"`D. `3.1 xx10^(-22) J// "molecule"`

What will be the temperature when the rms velocity is four times of that at 300 K ?A. 300 KB. 900 KC. 4800 KD. 1200 K

The volume of a gas is directly proportional to the temperature (constant `n,p`), if the temperature is expressed on the (i) thermodynamic scale ,(ii) Kelvin scale (iii) absolute scale , (iv) Celsius scaleA. `(i),(ii),(iii),(iv)`B. `(ii),(ii),(iii)`C. `(ii),(iii)`D. `(i),(ii)`

According to Avogadro,s law, `V = k_(4)T`. The value of `k_(4)` (proportionality constant) depends upon (i) temperature (ii) pressure (iii) volume (iv) nature of gasA. `(i),(ii),(iii)`B. `(ii),(iii),(iv)`C. `(i),(ii)`D. `(i),(ii),(iv)`

The ratio, `("rms velocity of" SO_(2))/("rms velocity of He")`, of sulpur dioxide and helium gases at `30^(@)C` is equal to:A. `4`B. `0.25`C. `0.10`D. `8`

A temperature at which `rms` speed of `SO_(2)` molecule is half of that of helium molecules at `300 K`A. `1200 K`B. `600 K`C. `800 K`D. `900 K`

The average velocity of gas molecules is `400 m s^(-1)`. Calculate their `rms` velocity at the same temperature.

At what temperature will the `rms` velocity of `SO_(2)` be the same as that of `O_(2) at 303 K`?A. `606 K`B. `403 K`C. `273 K`D. `303 K`

At what temperature, the rms velocity of `SO_(2)` be same as that of `O_(2)` at 303K ?A. 273 KB. 606 KC. 303 KD. 403 K

Sodium metal crystallizes in a body -centred cubic lattice with a unit cell edge of 4.29 Å. The radius of sodium metal is approximatelyA. 5.72 ÅB. 0.93 ÅC. 1.86 ÅD. 3.22 Å

A crystalline solid of a pure substance has a face-centred cubic structure with a cell edge of 400 pm. If the density of the substance in the crystal is `8 g cm^(-3)`, then the number of atoms present in 256g of the crystal is `N xx 10^(24)`. The value of `N` is

A 500 mL sample of a gas weighs 0.326 g at `100^(@)C` and 0.500 atm. What is the molecular mass of the gas

Cu metal crystallies in face centred cubic lattic with cell edge, a =361.6pm. What is the density of Cu crystal ? (Atomic mass of copper =63.5 amu, `N_(A)=6.023xx10^(23)`)

If a metal crystallises in a face-centred cubic structure with metallic radius of 25 pm, the number of unit cells in `1 cm^(3)` of lattice is :A. `2.828 xx 10^(28)`B. `1.414xx10^(28)`C. `1.414xx10^(24)`D. `2.828xx 10^(24)`

Two gases A and B having the same volume diffuse through a porous partition in 20 and 10 seconds respectively. The molecular mass of A is 49 u. Molecular mass of B will beA. 25.00 uB. 50.00 uC. 12.25 uD. 6.50 u

The density of a gas is found to be 3.43 g/litre at 300K and 1.00 atm pressure. Calculate the molar mass of the gas.

Copper crystallises in face-centred cubic lattice with a unit cell length of 361 pm. What is the radius of copper atom in pm?A. 108B. 128C. 157D. 181

In the equation PV=RT, the value of R will not depend on (one or more)A. the nature of the gasB. the temperature of the gasC. the pressure of the gasD. units of measurement.

Which the following represents the molar volume of the gas correctlyA. `22.4" L at "0^(@)C` and 1 atm pressureB. `22.7" L at "0^(@)C` and 1 bar pressureC. `24.8" L at "` SATP conditionsD. `22.5" L at "25^(@)C` and 1 bar pressure .

The density of lead is `11.35 "g cm"^(-3)` and the metal crystallizes with fcc unit cell. Estimate the radius of lead atom. (At. Mass of lead = `207 "g mol"^(-1)` and `N_A=6.02xx10^23 "mol"^(-1)`).

Graphite is a good conductor of electricity due to the presence of _____A. lone pair of electronsB. free valence electronsC. cationsD. anions

`2.9 g` of a gas at `95^(@) C` occupied the same volume as `0.184 g` of hydrogen at `17^(@)C` at same pressure What is the molar mass of the gas ? .A. `40 g mol^(-1)`B. `50 g mol^(-1)`C. `60 g mol^(-1)`D. `30 g mol^(-1)`

Calculate the density of silver which crystallizes in a face-centred cubic structure. The distance between the nearest silver atoms in this structure is 287 pm. (Molar mass of Ag =`107.87 " g mol"^(-1), N_A=6.02xx10^23 "mol"^(-1)`)

Which kind of defect is found in KCl crystal?A. FrenkelB. SchottkyC. LinearD. Impurity

Write the coordination numbers of cations and anions in the following ionic compounds : (a)Zinc blende (b)Fluorite

`3.7 gm` of gas at `25^(@)C` occupied the same volume as `0.184 gm` of hydrogen at `17^(@)C ` and at the same pressure. What is the molecular mass of the gas ?

The lattice site in a pure crystal cannot be occupied by _____A. moleculeB. ionC. electronD. atom

Niobium crystallizes in a body centred cubic structure. If density is `8.55 "g cm"^(-3)`, calculate atomic radius of niobium, given that its atomic mass is 93 u.

Schottky defect is observed in crystal when ______A. some cations move from their lattice sites to interstitial sitesB. equal number of cations and anions are missing from the lattice.C. some lattice sites are occupied by electronsD. some impurity is present in the lattice.

In the fluorite structure the coordination number of `Ca^(2+)` ion isA. 4B. 6C. 8D. 3