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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.
| 151. |
If a gas is allowe to expand at constant temperature,A. the pressure decreasesB. the kinetic energy of the molecules remains the sameC. kinetic energy decreasesD. no. of molecules of the gas incresaes. |
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Answer» Correct Answer - A::B are the correct options. |
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| 152. |
If a gas is allowed to expand at constant temperature, then :A. number of molecules of the gas decreasesB. the kinetic energy of the gas molecules decreasesC. the kinetic energy of the gas molecules increasesD. the kinetic energy of the gas molecules remains the same |
| Answer» Correct Answer - D | |
| 153. |
A sample of unknown gas is placed in a 2.5 L bulb at a pressure of 360 Torr and at a temperature of `22.5^(@)`C and Is found to weight 1.6616 g. What is the molecular weight of the gas?A. 80 gB. 64 gC. 55 gD. 34 g |
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Answer» Correct Answer - D `M=(wRT)/(pV)=(1.6616xx0.082xx295.5)/((360)/(760)xx2.5)=34" g"` |
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| 154. |
Ethane burns in oxygen according to the following equation: `2C_(2)H_(6)(g) + 7O_(2)(g) rarr 4CO_(2)(g) + 6H_(2)O(I)` `2.5 L` of ethane are burnal in excess of oxygen at `27^(@)C` and 1 bar pressure. Calculate how many litre of `CO_(2)` are formed at `50^(@)C` and `1.5` bar. |
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Answer» The equation for the combustion reaction is : `underset(2vol("Litre"))(2C_(2)H_(6)(g)) + 7O_(2)(g) + underset(4vol("Litre"))(4CO_(2(g)) + 6H_(2)O(l)` Under the given conditions i.e., at `27^(@)C` and 1 bar pressure `2L` of ethane evolve `CO_(2)(g) = 4L` `2.5 L` of ethan evolve `CO_(2)(g) = (4L) xx ((2.5 L)/(2L)) = 5.0L` The volume of `CO_(2)(g)` at `50^(@)C` and `1.5` bar pressure can be pressure can be calculated with the help of gas equation. `V_(1) = 5.0L , V_(2) = ?` `P_(1) = 1` bar , `P_(2) = 1.5` bar `T_(1) = (27+273) = 300 K` , `T_(2) = 50+273 = 323 K` According to gas equation : `(P_(1)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2))` or `V_(2) = (P_(1)V_(1)T_(2))/(P_(2)T_(1))` Substituting the values : `V_(2) = ((1 "atm" ) xx (5.0 L) xx (323 K))/((1.5 "atm") xx (3.00 K)) = 3.59 L` |
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| 155. |
The total pressure of a gaseous mixture of `2.8 g N_(2)`, `3.2 g O_(2)`, and `0.5 g H_(2)` is `4.5 atm`. Calculate the partial pressure of each gas. |
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Answer» Number of moles `=(Mass)/(Molar mass)` Moles of `N_(2)(n_(N_(2)))=(2.8)/(28)=0.1mol` Moles of `O_(2)(n_(O_(2)))=(3.2)/(32)=0.1 mol` Moles of `H_(2)(n_(H_(2)))=(0.5)/(2)=0.25 mol` `:. ` Total number of mole `=n_(N_(2))+n_(O_(2))+n_(H_(2))` `=0.1+0.1+0.25=0.45` Partial pressure a gas is `(Number of moles)/(Total number of moles)xxTotal pressure` `P_(N_(2))=(0.1)/(0.45)xx4.5atm=1.0 atm` `P_(O_(2))=(0.1)/(0.45)xx4.5 atm=1.0 atm` `P_(H_(2))=(0.25)/(0.45)xx4.5 atm=2.5 atm` |
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| 156. |
A gaseous mixture containing 8g of `O_(2)` and 227 mL of `N_(2)` at STPis enclosed in flask of 5 L capacity at `0^(@)C`. Find the partial pressure of each gas and calculate the total pressure in the vessel. |
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Answer» total `P=1.164` atm `, `p_(O_(2))=1.12` mm |
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| 157. |
If volume occupied by `CO_(2)` molecules is negligible, then calculate pressure `((P)/(5.277))` exerted by one mole of `CO_(2)` gas at `300 K`. `(a=3.592 atm L^(2)mol^(-2))`A. 7B. 8C. 9D. 3 |
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Answer» Correct Answer - B `(P+(an^(2))/(V^(2)))(V-nb)=nRT` `(P+(a)/(V^(2)))(V-b)=RT " " (because n = 1)` If b is negligible `P=(RT)/(V)-(a)/(V^(2))` The equation is quadratic in V, thus `V= (+ RT pm sqrt(R^(2)T^(2)-4aP))/(2P)` Since V has one value at given P and T, thus numerical value of discriminant = 0 `R^(2)T^(2)=4aP` `P=(R^(2)T^(2))/(4a)=((0.0821)^(2)(300)^(2))/(4xx3.592) therefore (P)/(5.277)=8` |
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| 158. |
In a closed vessel of 5 litres capacity, 1 g of `O_(2)` is heated from `300` to `600 K`. Which statement is not correct ?A. Pressure of the gas increasesB. The rate of collisions increasesC. The no. of moles of the gas increasesD. The energy of the gaseous molecules increases. |
