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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.
| 651. |
Surface tension of a liquid is a molecular phenomenon of liquids involving the force of cohesion among the liquid molecules. It is a scalar quantity and numerically equal to the surface energy. Numerically, it is proved that the potential enenergy of a liquid is maximum on the surface. Sparingly soluble salts and surface active substances decrease the surface tension of the liquid, however, the fairly soluble solute increase is independent of surface area but if depends on the intermolecular force and the temperature. Surface tension of a liquid is zero at:A. inversion temperatureB. boiling pointC. critical pointD. saturation point. |
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Answer» Correct Answer - C Surface tension of a liquid is zero at critical point. |
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| 652. |
Surface tension of a liquid is a molecular phenomenon of liquids involving the force of cohension along the liquid molecules. It is scalar quantity and is numberically equal to the surface energy. Numerically, it is proved that the potential soluble salts and surfce active substance. Sparingly soluble salts and surface acitve substances decrease the surface tension of the liquid. However, the fairly soluble solutes increase the surface tension of the liquid. Surface tension of a liquid is independent of surface area but it depends on the intemolecular forces and the temperature. Surface tension of a liquid does not depend on :A. temperatureB. intermolecular forcesC. surface areaD. solute dissolved in liquid. |
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Answer» Correct Answer - C Surface tension does not depend on surface area. |
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| 653. |
Surface tension of a liquid is a molecular phenomenon of liquids involving the force of cohension along the liquid molecules. It is scalar quantity and is numberically equal to the surface energy. Numerically, it is proved that the potential soluble salts and surfce active substance. Sparingly soluble salts and surface acitve substances decrease the surface tension of the liquid. However, the fairly soluble solutes increase the surface tension of the liquid. Surface tension of a liquid is independent of surface area but it depends on the intemolecular forces and the temperature. Which among the following is not the unit of surface tension ?A. `"dyne"//"cm"`B. `"newton"//m`C. `J//m^(2)`D. `erg//cm`. |
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Answer» Correct Answer - D It is not the unit of surface tension. |
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| 654. |
Why liquids have a definite volume but no definite shape? |
| Answer» This is because the intermolecular forces are stron enough to hold the molecules together but not strong enough to fix them into definite positions (as in solids). Instead, they posses fluidity, and hence, no definite shape. | |
| 655. |
At a particular temperature why is the vapour pressure of acetone less than that of ether? |
| Answer» This is because the intermolecular forces of attraction in acetone are stronger than those present in ether. | |
| 656. |
Which of the following are true ?A. In NaCl crystals , `Na^+` ions are present in all the octahedral voidsB. In ZnS (zinc blende) , `Zn^(2+)` ions are present in alternate tetrahedral voids .C. In `CaF_2, F^- ` ions occupy all the tetrahedral voidsD. In `Na_2O, O^(2-)` ions occupy half the octahedral voids. |
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Answer» Correct Answer - A,B,C Only (d) is not true because in `Na_2O, O^(2-)` ions are present in close packing and `Na^+` ions occupy all the tetrahedral voids. |
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| 657. |
An element (density `6.8 g cm^(-3)`) occurs in the BCC structure with cell edge of 290 pm. Calculate the number of atoms present in 200 g of the element. |
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Answer» Correct Answer - `24.12 xx10^23` atoms Calculate the atomic mass, M. Then M grams contains `6.023xx10^23` atoms. Calculate atoms present in 200 g. |
