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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.
| 1051. |
The ratio of root mean square velocity of average velocity of a gas molecule at a particular temperture isA. `1.085:1`B. `1:1.086`C. `2:1.086`D. `1.086:2` |
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Answer» Correct Answer - A |
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| 1052. |
The ratio of root mean square velocity of average velocity of a gas molecule at a particular temperture isA. `1.086 : 2`B. `1 : 1.086`C. `2 : 1.086`D. `1.086 : 1` |
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Answer» Correct Answer - D According to the kinetic molecular theory of gases , we have `u_(rms) : u_(aV) = sqrt((3RT)/(M)) : sqrt((8RT)/(pi M))` `= sqrt(3) : sqrt((8)/(pi))` ` = sqrt(3) : sqrt(2.54)` `= 1.732 : 1.595` ` = (1.732)/(1.595) : (1.595)/(1.595)` ` = 1.086 : 1` |
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| 1053. |
The ratio of root mean square velocity of average velocity of a gas molecule at a particular temperture isA. `1 : 1.086`B. `2 : 1.86`C. `1.086 : 1`D. `2.086 : 1` |
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Answer» Correct Answer - C Root mean square velocity `(V_(rms))=sqrt(3RT)/(M)` Average velocity `(V_(av))=sqrt(8RT)/(piM)` For a gas at temperature T, M is constant. `:.V_(rms)/V_(av)=sqrt(3Rt)/(M)xxsqrt(piM)/(8RT)` `(sqrt(3)xxsqrt(pi))/(sqrt(8))=(sqrt(3)xxsqrt(3.14))/(sqrt(8))=1.086/1` |
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| 1054. |
The term that is correct for the attractive forces present in a real gas in the van der Waals equation isA. `-nb`B. `-(an^(2))/(V^(2))`C. `(an^(2))/(V^(2))`D. `nb` |
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Answer» Correct Answer - C The total force of attraction on any molecule about to hit wall is proportional to the concentration of the neighboring molecules , `n//V` . However , the number of molecules about to hit the wall per unit wall area is also proportional to the concentration , `n//V`. Therefore , the force per unit wall area (pressure) is reduced from that assumed in the ideal gas law by a factor proportional to `n^(2)//V^(2)`. Letting `a` to be the proportionality constant, we can write the correction factor as `an^(2)//V^(2)` |
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| 1055. |
The ratio of root mean square velocity of average velocity of a gas molecule at a particular temperture isA. `1.086:1`B. `1:1.086`C. `2:1.086`D. `1.086:2` |
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Answer» `v_(rms)=sqrt((3RT)/(M))` `v_("average")=sqrt((8RT)/(piM))` `(v_(rms))/(v_("average"))=sqrt((3)/(8pi))=1.086` |
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| 1056. |
The term that is correct for the attractive forces present in a real gas in the van der Waals equation isA. `nb`B. `(an^(2))/(V^(2))`C. `-(an^(2))/(V^(2))`D. `-nb` |
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Answer» `(P+(an^(2))/(V^(2)))(V-nb)=nRT` The term that is correct for the attractive forces present in a real gas in the van der Waals equation is `an^(2)//V^(2)`. |
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| 1057. |
A certain mass of a gas occupies 39 mL at 760 mm pressure. What volume would it occupy if the pressure is raised to 780 mm provided that temperature remains constant ? |
| Answer» Correct Answer - C | |
| 1058. |
Calculate the volume in `mL` hydrogen peroxide labelled `10` volume required to liberate `600 mL` of oxygen at `27^(@)C` and `760 mm`. |