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Answer» Correct Answer - C Upon heating, the number of moles of a gas do not change. |
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| 159. |
The pressure exerted by `1 mol` of `CO_(2)` at `273 K` is `34.98 atm`. Assuming that volume occupied by `CO_(2)` molecules is negligible, the value of van der Waals constant for attraction of `CO_(2)` gas isA. `3.59 dm^(6) atm mol^(-2)`B. `2.59 dm^(6) atm mol^(-2)`C. `1.25 dm^(6) atm mol^(-2)`D. `1.59 dm^(6) atm mol^(-2)` |
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Answer» `[P+(a)/(V^(2))][V-b]=RT` `:. [P+(a)/(V^(2))]V=RT` or `V^(2)P-RTV+a=0` `V=(+-RT+-sqrt(R^(2)T^(2)-4Pa))/(2P)` Since, `V` is constant at given `P` and `T`, `V` can have only one value or discriminant `=0` `:. R^(2)T^(2)=4Pa` or `a=(R^(2)T^(2))/(4P)` `=((0.821)^(2)xx(273)^(2))/(4xx34.98)` `=3.59 dm^(6) atm mol^(-2)` |
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| 160. |
If the volume occupied by `CO_(2)` molecules is negligible, then calculate the pressure exerted by one mole of `CO_(2)` gas at 273 K (a=3.592 atm `"litre"^(-2)mol^(-2)`) |
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Answer» For 1 mole, `(P+(a)/(V_(2)))(V-b)=RT` If b is negligible, `(P+(a)/(V_(2)))V=RT" or " P=(RT)/(V)-(a)/(V_(2))" or "PV^(2)-RTV+a=0` This equation is quadratic in V. Hence, `V=(+RT+-sqrt(R^(2)T^(2)-4aP))/(2P)` As V can have only one value at P and T, hence we must have `R^(2)T^(2)-4aP=0 " or " R^(2)T^(2)=4 a P` or `" " P=(R^(2)T^(2))/(4a)=((0.0821" L atm "K^(-1))^(2)(273 K)^(2))/(4xx3.592" atm "L6(-2)mol^(-2))=34.96" atm"` |
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| 161. |
The vapour pressure of water at 300 K in a closed container is 0.4 atm. If the volume of the container is doubled, its vapour pressure at 300 K will beA. 0.8 atmB. 0.2 atmC. 0.4 atmD. 0.6 atm |
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Answer» Correct Answer - A::C::D |
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| 162. |
Statement-1. At zero degree Kelvin, the volume occupied by a gas is negligible. Statement-2. All molecullar motion ceases at 0 K.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. |
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Answer» Correct Answer - C Correct statement-2. Vibrational motion exists even at 0 K |
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| 163. |
Assertion: At constant temperature, if pressure on the gas is doubled, density is also doubled. Reason: At constant temperature, molecular mass of a gas is directly proportional to the density and inversely proportional to the pressureA. 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. |
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Answer» Correct Answer - A Reason is correct explanation for assertion. |
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| 164. |
The plot of volume versus pressure at constant temperature is a……………………… |
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Answer» Correct Answer - hyperbola |
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| 165. |
Pressure versus volume graph for real gas and are shown in figure. Answer the following question on the basis of this graph. (i) Interpret the behaviour of real gas with respect to ideal gas at low pressure. (ii) Interpret the behaviour of real gas with respect to ideal gas at high pressure. (iii) Mark the pressure and volume by drawing a line at the point where real gas behaves as an ideal gas. |
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Answer» (i) At low pressure, the two volumes almost coincides. The mean s that the real gas behaves like and ideal gas at low pressure. (ii) At high pressure, the volume of the gas is more as compared to that of an ideal gas. |
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| 166. |
If pressure of a gas is quadrupled and the temperature in degrees kelvin is doubled, the density of the will become …………….. Times |
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Answer» Correct Answer - 2 `d=(PM)/(RT)` `:.(d_(1))/(d_(2))=(P_(1))/(T_(1))xx(T_(2))/(P_(2))=(P_(1))/(P_(2))xx(T_(2))/(T_(1))=(1)/(4)xx(2)/(1)=(1)/(2)` `:. d_(2)=2d_(1)` |
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| 167. |
The linear plot of volume in litres versus temperature in degrees centigrade on extrapolation cuts the temperature axis at ………………… |
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Answer» Correct Answer - `-273^(@)C` |
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| 168. |
For each degree rise in temperature , the volume of a gas increases by ……. Of the original volume of the gas at `0^(@)C`.A. `273.15`B. `1//273.15`C. `(273.15)^(2)`D. `(1//273.15)^(2)` |