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| 658. |
A bottle of dry `NH_(3)` and another bottle of dry `HCl` connected through a long tube are opened simultaneously at both ends of the tube. The white ring `(NH_(4)Cl)` first formed will beA. AB. BC. CD. A, B &C simultaneously |
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Answer» Correct Answer - C `r prop (1)/(sqrt(M))` So, that weight of `NH_(3)` diffuses with faster rate. |
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| 659. |
Why is mercury used in thermometers ? |
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Answer» (i) Mercury is a liquid which can flow but does not stick to glass. (ii) It has high coefficient of expansion so that even a smal rise in temperature brings about sufficient expansion which can be detected in the capillary of the calibrated part of the thermometer. |
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| 660. |
Rate of diffustion of a gas is :A. directly proportional to its densityB. directly proportional to its molecular massC. directly proportional to the square root of its molar massD. inversely propertional to the squaure root of its molar mass. |
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Answer» Correct Answer - D is the correct answer. |
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| 661. |
The compressibility factor `(Z=PV//nRT)` for `N_(2)` at `223 K` and `81.06 MPa` is `1.95`, and at `373 K` and `20.265 MPa`, it is `1.10`. A certain mass of `N_(2)` occupies a volume of `1.0 dm^(3)` at `223 K` and `81.06 MPa`. Calculate the volume occupied by the same quantity of `N_(2)` at `373 K` and `20.265 MPa`. |
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Answer» For `T=223 K`, `P=81.06 MPa`, `Z=1.95` and `V=1.0 dm^(3)=10^(3) cm^(3)`, we have `n=(PV)/(nRT)=(81.06xx10^(3))/(1.95xx8.314xx223)=22.42 mol` Now, at `T=373 K`, `P=20.65 MPa`, `Z=1.10`, the volume occupied will be `V=(ZnRT)/(P)=(1.10xx22.42xx8.314xx373)/(20.265)=3774.0 cm^(3)` `:. V=3.774 dm^(3)` |
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| 662. |
The edge length of face centred cubic cell of an ionic substance is 508pm. If the radius of cation is 110 pm, the radius of anion is :A. 618 pmB. 144 pmC. 288 pmD. 398 pm |
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Answer» Correct Answer - B Edge length `=2(r^(+)+r^(-))` `r^(-)=144` pm |
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| 663. |
At very high pressure, the van der Waals equation reduces toA. `PV=RT+Pb`B. `PV=(aRT)/(V^(2))`C. `P=(RT)/(V-b)`D. `PV=RT-(a)/(V)` |
| Answer» `(P+(an^(2))/(V^(2)))(V-nb)=nRT`, at high pressure `P gt gt (an^(2))/(V^(2))` | |
| 664. |
A bulb A containing gas at 1.5 bar pressure was connected to an evacuated vessel of 1.0 `dm^(3)` capacity through a stopcock. The final pressure of the system dropped to 920 mbar at the same temperature. What is the volume of the container A? |
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Answer» `V_(1)=1.0dm^(3)` `V_(2)=(1.0+V)dm^(3)` `P_(1)=1.5` bar `P_(2)=920` mbar`=0.92` bar Apply `P_(1)V_(1)=P_(2)V_(2)` `(1.5)(1.0)=(1.0+V)(0.92`bar) `1.5=0.92+0.92V` or `0.92=1.5-0.92=0.58` `thereforeV=0.58//0.92=0.6304dm^(3)` |
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| 665. |
A quantity of heat is confined in a chamber of constant volume. When the chamber is immersed in a bath of melting ice, the pressure of the gas is `1000 torr`. Final temperature when the pressure manometer indicates an absolute pressure of `400 torr` isA. `109K`B. `273K`C. `373K`D. `0 K` |
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Answer» Melting point of ice `=273 K` `(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))` `(1000)/(273)=(400)/(T_(2))` (at constant `V`) `T_(2)=109.2` |
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| 666. |
when a ship is sailing in Pacific Ocean where temperature is `23.4^(@)C`, a ballon is filled with 2.0 L of ship reaches Indian Ocean where temperature is `26.1^(@)C` ? |
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Answer» Given `T_(1)=23.4^(@)C=296.55K` `V_(1)=2.0L` `T_(2)=26.1^(@)C=299.25K` `V_(2)=?` `(V_(1))/(T_(1))=(V_(2))/(T_(2))` `thereforeV_(2)=(V_(1))/(T_(1))xxT_(2)=(2.0xx299.25)/(296.55)` `thereforeV_(2)=2.018L` |
