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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.
| 801. |
What is the pressure of `HCl` gas at `- 40^(@)C` if its density is `8.0 kg m^(-3) ? (R = 8.314 J K^(-1) mol^(-1))` |
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Answer» Equation for ideal gas, `PV = (w)/(M) RT` or `P = (w)/(V) xx (RT)/(M)` `= d xx (RT)/(M) ((w)/(V) = " density " = d)` Given, `d = 8.0 kg m^(-3), R = 8.314 K^(-1) mol^(-1)` `T = - 40 + 273 = 233 K` and `M = 36.5 g mol^(-1) = 36.5 xx 10^(-3) kg mol^(-1)` Substituting the values in the above equation, `P = (8.0 xx 8.314 xx 233)/(36.5 xx 10^(-3)) = 424.58 xx 01^(3) Pa` |
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| 802. |
Which of the following statements are correct ?A. The coordination number of each type of ion in CsCl crystal is 8B. A metal that crystallizes in bcc structure has coordination number of 12C. A unit cell of an ionic crystal shares some of its ions with other unit cells.D. The length of the edge of unit cell of NaCl is 552 pm (`r_(Na^+)` =95 pm, `r_(Cl^-)`=181 pm) |
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Answer» Correct Answer - A,C,D (d) is correct because length of unit cell `=2(r_(Na^+)+r_(Cl^-))`=2(95+181) pm (a) and (c ) are also corect Only (b) is wrong because for a metal with bcc structure, coordination number =8 |
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| 803. |
In the rock salt structure, the number of formula units per unit cell is equal to :A. 2B. 3C. 8D. 4 |
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Answer» Correct Answer - D In rock salt structure, there are four cations and four anions hence Z=4 |
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| 804. |
KBr has NaCl type sturcture . Its density is 2.75 g `cm^(-3)` . Edge length of unit cell will be: (Molar mass of KBr =119 g `"mol"^(-1)`)A. `3.3xx10^(-8)` cmB. `6.6 xx 10^(-8)` cmC. `9.9xx10^(-8)` cmD. `1.6xx10^(-8)` cm |
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Answer» Correct Answer - B `a=[(ZM)/(rhoN_(A))]^(1//3)=[(4xx119)/(2.75xx6.023xx10^(23))]^(1//3)=6.6xx10^(-8)` cm |
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| 805. |
The density of KBr is `2.75 g//cm^3` . The length of the unit cell of 654 pm. Atomic mass of K=39, Br=80. Then what is true about the predicted nature of the solid ?A. It has `4 K^+` and `4 Br^-` ions per unit cellB. It is face-centredC. It has rock-salt type structureD. It can have Schottky defects |
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Answer» Correct Answer - A,B,C,D `Z=(rhoxxa^3xx10^(-30)xxN_0)/M` `=(2.75xx(654)^3xx10^(-30)xx6.02xx10^23)/119=4` Thus, it has 4 formula units per unit cell. |
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| 806. |
The density of a certain solid AB (formula mass =119) is 2.75 `g//cm^(3)` .The edge of the unit cell is 654 pm long .What is/are true about the solid AB ?A. It has bcc unit cellB. There are four constituents per unit cellC. unit cell constituted by anions is fccD. structure is similar to ZnS |
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Answer» Correct Answer - B::C `Z=(a^(3)xxrhoxxN)/(M)` `=((654xx10^(-10))^(3)xx2.75xx6.023xx10^(23))/(119)~~4` |
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| 807. |
A unit of cell of sodium chloride has four formula units. The edge length of the unit cell is `0.564 nm`. What is the density of sodium chloride? |
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Answer» `rho=(ZM)/(a^(3)N)=(4xx58.5)/((5.64xx10^(-8))^(3)xx6.023xx10^(23))` `=2.16 g cm^(-3)` |
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| 808. |
The density of potassium bromide crystal is `2.75 g cm^(-3)` and the length of an edge of a unit cell is 654 pm. The unit cell of KBr is one of three types of cubic unit cells. How many formula units of KBr are there in a unit cell ? Does the unit cell have a NaCl or CsCl structure ? |