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Answer» Correct Answer - B If the pressure is held constant , the ratio of the volume `(V_(2))` of a given mass of air at `100^(@) C` to its volume `(V_(1))` at `0^(@)C` is constant and is independent of the initial volume. Experiments show that this ratio is `1.366`,i.e., `(V_(2))/(V_(1)) = 1.366` `V_(2) = 1.366 V_(1) = V_(1)(1 + 0.366)` `= V_(1) + 0.366 V_(1)` Thus implies that the volume of the gas at `100^(@)C` is greater than its volume at `0^(@)C` by `0.366` times its volume at `0^(@)C`. Since this expansion is for `100^(@)` change in temperature and since this expansion is found to be uniform, it follows that the expansion per unit degree will be by `(0.366//100)V_(1)`or `(1//273)V_(1)`. |
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| 169. |
Assertion: The numerical values of `P_(c ), V_(c )` and `T_(c )` are `(a)/(27b^(2)), 3b and (8 a)/(27 Rb)` respectively. Reason: The compressibility factor `Z` at critical conditions is `3//8`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 - C `(RT_(c ))/(P_(c ).V_(c ))=(Rxx8axx27b^(2))/(27RTxxaxx3b)=(8)/(3)` |
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| 170. |
A gas in an open container is heated from `27^(@)C` to `127^(@)C` The fraction of the original amount of gas remaining in the container will be .A. `3//4`B. `1//2`C. `1//4`D. `1//8` |
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Answer» `PV=nRT` `n_(1)T_(1)=n_(2)T_(2)` `n_(2)=(T_(1))/(T_(2))=(3)/(4)` |
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| 171. |
Pressure remaining the same, the volume of a given mass of an ideal gas increases for every degree centigrade rise in temperature by define fraction of its volume atA. `0^(@)C`B. Its critical temperatureC. Absolute zeroD. Its Boyle temperature |
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Answer» Correct Answer - A `V_(t)=V_(o)(1+alpha_(u)t)` `because (V_(2)-V_(1))=DeltaV=V_(o)alpha(t_(2)-t_(1))` if `t_(2)-t_(1)=1^(@)` then `DeltaV=alphaV_(o)` For every `1^(@)C` increase in temperature, the volume of a given mass of an ideal gas increased by a definite fraction. |
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| 172. |
Which of the following statements is not correct about the three states of matter, i.e., solid, liquids and gas?A. Molecules of solid posses least energy whereas those of a gas posses highest energy.B. The density of a solid is highest whereas that of gases is lowest.C. Gases like liquids posses definite volumes.D. Molecules of a solid possess vibratory motion. |
| Answer» Gases and liquid do not posses definite volume. | |
| 173. |
Ammonia gas at 76 cm Hg pressure was connected to a manometer. After sparking in the flask, ammonia is partially dissociated as follows : `2NH_(3) (g) hArr N_(2) (g) + 3H_(2) (g)` the level in the mercury column of the manometer was found to show the difference of 18 cm. What is the partial pressure of `H_(2) (g)` at equilibrium ?A. 18 cm HgB. 9 cm HgC. 27 cm HgD. 24 cm Hg |
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Answer» Correct Answer - C `{:(,2NH_(3) (g),hArr,N_(2) (g) ,+ 3H_(2) (g)),(t_(0),76,,0," "0),(t_(eq),76 - 2x,,x," "3x):}` Total pressure after dissociation `= 76 - 2x + x + 3 x = 76 + 2x` `:.` Increase in pressure `= 2x = 18` `x = 9 cm` Partial pressure of `H_(2) = 3 x = 27 cm` |
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| 174. |
Name a compound having body centred cubic unit cell crystal lattice |
| Answer» Correct Answer - CsCl | |
| 175. |
At very low pressures , all real gases `(N_(2),H_(2),O_(2),etc.)` haveA. `Z gt 1`B. `Z lt 1`C. `Z = 1`D. `Z ~~ 1` |
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Answer» Correct Answer - D At high pressure, all the real gases have `Z gt 1` as these gases are less compressible than an ideal gas. On account of the repulsive forces, these are more difficult to compress. At intermediate pressure, most gases have `Z lt 1`. Thus, real gases show ideal behavior when the volume occupied by the gas is large (at low pressure), so that the volume of the molecules can be neglected in comparision to it. In other words , the behavior of the real gas becomes more ideal when the pressure is very low. |
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| 176. |