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Answer» `10V` of `H_(2)O_(2)` `1V` of `H_(2)O` gives `10V` of `O_(2)` at `STP`. `1 mL` of `10 V H_(2)O_(2)` gives `10 ml` of `O_(2)`. Volume of `O_(2)` required at `NTP` `{:(V_(1)= 600, P_(1)=760 mm),(V_(2)=?, P_(2)=760 mm),(T_(1)=300K, T_(2)=273K):}` `(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))` `V_(2)=(P_(1)V_(1)T_(2))/(P_(2)T_(1))=600xx(273)/(300)=546 mL` `10 V` of `H_(2)O_(2)` produces `10 mL` of `O_(2)` at `STP` for `1 mL` of its volume. Hence, for `546 mL` of `O_(2)` at `NTP`, the volume of `10 V` of `H_(2)O_(2)` required `=(546xx1)/(10)=54.6 mL` (Alternative method) Volume of `O_(2)` at `STP=?` `600 mL` of `O_(2)` at `27^(@)C(273+27=300 K)` and `760 mm` or `1 atm` `(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))` `{:(P_(1)= 1 atm, P_(2)=1 atm),(V_(1)=600 mL, V_(2)=?),(T_(1)=300K, T_(2)=273K):}` `(1xx600)/(300)=(1xxV_(2))/(273)` `V_(2)=(600xx273)/(300)=546 mL` Volume strength of `H_(2)O_(2)xx` Volume of `H_(2)O_(2)` `=` volume of `O_(2)` at `STP` `10 Vxx x=546 mL` `x=(546)/(10)=54.6 mL` Volume of `H_(2)O_(2)=54.6 mL` |
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| 1059. |
For gaseous reactions, the rate is expressed in terms of `dP//dt` instead of `dc//dt` or `dn//dt` (where `c` is the concentration and `n` the number of `mol`). What is the relation among these expresisons ?A. `(dC)/(dt)=1/V((dn)/(dt))=1/(RT)=((dp)/(dt))`B. `(dC)/(dt)=((dn)/(dt))=((dp)/(dt))`C. `(dC)/(dt)=1/V((dn)/(dt))=V/(RT)((dp)/(dt))`D. none of the above is correct |
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Answer» Correct Answer - A |
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| 1060. |
For a monatomic gas, kinetic energy `= E`. The relation with `rms` velocity isA. `u = ((2E)/(m))^(1//2)`B. `u = ((E)/(2m))^(1//2)`C. `u = ((3E)/(2m))^(1//2)`D. `u = ((E)/(3m))^(1//2)` |
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Answer» Correct Answer - A According to the kinetic gas eqauation , `pV = (1)/(3) mnc^(2)` For a single molecule of mass `m , n = 1` . Thus , `u_(rms) = sqrt((3 pV)/(m))` (1) According to gas equation , for one mole of gas , we have `pV = RT` For a single molecule , we have `pV = (R )/(N_(0))T = kT` (2) lt brgt where `k` is called the Boltzmann constant. Average kinetic energy per molecule `(E)` is given as `E = (3)/(2) kT` or `kT = (2)/(3) E` (3) Combining Eqs. (1),(2), and (3) , we have ` u_(rms) = sqrt((3 xx (2)/(3)E)/(m)) = sqrt((2E)/(m))` |
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| 1061. |
In the following figure, when the two stopcocks are opened, the total pressure inside the flask will be A. 1.41 atmB. 2.41 atmC. 3.41 atmD. 1.12 atm |
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Answer» Correct Answer - A |
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| 1062. |
For a monatomic gas, kinetic energy `= E`. The relation with `rms` velocity isA. `u=((2E)/m)^(1//2)`B. `u=((3E)/(2m))^(1//2)`C. `u=(E/(2m))^(1//2)`D. `u=(E/(3m))^(1//2)` |
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Answer» Correct Answer - A |
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| 1063. |
The total energy of `1mol` of an ideal monatomic gas at `27^(@)C` is……. . |
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Answer» Correct Answer - 900 |
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| 1064. |
The root mean square speed of one mole of a monoatomic gas having molecular mass `M` is `u_(rms)` The relation between the average kinetic energy `(E)` of the gas and `u_9rms)` is .A. `U_(rms) = sqrt((3E)/(2M))`B. `U_(rms) = sqrt((2E)/(3M))`C. `U_(rms) = sqrt((2E)/(M))`D. `U_(rms) = sqrt((E)/(3M))` |
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Answer» Correct Answer - C `U_(rms)=sqrt((3RT)/(M))` `E=3/2RT` or `RT=(2E)/(3)` `U_(rms)=sqrt((3xx2E)/(3//m))=sqrt((2E)/(M))` |
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| 1065. |