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Answer» We know that, `rho=(N)/(a^(3))[(M)/(N_(A))]` `N=(rhoxxa^(3)xxN_(A))/(M)` `=(2.75xx(654xx10^(-10))^(3)xx6.023xx10^(23))/(119)=3.89 ~~4` Number of mass points per unit cell =4 It is NaCl type crystal, i.e., fcc structure. |
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| 809. |
`AB` crystallises in a bcc lattice with edge length `qa` equal to `387` pm`.The distance between two oppositely chariged ions in the lattice isA. 335 pmB. 250 pmC. 200 pmD. 300 pm |
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Answer» Correct Answer - A Distance between two oppositely charged ions `(r^(+)+r^(-))=(asqrt(3))/(2)=(387xxsqrt(3))/(2)` `=335.14` pm |
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| 810. |
Equal masses of helium and oxygen are mixed in a container at `25 C^(@)` . The fraction of the total pressure exerted by oxygen in the mixture of gases isA. 1/3B. 2/3C. 1/9D. 4/9 |
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Answer» Correct Answer - C Mole fraction of `O_(2)=((w)/(32))/((w)/(32)+(w)/(4))=(w)/(32)xx(32)/(9w)=(1)/(9)` partial pressure of oxygen `=Pxx x_(O_(2))` `=P_("total")xx(1)/(9)` or `(1)/(9)P_("total")` |
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| 811. |
Which of the following statement is/are true? According to kinetic theory of gaseA. Collisions are always elastic.B. Heavier molecules transfer more momentum to the wall of the container.C. Only a small number of molecules have very high velocity.D. Between collisions, the molecules move in straight lines with constant velocities. |
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Answer» (`a`) True (`b`) `P=sqrt(2m) KE`, `KE` is same at same temperature. So, pressure is high, if `m` is higher. (`c`) True (`d`) True |
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| 812. |
The gas constant has unitsA. `L atm K^(-1)mol^(-1)`B. `L atm^(-1) K^(-1)mol^(-1)`C. `atm cm^(3)K^(-1)mol^(-1)`D. `ergK^(-1)` |
| Answer» `R=(PV)/(nT) :. R(unit)=L atm K^(-1) mol^(-1)` and `atm cm^(3) K^(-1) mol^(-1)` | |
| 813. |
In van der Waals equation of gases, the kinetic equation for gas is modified with respect toA. Repulsive forcesB. Attractive forces between the gaseous moleculesC. Actual volume of the gasD. Pressure of the molecules |
| Answer» Volume and pressure of gas | |
| 814. |
The density of a gas at `27^(@)C` and `1 atm` is `d`. Pressure remaining constant, at which of the following temperture will its density become `0.75 d`?A. `20^(@)C`B. `30^(@)C`C. `400 K`D. `300 K` |
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Answer» `(d)_(1)=d` `(d)_(2)=0.75 d` `P_(1)= 1 atm`, `T_(1)=27^(@)C=300 K` `P_(2)=1 atm`, `T_(2)=?` `((d_(1))/(d_(2)))=(P_(1)xxT_(2))/(T_(1)xxP_(2))` or `(T_(2))/(T_(1))` `:. T_(2)=(d_(1))/(d_(2))xxT_(1)=(d)/(0.75d)xx300=400K` |
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| 815. |
What is the difference between barometer and manometer? |
| Answer» Gas pressure is due to bombardment of gas molecules against the walls of the container. | |
| 816. |
One mole of which of the following will have `22.7 L` at `STP` (` 1` bar, `273.15 K`)?A. `SO_(2)`B. `He`C. `H_(2)O`D. `C Cl_(4)` |
| Answer» Correct Answer - `SO_(2)` gas | |
| 817. |
`3.2 g` oxygen is diffused in `10 min`. In similar conditions, `2.8 g` nitrogen will diffuse inA. `9.3 min`B. `8.2 min`C. `7.6 min`D. `11.8 min` |
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Answer» Rate of diffusion `prop sqrt((1)/(M))` `:. ((1)/(t_(O_(2))))/((1)/t_(N_(2)))=sqrt((M_(N_(2)))/(M_(O_(2))))=sqrt((2.8)/(3.2))=0.93` or `(t_(N_(2)))/(t_(O_(2)))=0.93` `:. T_(N_(2))=0.93xx10=9.3 min` |
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| 818. |
The kinetic theory of gases predicts that total kinetic energy of a gaseous assembly depends onA. Pressure of the gasB. Temperature of the gasC. Volume of the gasD. Pressure, temperature, and volume of the gas |