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| 667. |
The electrical resistivity of a semiconductor :A. increases with temperatureB. decreases with temperatureC. increases at low temperature and then decreaseD. does not change with temperature |
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Answer» Correct Answer - B Resistivity of semiconductors decreases with increase in temperature because at high temperature , the number of current carriers increases. |
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| 668. |
Which of the following, when doped into ultrapure germanium, will convert it into a p-type semiconductor ?A. CB. AsC. InD. Na |
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Answer» Correct Answer - C Doping of lower group impurity develops p-type semiconductor. |
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| 669. |
`1 g` of an alloy of `Al` and `Mg` reacts with excess `HCl` to form `AlCl_(3)`, `MgCl_(2)`, and `H_(2)`. The evolved `H_(2)` collected over mercury at `0^(@)C` occupied `1200 mL` at `699 mm Hg`. What is the composition of alloy? |
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Answer» `Al+3HCltoAlCl_(3)+(3)/(2)H_(2)` `Mg+2HCltoMgCl_(2)+H_(2)` Let the mass of `Al` be `x g`. Then the mass `Mg` is `(1-x)g`. Also moles of `H_(2)=(PV)/(RT)=((699)/(760)xx1.2)/(0.0821xx273)=0.0492` `implies ` Moles `H_(2)=(3)/(2)xx(x)/(27)+((1-x))/(24)xx1=0.0492` `0.0555x+0.042(1-x)=0.0492` `implies 0.0135x=7.24xx10^(-3)` `implies x=0.536 g` `implies %Al=53.6%`, `%Mg=46.4%` |
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| 670. |
Assertion Ammonia has higher critical temperature than `CO_(2)` Reason Ammonia forms intermolecular hydrogen bonds while `CO_(2)` does not. |
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Answer» Correct Answer - a |
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| 671. |
Equation of state of an ideal gas isA. `pV = RT//n`B. `pV = nRT`C. `pn = VRT`D. `pT = nRV` |
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Answer» Correct Answer - B The ideal gas equation decribes the the state of an ideal gas because it is a relation between four state variables: `p,V,n`,and `T`. |
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| 672. |
What is the relationship between the average velocity `(v)`, root mean square velocity `(u)` and most probable velocityA. ?B. `a:v:u::1:1.128:1.224`C. `a:v:u::1.128:1.224`D. `a:v:u::1.124:1.228:1` |
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Answer» Correct Answer - A `V_(av):V_(rms):V_("most probable")=V:U:a` `sqrt((8RT)/(pi M)):sqrt((3RT)/(M)):sqrt((2RT)/(M))` `a:V:U=sqrt(2): sqrt((8)/(pi)):sqrt(3)= 1:1.28:1.224` |
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| 673. |
For the non-zero value of the force of attraction between gas molecules, gas equation will beA. `PV= nRT-(n^(2)a)/(V)`B. `PV=nRT+nbP`C. `PV=nRT`D. `P=(nRT)/(V-B)` |
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Answer» Correct Answer - A `(P+(an^(2))/(V^(2))) (V)=nRT or PV= nRT-(n^(2)A)/(V)` |
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| 674. |
Why LiCl acquires pink colour when heated in Li vapours ? |
| Answer» This is because on heating electrons are trapped at the anionic sites forming F-centres which absorb energy from light and radiate pink colour. | |
| 675. |
Absolute temperature is the temperature at whichA. all molecular motion ceasesB. volume ceases to zeroC. mass becomes zeroD. none of these. |
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Answer» Correct Answer - A is the correct answer. |
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| 676. |
Pressure of 1 g of an ideal gas A `27^(@)C` is found to be 2 bar. When 2 g of another ideal gas B is introduced in the same flask at same temperature, the pressure becomes 3 bar. Find the relationship between their molecular masses. |
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Answer» Suppose molecular masses of A and B `M_(A)` and `M_(B)` respectively. Then their number of moles will be `m_(A)=(1)/(M_(A))," "n_(B)=(2)/(M_(B))` `P_(A)=2"bar", P_(A)+P_(B)=3"bar",i.e.,P_(B)=1"bar"` Applying the relation, PV=nRT `P_(A)V=n_(A)RT,P_(B)V=n_(B)RT` `:." " (P_(A))/(P_(B))=(n_(A))/(n_(B))=(1//M_(A))/(2//M_(B))=(M_(B))/(2M_(A))` `"or " (M_(B))/(M_(A))=2xx(P_(A))/(P_(B))=2xx(2)/(1)=4" " or " " M_(B)=4M_(A)`. |