Compressibility factor `Z=(PV)/(RT)`. Considering ideal gas, real gas, and gases at critical state, answer the following questions: The cpmpressibility factor of an ideal gas isA. `0`B. `1`C. `2`D. `3` |
| Answer» `Z=(PV)/(RT) :. Z_(ideal)=1` | |
| 1066. |
1 L of air weighs 1.293 g at `0^(@)C` and 1 atm pressure. At becomes `30 dm^(3)` . To what temperature the gas must be raised to accomplish the change ? |
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Answer» Given `V_(1)=1L` `T_(1)=0^(@)C` or `273.15K` `P_(1)=1atm` `V_(2)=1L` `T_(2)=?` `P_(2)=1atm` `thereforeT_(2)T_(1)//V_(1)=79.9^(@)C` |
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| 1067. |
If Z is a compressibility factor, van der Waals equation at low pressure can be written asA. `Z=1+(Pb)/(RT)`B. `Z=1+(RT)/(Pb)`C. `Z=1-(a)/(VRT)`D. `Z=1-(Pb)/(RT)` |
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Answer» Correct Answer - C For 1 mole of a real gas, van der Waals equation is `(P+(a)/(V^(2)))(V-b)=RT` At low pressure, V is large and therefore, b can be neglected in comparison to V so that `V-b~-V`. Hence, `(P+(a)/(V^(2)))V=RT` or `PV+(a)/(V)=RT" or " PV=RT-(a)/(V)` or `(PV)/(RT)=1-(a)/(VRT)" or " Z=1-(a)/(VRT)` |
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| 1068. |
Compressibility factor (Z) for a van der Waals real gas at critical point isA. 1B. `3//8`C. `9//8`D. `8//9` |
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Answer» Correct Answer - B |
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| 1069. |
For a monatomic gas, kinetic energy `= E`. The relation with `rms` velocity isA. `u=((2E)/(m))^(1//2)`B. `u=((3E)/(2m))^(1//2)`C. `u=((E)/(2m))^(1//2)`D. `u=((E)/(3m))^(1//2)` |
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Answer» `v_(rms)=sqrt((3RT)/(M))` `KE(E) "per mole"=(3)/(2)RT` `RT=(2)/(3)KE` `v_(rms)=sqrt((3)/(M)xx(2)/(3)KE)` `=sqrt((2KE)/(M))=((2KE)/(M))^(1//2)` |
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| 1070. |
Using van der Waals equation `(P+(a)/(V^(2)))(V-b)=RT`, answer the following questions: At high pressure, the van der Waals equation gets reduced toA. `(P+(a)/(V^(2)))V=RT`B. `P(V-b)=RT`C. `PV=RT`D. `(P+(a)/(V^(2)))(V-b)=RT` |
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Answer» At high pressure `V gt gt b`. Hence, `(P+(a)/(V^(2)))V=RT` |
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| 1071. |
Using van der Waals equation `(P+(a)/(V^(2)))(V-b)=RT`, answer the following questions: The term that accounts for effective volume in the van der Waals equation for non-ideal gas isA. `RT`B. `V-b`C. `(P+(a)/(V^(2)))`D. `RT^(-1)` |
| Answer» Correct Answer - V-b | |
| 1072. |
Using van der Waals equation `(P+(a)/(V^(2)))(V-b)=RT`, answer the following questions: The term that accounts for intermolecular forces in the van der Waals equation for non-ideal gas isA. `RT`B. `V-b`C. `(P+(a)/(V^(2)))`D. `RT^(-1)` |
| Answer» Correct Answer - `(P+(a)/(V^(2)))` | |
| 1073. |
Assertion: In van der Waals equation `(P+(a)/(V^(2)))(V-b)=RT` pressure correction `(a//V^(2))` is due to the force of attraction between molecules. Reason: Volume of gas molecule cannot be neglected due to force of attraction.A. If both (`A`) and (`R`) are correct and (`R`) is the correct explanation of (`A`).B. If both (`A`) and (`R`) are correct, but (`R`) is not the correct explanation of (`A`).C. If (`A`) is correct, but (`R`) is incorrect.D. If (`A`) is incorrect, but (`R`) is correct. |
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Answer» Correct Answer - A::B::C::D Volume is independent of the force of attraction. |
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| 1074. |