| Answer» `KE=(3)/(2)RT` or `(3)/(2)PV` | |
| 819. |
Assertion : Molar volume of an ideal gas at 273 . 15 K and 1 bar is 22.4 L. Reason : Volume of a gas is inversely proportional to temperature .A. If both assertion and reason are true and reason is the correct explanation of assertion .B. If both assertion and reason are true but reason is not the correct explanation of assertion .C. If assertion is true but reason is false .D. If both assertion and reason are false . |
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Answer» Correct Answer - D Molar volume of an ideal gas at 273.15 K and 1 atm is 22.4 L whereas molar volume of an ideal gas at 273.15 K and 1 bar is 22.7 L. Also, the volume of a fixed mass of a gas is directly proportional to its absolute temperature at constant pressure. |
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| 820. |
Why do we add 273 to the temperature while dealing with problems on gas equation? |
| Answer» it cuts the volume axis at `0^(@)C` but it cuts the temperature axis at `-273.15^(@)C` i.e., absolute zero. | |
| 821. |
At `STP`, the order of mean square velocity of molecules of `H_(2)`, `N_(2)`, `O_(2)`, and `HBr` isA. `H_(2) gt N_(2) gtO_(2) gt HBr`B. `HBr gt O_(2) gt N_(2) gt H_(2)`C. `HBr gt H_(2) gt O_(2) gt N_(2)`D. `N_(2) gt O_(2) gt H_(2) gt HBr` |
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Answer» Correct Answer - A `U_(rms)= sqrt((3RT)/(M_(w)))` |
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| 822. |
At `STP`, the order of mean square velocity of molecules of `H_(2)`, `N_(2)`, `O_(2)`, and `HBr` isA. `H_(2)gtN_(2)gtO_(2)gtHBr`B. `HBrgtO_(2)gtM_(2)gtH_(2)`C. `HBrgtH_(2)gtO_(2)gtN_(2)`D. `N_(2)gtO_(2)gtH_(2)gtHBr` |
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Answer» `mu_(rms)=sqrt((3 RT)/(M))` or `mu_(rms) prop sqrt((1)/(M))` Hence, `HegtN_(2)gtO_(2)gtHBr` |
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| 823. |
At `STP`, the order of mean square velocity of molecules of `H_(2)`, `N_(2)`, `O_(2)`, and `HBr` isA. `H_(2)ltN_(2)ltO_(2)ltHBr`B. `HBrltO_(2)ltN_(2)ltH_(2)`C. `H_(2)ltN_(2)=O_(2)ltHBr`D. `HBrltO_(2)ltH_(2)ltN_(2)` |
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Answer» Correct Answer - B The root mean square velocity of gas molecule at STP given by `U_(rms)=sqrt((3RT)/(M))` `therefore` As the molar mass of gas increases, then `U_(rms)` will decrease, so the order of `U_(rms)` of these gases is `HBr ltO_(2) lt N_(2) lt H_(2)` |
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| 824. |
Van der Waals real gas acts an ideal gas at which conditions?A. high temperature, low pressureB. low temperature, high pressureC. high temperature, high pressureD. low temprature, low pressure |
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Answer» Correct Answer - A a real gas acts as ideal gas under high temperature and low pressure. |
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| 825. |
It P is the pressure and `rho` is the density of a gas, then P and `rho` are realted as :A. `P prop rho`B. `P prop rho^(2)`C. `P prop 1//rho`D. `P prop 1//rho^(2)` |
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Answer» Correct Answer - A is the correct answer. |
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| 826. |
A cotainer contains CO and `H_(2)` in equal molar concentrations at the same temperature. What is the ratio of the average molar kinetic energy of the two gases ?A. Depends upon volumeB. `1 : 1`C. `KE_(CO) gt KE_(H_(2))`D. `KE_(H_(2)) gt KE_(CO)` |
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Answer» Correct Answer - B Molar kinetic energies are also in the same ratio `(1 : 1)` |
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| 827. |