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| 677. |
A highly viscous liquid was heated from `10^(@)C` to `14^(@)C`. The per cent decrease in viscosity will be about |
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Answer» Correct Answer - 8 For every degree rise of temperature, viscosity decreases by about 2%. Hence for `4^(@)c` rise of temperature viscosity will decrease by about 8%. |
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| 678. |
At `0^(@)C` the density of a gaseous oxide at 2 bar is same as that of nitrogen at 5 bar What is the molecular mass of the oxide ? . |
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Answer» The density `(rho)` of a gas `= PM//RT` Here R and T are the constants for the gases For nitrogen, `P = 5 "bar", M = 28 g mol^(-1)` `:. rho_(N_(2)) = (PM)/(RT) = ((5 "bar") xx (28 g mol^(-1)))/(R xx T)` For gaseous oxide, `P = 2 "bar" , M = ?` `:. rho_("oxide") = (PM)/(RT) = ((2 "bar") xx M)/(R xx T)` According to available data , `rho_(N_(2)) = rho _("oxide")` or `(5 "bar") xx (28 g mol^(-1)) = (2 "bar") xx M` `:. M = ((5 "bar") xx (28 g mol^(-1)))/((2 "bar")) = 70 g mol^(-1)`. |
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| 679. |
At `0^(@)C`, the density of a gaseous oxide at 2 bar is same as that of nitrogen at 5 bar. What is the molecular mass of the oxide ? |
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Answer» Using the expression, `d=(MP)/(RT)`, at the same temperature and for same density, `underset(("Gaseous oxide"))M_(1)P_(1) " "= " " underset((N_(2))M_(2)P_(2)` (as R is constant) `:. " " M_(1)xx2=28xx5 ("Molecular mass of "N_(2)=28 u)` or `" " M_(1)=70 u` |
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| 680. |
The average speed of gas molecules is equal toA. `((8 RT)/(pi M))^(1//2)`B. `((3 RT)/( M))^(1//2)`C. `((2 RT)/( M))^(1//2)`D. `(( RT)/( M))^(1//2)` |
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Answer» Correct Answer - A If there are `n` number of molecules in a sample and their individual speeds are `u_(1),u_(2),….u_(n)`, then the average speed of molecules `u_(av)` can be calculated as follows: `u_(av) = (u_(1) + u_(2) + … + u_(n))/(n)` According to the Maxwell-Boltzmann distribution of speeds, it is given as `u_(av) = sqrt((8 RT)/(pi M))` Option (2) is root mean square speed and option (3) is most probable speed. |
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| 681. |
To determine the value of `R`, which of the `PV` value is considered to be equal for every gas at `273 K`?A. `underset(P rarr 1 atm)lim (PV_(m))`B. `underset(P rarr 0)lim (PV_(m))`C. `underset(P rarr oo)lim (PV_(m))`D. `underset(Vrarr 0)lim (PV_(m))` |
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Answer» Correct Answer - B `lim(pV_(m))= 2271.099KP_(a) dm^(3)` is same for all gases at `273K` |
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| 682. |
……….. Between amino acid submits is very important in establishing the three-dimensional structure of proteins.A. Dispersion forcesB. Dipole-induced dipole forcesC. Dipole-dipole forcesD. Hydrogen bonding |
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Answer» Correct Answer - D In amino acids, `RCH(NH_(2)) COOH` , hydrogen bonding exists among molecules because `H` is bonded to very electronegative `(O and N)`. |
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| 683. |
The plot of `pV` versus `p` is a straight line for …….. At `273 K`. (i) `H_(2)` (ii) `CO` (iii) `CH_(4)` (iv) `He`A. `(i),(ii),(iii),(iv)`B. `(i),(iv)`C. `(ii),(iii)`D. `(ii),(iii),(iv)` |
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Answer» Correct Answer - B At constant temperature, the plot of `pV vs p` for real gases is not a straight line with zero slope. There is a singinificant deviation from ideal behavior. Two types of curves are seen. In the curves for `H_(2)` and `He` , as the pressure increases, the value of `pV` also increases. The second type of plot is seen in the case of other gases like `CO` and `CH_(4)`. In these plots, first there is a negative deviation from the ideal behaviour , the `pV` values decreases with increase in pressure and reaches a minimum value characteristic of a gas. After that, the `pV` value starts increasing. The curve then crosses the line for ideal gas and after that shows positive deviation continuously. |