Assertion: All molecules of an ideal gas more with the same speed. Reason: There is no attraction between the molecules in an ideal gas.A. If both (`A`) and (`R`) are correct and (`R`) is the correct explanation of (`A`).B. If both (`A`) and (`R`) are correct, but (`R`) is not the correct explanation of (`A`).C. If (`A`) is correct, but (`R`) is incorrect.D. If (`A`) is incorrect, but (`R`) is correct. |
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Answer» Correct Answer - A::B::C::D Speed of gases depend upon molecular mass of gas. Therefore, all ideal gas does not move with same speed. |
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| 1075. |
KF has NaCl structure. If the distance between `K^+` and `F^-` is 269 pm, find the density of KF (`N_A=6.02 xx 10^23 mol^(-1)`, atomic masses K=39, F = 19 amu) |
| Answer» Correct Answer - `2.48 " g cm"^(-3)` | |
| 1076. |
Copper crystallises into a fee lattice. Its edge length is `3.62 xx 10^(-8)` cm. Calculate the density of copper (atomic mass of Cu=63-5 u, `N_A= 6-022 xx 10^(23) mol^(-1)`). |
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Answer» Correct Answer - `8.9 "g cm"^(-3)` For fcc, `r=a/(2sqrt2) or a=2sqrt2r=2xx1.414xx128` pm =362 pm `rho=(ZxxM)/(a^3xxN_0)=(4xx63.5 "g mol"^(-1))/((362xx10^(-10)cm)^3xx6.02xx10^23 mol^(-1))=8.9 "g cm"^(-3)` Alternatively, `rho="Mass of unit cell "/"Volume of unit cell "=(4xx(63.5 "g mol"^(-1)//6.02xx10^23 mol^(-1)))/((362xx10^(-10)cm)^3)=8.9 g cm^(-3)` |
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| 1077. |
Assertion: Compressibility factor for hydrogen varies with pressure with positive slope at all pressures. Reason: Event at low pressures, repulsive forces dominate hydrogen gas.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 In case of `H_(2)`, compressibility factor increases with the pressure. At `273 K, Z gt1` which shows that it is difficult to compress the gas as compared to ideal gas. In this case repulsive force dominate. |
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| 1078. |
(`a`) One mole of nitrogen gas at `0.8 atm` takes `38 s` to diffuse through a pinhole, whereas one mole of an unknown compound of xenon with fluorine at `1.6 atm` takes `57 s` to diffuse through the same hole. Calculate the molecular formula to the compound. (`b`) The pressure exerted by `12 g` of an ideal gas at temperature `t^(@)C` in a vessel of volume `V litre` is `1 atm`. When the temperature is increased by `10^(@)C` at the same volume, the pressure increases by `10%`. Calculate the temperature `t` and volume `V`. (Molecular weight of the gas is `120`.) |
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Answer» (`a`) The rate of diffusion depends on the following facotrs `r prop p` `r prop sqrt(1//M)` Taking these together, we get `(r_(2))/(r_(1))=(p_(2))/(p_(1))((M_(1))/(M_(2)))^(1//2)` Since `r prop 1//t`, we can write `(t_(1))/(t_(2))=(p_(2))/(p_(1))((M_(1))/(M_(2)))^(1//2)` or `M_(2)=((p_(2))/(p_(1))xx(t_(2))/(t_(1)))^(2)M_(1)` Identifying the subscript `1` with nitrogen and `2` with unknown gas, we get `M_(2)=((1.6)/(0.8)xx(57)/(38))^(2)(28 g mol^(-1))=252 g mol^(-1)` Let the molecular formula of the unknown compound be `XeF_(n)`. We will have `M_(Xe)+nM_(F)=252 g mol^(-1)` i.e., `[131+n(19)] g mol^(-1)=252 g mol^(-1)` or `n=(252-131)/(19)=6.36=6` Hence, the molecular formula of the gas is `XeF_(6)`. (`b`) From the expression `pV=nRT`, we can write `(P_(1))/(P_(2))=(T_(1))/(T_(2))` (since `V` and `n` are constants) For the given data, we get `(1 atm)/(1.1 atm)=((273+t//^(@)C)K)/((273+t//^(@)C+10)K)` or `283+t//^(@)C=1.1(273-t//^(@)C)` or `t//^(@)C=(283-1.1xx273)/(0.1)=-173` or `t=-173^(@)C` For the given system `n=(12 g)/(120 g mol^(-1))=0.1 mol` `T=(273-173 K)=100 K` Hence, `V=(nRT)/(P)` `((0.1 mol)(0.082 L atm K^(-1)mol^(-1))(100 K))/(1 atm)` `=0.82 atm` |
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| 1079. |