The figure given below shows three glass chambers that are connected by valves of negligible volume. At the outset of an experiment, the valves are closed and the chambers contain the gases as detailed in the diagram. All the chambers are at the temperature of `300 K` and external pressure of `1.0 atm`. What will be the work done by `N_(2)` gas when valve `2` is opened and value `1` remains closed?A. `8.2 L atm`B. `-8.2 atm`C. `0`D. `3.28 L atm` |
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Answer» Work done during the expansion of `N_(2)` at `0.82 atm` to chamber `B` at `P=0` `:. W=PV=0xx4=0` |
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| 828. |
A liquid distributed by stirring comes to rest after some time due to its property ofA. viscosityB. surface tensionC. compressibilityD. volatility |
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Answer» Correct Answer - A Viscosity is the resistance to the flow of a liquid . Strong intermolecular forces between molecules hold them together and resist the movement of layers past one another. |
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| 829. |
In diamond, the coordination number of carbon is:A. four and its unit cell has eight carbon atomsB. four and its unit cell has six carbon atomsC. six and its unit cell has four carbon atomsD. four and its unit cell has four carbon atoms |
| Answer» (a) The space lattice of diamond is fcc and has tetrahedral geometry. Thus, the convertional uit cube contains eight atoms with CN foiur. | |
| 830. |
Positive deviation from ideal behaviour takes place because ofA. finite size of atoms and `pV//nRT lt 1`B. molecular interaction between atoms and `pV//nRT gt 1`C. finite size of atoms and `pV//nRT gt 1`D. molecular interaction between atoms and `pV//nRT lt 1` |
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Answer» Correct Answer - C According to the van der Waals equation , we have `(p + (an^(2))/(V^(2))) (V - nb) = nRT` When molecular interaction (i.e., attraction ) is insignificant , then the `an^(2)//V^(2)` may be neglaected relative to `p` to give `p(V - nb) = RT` or `pV - npb = nRT` or `pV = nRT + npb` Dividing both sides by `nRT` , we have `(pV)/(nRT) = 1 + npb` Thus , `pV//nRT gt 1` This happens whenn the factor `b` which measures the size of particles is significiant. |
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| 831. |
Which of the following is not true for liquids ?A. They have no definite shape (assume shapes of containers).B. They have definite volume (are only very slightly compressible).C. They have high density .D. They are not fluids. |
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Answer» Correct Answer - D In the liquid state, molecules are held by one or more types of attractive forces. The forces of attraction among the particles are great enough that disordered clustering occurs. In a liquid , the molecules are held so close together that very little of the volume occupied by a liquid is empty space. As a result , it is very hard to compress a liquid than a gas. Also , liquids are much denser than gases under normal conditions. A liquid also has a definite volume , since molecules in a liquid do not break away from the attractive forces (except evaporation). However , the molecules in the liquid state are able to slide past one another freely (thus, they are called fluids). As a result , a liquid can flow , can be poured , and assumes the shape of its container up to the volume of the liquid. |
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| 832. |
Correct gas equation is :A. `(V_(1)T_(2))/(p_(1))=(V_(2)T_(1))/(p_(2))`B. `(p_(1)T_(1))/(V_(1))=(p_(2)V_(2))/(T_(2))`C. `(p_(1)V_(1))/(p_(2)V_(2))=(T_(1))/(T_(2))`D. `(V_(1)V_(2))/(T_(1)T_(2))=p_(1)p_(2)` |