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| 684. |
Which one of the following statement is not true about the effect of an increase in temperature on the distribution of molecular speed of gas ? .A. The most probable velocity increasesB. The friction of the molecules with the most probable speed increasesC. The distribution becomes broaderD. The area under distribution curve remains the same as under lower pressure. |
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Answer» Correct Answer - B By increasing the temperature, the fraction of the molecules with speed most probable decreases. |
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| 685. |
For two gases `A` and `B` with molecular weights `M_(A)` and `M_(B)`, respectively, it is observed that at a certain temperature `T`, the mean velocity of `A` is equal to the `V_(rms)` of `B`. Thus, the mean velocity of `A` can be made equal to the mean velocity of `B`, ifA. `P` is lowered to a temperature `T_(2)` and `T_(2) lt T and Q` is maintained at temperature `T`B. `P` is at a temperature `T and Q` at a temperature `T_(2)` where `T gt T_(2)`C. both `p` and `Q` are raised to higher temperatureD. both `P and Q` are placed at lower temperature |
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Answer» Correct Answer - A Average velocity `= sqrt((8RT)/(pi M))` Root mean sqyare velocity `= sqrt((3RT)/(pi M))` `rArr` Average velocity of `P` at temperature `T= u_(av(p))` `=sqrt((8RT)/(pi M_(P)))` Root mean square velocity `Q` at temperature `T` `=sqrt((3RT)/(M_(Q)))= u_(RMS(Q))` `implies sqrt((8RT)/(piM_(P)))=sqrt((3RT)/(M_(Q)))implies sqrt((8xx7)/(22M_(p)))=sqrt((3)/(M_(Q)))` `implies (56)/(22M_(p))=(3)/(M_(Q))implies (56)/(66)M_(Q)` Let the temperature of `Q` be maintained at `T` and that of `P` be changed to `T_(2)` `implies` Now `RMS` of both `P and Q` should be equal `implies sqrt((3RT_(2))/(M_(P)))=sqrt((3RT)/(M_(Q)))implies (T_(2))/(M_(P))=(T)/(M_(Q))` Since `M_(P)=(56)/(66)M_(Q)` `implies (66T_(2))/(56M_(Q))=(T)/(M_(Q))implies 66T_(2)= 56T` `T_(2)= (56)/(66) T, T_(2) lt T` |
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| 686. |
Which of the two gases, ammonia and hydrogen chloride, will diffuse faster and by what factor? |
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Answer» `r_(NH_(3))//r_(HCl)=(M_(HCl)//M_(NH_(3)))^(1//2)` `=(36.5//17)^(1//2)=1.46` or `r_(NH_(3))=1.46r_(HCl)` Thus, ammonia will diffuse `1.46` times faster than hydrogen chloride gas. |
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| 687. |
The molecules of a gas `A` travel four times faster than the molecules of gas `B` at same temperature. The ratio of molecular weights `(M_(A)//M_(B))` isA. `1//16`B. `4`C. `1//4`D. `16` |
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Answer» Correct Answer - A `(r_(1))/(r_(2))=(x_(1))/(x_(2))=sqrt((M_(2))/(M_(1)))=4` `(M_(2))/(M_(1))=16 (M_(1))/(M_(2))=(1)/(16)` |
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| 688. |
A porpous cup is filled with `H_(2)` gas at the atmospheric pressure and is connected to a thin glass tube a vertical position. The second end of the tube is immersed in water below it. After some time, water rises in the glass tube. Explain giving reasons. |
| Answer» Pressure is exerted due to bombardment of gaseous molecues with the wall of container. | |
| 689. |
The weight of `350mL` of a diatomic gas at `0^(@)C` and 2 atm pressure is `1 g`. The weight in g of one atom at `NTP` is:A. `2.64xx10^(-23)g`B. `2.64xx10^(-22)g`C. `5.28xx10^(-23)g`D. `0.82xx10^(-22)g` |
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Answer» Correct Answer - A `PV(w)/(M_(w))RT` `M_(w) =(wRT)/(PV)implies (1xx0.082xx273)/(2xx0.350) M_(w)= 32` |
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| 690. |
At high temperature and low pressure the van der Waals equation is reduced to .A. `(p + (a)/(V_(m)^(2))) (V_(m)) = RT`B. `pV_(m) = RT`C. `p(V_(m) - b) = RT`D. `(p + (a)/(V_(m)^(2))) = (V_(m) - b) = RT` |