One mole of nitrogen gas at `0.8atm` takes `38s` to diffuse through a pinhole, while `1 mol` of an unknown fluoride of xenon at `1.6 atm` takes `57 s` to diffuse through the same hole. Calculate the molecular formula of the compound. |
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Answer» The available data is : `P_(N_(2)) = 0.8` atm, `t_(N_(2)) = 38 s, P_(g) = 1.6` atm , `t_(g) = 57 s` Rate of diffusion `(r) prop (P)/(sqrt(M))` `:. (r_(N_(2)))/(r_(g)) = (P_(N_(2)))/(P_(g)) xx sqrt((M_(g))/(M_(N_(2))))` Since rate of diffusion `(r) prop 1/t`. `:. (t_(g))/(t_(N_(2))) = (P_(N_(2)))/(P_(g)) xx sqrt((M_(g))/(M_(N_(2))))` On substituting the value, `57/58 = (0.8)/(1.6) xx sqrt((M_(g))/(28))` or `sqrt((M_(g))/(28)) = (57)/(38) xx 2 = 3` Squaring both sides, `(M_(g))/(28) = 9` or `M_(g) = 28 xx 9 = 252` Let the molecular formula of compound of `Xe` be `Xe F_(n)`. Molecular mass of `XeF_(n) = 132 + n xx 19` Comparing (i) and (ii) `131 + n xx 19 = 252` or `n = (252-131)/(19) = 121/19 = 6.3` or `6` `:.` Molecular formula `= XeF_(6)`. |
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| 1080. |
5 g of an unknown hgas has pressure P at temperature TK in a vessel. On increasing the temperature by `50^(@)C`, 1g of the was given out to maintain the pressure P. What was the original temperature of the gas ? |
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Answer» Initially, `T_(1)=T,P_(1)=P,V_(1)=V `(Volume of the vessel), `n_(1)=(5)/(M)`mole Finally, `T_(2)=(T+50)K, P_(2)=P, V_(2)=V, n_(2)=(5-1)/(M)=(4)/(M)` mole (Increase of `50^(@)C`=increase of 50 K because size of `1^(@)C`=1 K) Applying PV=nRT `n_(1)T_(1)=n_(2)T_(2)` `(5)/(M)xxT=(4)/(M)(T+50)" or "5T=4T+200" or T=200 K` |
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| 1081. |
If tempetature changes form `27^(@)C` to `127^(@)C`, the relative percentage change in rms velocity isA. 1.56B. 2.56C. 15.6D. 82.4 |
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Answer» Correct Answer - C `c=sqrt((3RT)/(M))` At `27^(@)C, " "c=sqrt((3Rxx300)/(M))=sqrt(300)x` `=17.3x" " (x=sqrt((3R)/(M)))` At `127^(@)C` `c=sqrt((3Rxx400)/(M))=sqrt(400)x=20x` `:.` Increase`=20x-17.3x=2.7x` % increase`=(2.7x)/(17.3x)xx100=15.6` |
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| 1082. |
What is the effect of temperature on surface tension and viscosity? |
| Answer» Both decrease with increase in temperature. | |
| 1083. |
Why vegetables are cooked with difficulty at a hill station? |
| Answer» The atmospheric pressure is less and so the boiling point is lowered. | |
| 1084. |
A bottle contaning ammonia and a bottle containing hydrogen chloride are connected through along tube are opened simultaneously at both ends. The white ammonium chloride first formed will be :A. at the centre of the tubeB. near the hydrogen chloride tubeC. near the ammonia tubeD. throughout the length of the tube. |
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Answer» Correct Answer - B `NH_(3)` diffuse at a faster rate than `HCl` gas because the density of `NH_(3)` is less than that of `HCl` gas. |
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| 1085. |
A bolttle of dry ammonia and a bottle of dry hydrogen chloride connected through a long tube are opened simultaneously at both ends, The white ammonium chloride ring first formed will beA. at the centre of the tubeB. near the hydrogen chloride bottleC. near the ammonia bottleD. throughout the length of the tube. |
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Answer» Correct Answer - B `r prop 1//sqrt(M)`. Molecular mass of HCl (36.5) is greater than that of `NH_(3)`(17), hence HCl will diffuse slowly, i.e., the ring will be formed near the HCl bottle. |
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