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Answer» Correct Answer - C If temperature, volume and pressure of fixed amount (say n mole) of a gas vary from `T_(1),V_(1) and p_(1)" to " T_(2).V_(2)` and `p_(2)` respectively. Then, ideal gas equation for two states can be written as `p_(1)V_(1)=nRT_(1)` `"or "(p_(1)V_(1))/(T_(1))=nR" ...(i)"` `p_(2)V_(2)=nRT_(2)` `"or "(p_(2)V_(2))/(T_(2))=nR" ...(ii)"` On combining Eqs. (i) and (ii), we get `(p_(1)V_(1))/(T_(1))=(p_(2)V_(2))/(T_(2))` `"or "(p_(1)V_(1))/(p_(2)V_(2))=(T_(1))/(T_(2))` So, it called combined gas equation. Hence, (c) answer is correct. |
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| 833. |
Which of the following is correct regarding the liquid state? (i) A liquid can exist only between the boiling and melting point of a substance . (ii) Liquids diffuse slowly through other liquids. (iii) liquids consist of disordered clusters of particles that are quite close together, particles have random motion in three dimensions. (iv) A liquid resembles a gas near the critical temperature and resembles a solid near the melting point of the substance.A. `(i),(ii),(iii)`B. `(i),(ii),(iii),(iv)`C. `(i),(iii),(iv)`D. `(ii),(iii)` |
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Answer» Correct Answer - B A substance exists in the liquid state above its melting point and below its boiling point. Liquids diffuse into other liquids with which they miscible. For example, when a drop of red food coloring is added to a glass of water , the water becomes red throughout after diffusion is complete. Thus , the miscibility of two liquids refers to their ability to mix and produce a homogeneous solution . The natural diffusion rate is slow at normal temperatures because intermolecular distances are small , hence , the molecules (in the liquid state) undergo a large number of collisions with the neighboring molecules. Because the average separations among the particles in liquids are far less than those in gases , teh densities of liquids are much higher than the densities of gases. Because liquids are denser than gaes , the collision rate among molecules is much higher in the liquid phase than in the gas phase. |
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| 834. |
The ability of a substance to exist in two or more crstaline forms knows as:A. isomerismB. polymorphismC. isomorphismD. amorphism |
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Answer» Correct Answer - B Some substance adopt different structural arrangement under different conditions. Such arrangements are called polymorphs and this phenomenon is called polymorphism. |
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| 835. |
The composition of a sample of wustite is `Fe_(0.93)O_(1.00)`. What percentage of the iron is present in the form of Fe(III) ? |
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Answer» The composition is `Fe_0.93 O_1.00` instead and FeO because some `Fe^(2+)` ions have been replaced by `Fe^(+)` ions. Let us first calculate the number of `Fe^(2+)` and `Fe^(3+)` ions present. The formula `Fe_0.93 O_1.00` implies that 93 Fe atoms are combined with 100 O-atoms. Out of 93 Fe atoms , suppoe Fe atoms present as `Fe^(3+)=x`. Then `Fe^(2+)=93-x`.As the compound is neutral, total charge on `Fe^(2+)` and `Fe^(3+)` ions =total charge on `O^(2-)` ions. Thus, `3 xx x +2 (93-x)=2 xx 100` or 3x+ 186-2x =200 or x=14 , i.e., `Fe^(3+)=14` Hence, `Fe^(2+)` =93-14=79 Thus, out of 93 Fe atoms , Fe present as `Fe^(3+)`=14 `therefore` % age of Fe present of Fe (III) =`14/93xx100=15%` |
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| 836. |
Which of the following equations represents the compressibility factor?A. `Z = (nRT)/(pV)`B. `Z = (pV)/(nRT)`C. `Z = (pV)/(nR)`D. `Z = (pV)/(nT)` |
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Answer» Correct Answer - B The deviation from ideal behaviour can be measured in terms of the compressibility factor `(Z)` which is the ratio of the product `pV` to `nRT`. Mathematically, `Z = (pV)/(nRT)` |
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| 837. |