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Answer» Correct Answer - B At high temperature and low pressure, van der Walls equation is reduced to ideal gas equation. `PV = nRT` `PV = RT` (For `1` mole of gas) |
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| 691. |
The temperature of a given mass of air was changed from `15^(@)C` to `-15^(@)C`. If the pressure remains unchanged and the initial volume was `100 mL`, what should be the final volume? |
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Answer» Given `T_(1)=15^(@)C` or `15+273.15 K=288.15 K` `V_(1)=100 mL` `T_(2)= -15^(@)C=273.15-15=258.15 K` `V_(2)=?` We know `(V_(1))/(T_(1))=(V_(2))/(T_(2))` ` :. V_(2)=(V_(1))/(T_(1))xxT_(2)=(100xx258.15)/(288.15)=89.58 mL` |
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| 692. |
A sample of gas occupies `100 mL` at `27^(@)C` and `740 mm` pressure. When its volume is changed to `80 mL` at `740 mm` pressure, the temperature of the gas will beA. `21.6^(@)C`B. `240^(@)C`C. `-33^(@)C`D. `89.5^(@)C` |
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Answer» `V_(1)=100 mL`, `T_(1)=27^(@)C=300 K` `P_(1)=740` `V_(2)= 80mL` `T_(2)=?` `P_(2)=740 mm` `(V_(1))/(T_(1))=(V_(2))/(T_(2))` `:. T_(2)=(V_(2)xxT_(1))/(V_(1))=(80xx300)/(100)=240 K` or `T_(2)= -33^(@)C` |
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| 693. |
The volume of `10 "moles"` of an ideal gas at `10 atm` and `500 K` isA. `82 L`B. `41 L`C. `20.5 L`D. `(82)/(3)L` |
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Answer» Correct Answer - B `n= 10 mol e` `P=10 atm` `T= 500K` `R= 0.0821 L- atm K^(-1)mol^(-1)` `PV= nRT` `10xxV= 10xx0.082xx500` `V= 41L` |
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| 694. |
If the density of a gas at the sea level at `0^(@)C` is `1.29 kg m^(-3)`, what is its molar mass? (Assume that pressure is equal to `1"bar"`.) |
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Answer» `pV_(m)=RT` or `pM//d=RT` or `M=dRT//p` `=(1.29kgm^(-3)xx8.314 N mK^(-1)mol^(-1)xx273.15 K)/(1.0xx10^(5) N m^(-2)(or Pa))` `=(1.29xx8.314xx273.15 kg mol^(-1))/(1xx10^(5))` `0.0293 kg mol^(-1)` or molar mass is `29.3 g mol^(-1)` |
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| 695. |
Assertion:Pressure is exerted by gas in a container with increasing temperature of the gas. Reason: With the rise in temperature, the average speed of gas molecules increases.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 When the temperature, increases, the average speed of a gas molecules increases and by this increase the pressure of gas is also increased. |
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| 696. |
Heat capacity of a diatomic gas is higher than that of a monoatomic gas. |
| Answer» Correct Answer - True | |
| 697. |
Calculate the number of nitrogen molecules present in `2.8 g` of nitrogen gas. |
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Answer» Number of moles of nitrogen `=2.8 g// 28 g mol^(-1)=0.1 mol` Number of nitrogen molecules `=0.1 molxx6.022xx10^(23) mol^(-1)` `=6.022xx10^(22)` |
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| 698. |
A drop of liquid takes nearly a spherical shape because . |
| Answer» Because of the surfac tension, a liquid tends to posses minimum surface area. Since for a given volume, sphere has the minimum surface area therefore, drop of the liquid assumes spherical shape. | |
| 699. |
For the equation `N_(2)O_(5)(g)=2NO_(2)(g)+(1//2)O_(2)(g)`, calculate the mole fraction of `N_(2)O_(5)(g)` decomposed at a constant volume and temperature, if the initial pressure is `600 mm Hg` and the pressure at any time is `960 mm Hg`. Assume ideal gas behaviour. |
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Answer» If `p` is the partial pressure of `N_(2)O_(5)` that has decomposed, then `underset(600 mm Hg-p)(N_(2)O_(2)(g))tounderset(2p)(NO_(2)(g))+underset(p//2)((1)/(2)O_(2)(g))` Pressure at any time `=(600 mm Hg-p)+2p+p//2` `=600 mm Hg+(3//2)p` Equating this tp `960 mm Hg`, we get `p=(2)/(3)(960-600)mm Hg=244 mm Hg` The mole fraction of `N_(2)O_(5)` decomposed would be `x=(244 mm Hg)/(600 mm Hg)=0.407` |
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| 700. |
The gas above `T_(c )` cannot be liquefied. |
| Answer» Correct Answer - True | |