Examine the given defective crystal `{:(A^+, B^- , A^+ , B^- , A^+),(B^- , O, B^- , A^+ , B^-),(A^+, B^- , A^+ , O, A^+),(B^-, A^+, B^-, A^+, B^- ):}` Answer the following questions : (i)What type of stoichiometric defect is shown by the crystal ? (ii)How is the density of the crystal affected by this defect ? (iii)What type of ionic substances show such defect ? |
| Answer» (i)Schottky defect ,(ii) Density of the crystal decreases, (iii)This type of defect is shown by ionic compounds in which the ions have high coordination number and there is small difference in the size of cations and anions | |
| 838. |
The kinetic energy of `4 mol` of nitrogen gas at `127^(@) C` is …….. `cal(R = 2 cal mol^(-1) K^(-1))`A. `4400`B. `3200`C. `4800`D. `1524` |
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Answer» Correct Answer - C According to the kinetic molecular theory , kinetic energy `(KE)` for `n` moles is given as `KE = (3)/(2) nRT` `= (3)/(2) (4 "mol") (2 "cal mol"^(-1) K^(-1)) (400 K)` `= 4800 cal` |
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| 839. |
Kinetic theory of gases is a generalization offered by Maxwell. Boltzmann, Clausius, etc. to explain the behaviour of ideal gases. This theory assumes that ideal gas molecules neither attract nor repel each other. Average kinetic energy of gas molecules is directly proportional to the absolute temperature. A gas equation called kinetic gas equation was derived on the basis of kinetic theory `PV = (1)/(3) mv^(2)` The kinetic energy of 2.8 g of nitogen gas at `127^(@)C` is nearly :A. `2 xx 249.3 J`B. `2 xx 200.4 J`C. `2 xx 2.5 J`D. `20.5 J xx 2` |
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Answer» Correct Answer - A `KE = (3)/(2) nRT = (3)/(2) xx 0.1 xx 8.314 xx 400 = 498.84 J` |
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| 840. |
Maximum deviation from ideal gas is expected fromA. `N_(2) (g)`B. `CH_(4) (g)`C. `NH_(3) (g)`D. `H_(2)(g)` |
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Answer» Correct Answer - C Due to `H`-bonding, attractive forces are stronger in `NH_(3)`. So it can be easily liquefied and easily liquefiable gases show maximum deviation from the ideal gas. |
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| 841. |
When `100 ml` sample of methane and ethane along with excess of `O_(2)` is subjected to electric spark, the contraction in volume was observed to be `212 ml`. When the resulting gases were passed through `KOH` solution, further contraction in volume wasA. `60 ml`B. `96 ml`C. `108 ml`D. `124 ml` |
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Answer» Correct Answer - D `{:(CH_(4)(g),+,2O_(2)(g),rarr,CO_(2)(g),+,2H_(2)O(l)),(x,,2x,,x,,):}` `C_(2)H_(6)(g)+(7)/(2)O_(2)(g)rarr2CO_(2)(g)+3H_(2)O(l)` `(100-x)(7)/(2)(100-x) 2(100-x)` Total contraction in volume `2x+2.5 (100-x)= 212` `2x-2.5x= 212-250` `x=38xx2= 76` Therefore, contraction in volume when passed through `KOH` solution `=` Volume of `CO_(2)` evolved `x+2(100-x)= 124ml` |
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| 842. |
A constant volume and temperature conditions, the rate of diffusion `D_(A)` and `D_(B)` of gases `A` and `B` having densities `rho_(A) and rho_(B)` are related by the expressionA. `D_(A)=[D_(B).(rho_(A))/(rho_(B))]^(1//2)`B. `D_(A)=[D_(B).(rho_(B))/(rho_(A))]^(1//2)`C. `D_(A)=D_(B) ((rho_(A))/(rho_(B)))^(1//2)`D. `D_(A)=D_(B) ((rho_(B))/(rho_(A)))^(1//2)` |
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Answer» Correct Answer - D `(D_(A))/(D_(B))=sqrt((rhoB)/(rho_(A)))=[(rho_(B))/(rho_(A))]^((1)/(2)) :. D_(A)=D_(B)((rho_(B))/(rho_(B)))^((1)/(2))` |
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| 843. |
Given below are the critical temperatures of a few gases . When the gases are started cooling , which gas will liquefy first and which will liquefy in the end ? A. `N_(2)` will liquefy first and `NH_(3)` at last .B. `NH_(3)` will liquefy first and `CO_(2)` at last .C. `NH_(3)` will liquefy first and `N_(2)` at last .D. `CO_(2)` will liquefy first and `NH_(3)` at last . |
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Answer» Correct Answer - C Higher the critical temperature, faster is the liquefaction of the gas. Hence, `NH_(3)` will liquefy first and `N_(2)` at last. |
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| 844. |
Dipole-induced dipole interaction are present in which of the following pairsA. `Cl_(2)` and `C Cl_(4)`B. `HCl` and `He` atomsC. `SiF_(4)` and `He` atomsD. `H_(2)O` and alcohole |
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Answer» Correct Answer - B Dipole-induced dipole interactions are present in the pair in which the first species is polar and the other is non-polar. Both `CL_(2)` and `C CL_(4)` are non-polar so there exists induced dipole-induced dipole interaction in between them. The same is true for `SiF_(4)` and `He` atoms pair. In `H_(2)O` and alcohol, both are polar molecule, so there exists dipole-dipole interaction in between them. `HCl` is a polar molecule, whereas `He` atoms are non-polar so in between them dipole-induced dipole interactions exist. |
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| 845. |
Densities of two gases are in the ratio `1: 2` and their temperatures are in the ratio `2:1`, then the ratio of their respective pressure isA. `1:1`B. `1:2`C. `2:1`D. `4:1` |
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Answer» Correct Answer - A `(d_(1))/(d_(2))=(1)/(2), T_(1)/(T_(2))=(2)/(1)` `:. (P_(1))/(P_(2))=(V_(2))/(V_(1))xx(T_(1))/(T_(2))=(T_(1)d_(1))/(T_(2).d_(2))` `(P_(1))/(P_(2))=(2)/(1).(1)/(2)=(1)/(1)` |
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| 846. |
Which of the following is/are incorrect regarding the universal gas constant `(R )`?A. `R` is independent of pressureB. `R` is independent of temperatureC. `R` is independent of volume of gasD. `R` is dependent on nature of gas |
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Answer» Correct Answer - D `R` is dependent only on units of measurement of pressure and volume |
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| 847. |
A gaseous mixture was prepared by taking equal moles of `CO` and `N_(2)`. If the total pressure of the mixture was found to be `1` atomosphere, the partical pressure of the nitrogen `(N_(2))` in the mixture isA. `1 atm`B. `0.9 atm`C. `0.8 atm`D. `0.5 atm` |
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Answer» Correct Answer - D `P_(CO)=P_(N_(2))` As `P_(CO)+P_(N_(2))` `2P_(N_(2))= 1 atm P_(N_(2))=0.5 atm` |
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| 848. |
A gas such as carbon monoxide would be most likely to obey the ideal gas law atA. high temperature and high pressureB. low temperature and low pressureC. high temperature and low pressureD. low temperature and low pressure |
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Answer» Correct Answer - C Real gases show ideal gas behaviour at high temperature and low pressures. |
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| 849. |
A gas such as carbon monoxide would be most likely to obey the ideal gas law atA. low temperature and high pressuresB. high temperature and high pressuresC. low temperature and low pressuresD. high temperatures and low pressures |
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Answer» Correct Answer - D Real gases show ideal gas behaviour at high temperature and low pressures. |
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| 850. |
Domianance of strong repulsive forces among the molecules of the gas (Z=compressibility factor)A. depends on Z and indicated by Z=1B. depends on Z and indicated by `Z gt 1`C. depends on Z and indicated by `Z gt 1`D. is independent of Z |
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Answer» Correct Answer - B |
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