Two gases `A` and `B` present separately in two vessels `X` and `Y` at the same temperature with molecular weights `M` and `2M` respectively are effused out. The orifice in vessel `X` is circular while that in `Y` is a square. If the radius of the circular orifice is equal to that of the length of the square orifice the ratio of rates of effusion of gas `A` to that of gas `B`A. `sqrt(2)pi`B. `sqrt((pi)/(2))`C. `2 pi`D. `sqrt((2)/(pi))` |
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Answer» Correct Answer - A `(r_(x))/(r_(y))=sqrt((2M)/(M))xx(A_(x))/(A_(y))` where `A_(x)` and `A_(y)` are the area of cross section of orifice. `:. (r_(x))/(r_(y))=sqrt(2)xx(pi r^(2))/(l^(2))r`=radius of a radius orifice `l=` length of square orifice. |
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| 177. |
What is the dominant intermolecular forces or bond that must be overcome in converting liquid `CH_(3)OH` to gas ?A. Dipole-dipole interactionB. Covalent bondC. London forcesD. Hydrogen bonding |
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Answer» Correct Answer - D The molecules of `CH_(3)OH` are held together by intermolecular hydrogen bonding because `H` is directly attached to the highly electronegative `O` atom. |
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| 178. |
Which of the following metal oxides is anti-ferromagnetic in nature ?A. `MnO_2`B. `TiO_2`C. `VO_2`D. `CrO_2` |
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Answer» Correct Answer - A `MnO_2`=Antiferromagnetic, `TiO_2`=Diamagnetic , `VO_2`=Paramagnetic , `CrO_2`=Ferromagnetic. |
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| 179. |
What happens when ferrimagnetic `Fe_3O_4` is heated to 850 K and why ? |
| Answer» Ferrimagnetic `Fe_3O_4` on heating to 850 K becomes paramagnetic . This is due to greater alignment of domains (spins ) in one direction on heating | |
| 180. |
Assertion. The sum of the radii of `Na^+` and `Cl^-` ions in NaCl crystal is 281 pm. Hence, edge of the unit cell is 281 pm Reason. Edge of the unit cell is the distance between the centres of `Na^+` and `Cl^-` ions touching each other .A. If both assertion and reason are true, and reason is the true explanation of the assertion.B. If both assertion and reason are true, but reason is not the true explanation of the assertion.C. If assertion is true , but reason is falseD. If both assertion and reason are false. |
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Answer» Correct Answer - D Correct A . The edge of NaCl unit cell is the distance from `Na^+` to next `Na^+` ion or from `Cl^-` ion to next `Cl^-` ion. Correct R. Edge of the unit cell of NaCl is double the distance between the centre of `Na^+` and `Cl^-` ions touching each other. |
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| 181. |
If the distance between `Na^(+)` and `Cl^(-)` ions in NaCl crystals is 265 pm, then edge length of the unit cell will be ?A. 265 pmB. 530 pmC. 795 pmD. 132.5 pm |
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Answer» Correct Answer - B In NaCl : Edge length `=2xx` distance between `Na^(+)` and `Cl^(-)` ions `=2xx265=530` pm |
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| 182. |
Consider and ideal gas at same temperature, separated initially as shown below in the diagram When the valve is opened, the equilibrium pressure is found to be `20//7` atmosphere. What was the initial pessure in the smaller flask? |
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Answer» Correct Answer - 5 |
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| 183. |
The pyknometric density of sodium chloride crystal is `2.165xx10^3 " kg m"^(-3)` while its X-ray density is `2.178xx10^3 " kg m"^(-3)` . The fraction of the unoccupied sites in sodium chloride crystal isA. 5.96B. `5.96xx10^(-2)`C. `5.96xx10^(-1)`D. `5.96xx10^(-3)` |
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Answer» Correct Answer - D Molar volume from pyknometric density =`M/(2.165xx10^3)m^3` Molar volume from X-ray density `=M/(2.178xx10^3)m^3` `therefore` Volume unoccupied `=M/10^3(1/2.165-1/2.178)m^3=(0.013Mxx10^(-3))/(2.165xx2.178)` `therefore` Fraction unoccupied `=((0.013Mxx10^(-3))/(2.165xx2.178))//((Mxx10^(-3))/2.165)=5.96xx10^(-3)` |
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| 184. |
What type of crystal defect is produced when sodium chloride is doped with `MgCl_2` ? |
| Answer» It is called impurity defect. A cation vacancy is produced . A substitutional solid solution is formed (because 2 `Na^+` ions are replaced by one `Sr^(2+)` ion in the lattice site. ) | |
| 185. |
Why does Frenkel defect not change the density of AgCl crystals ? |
| Answer» Due to Frenkel defect, as no ions are missing from the crystal as a whole , so there is no change in density | |
| 186. |
An ideal gas obeying the kinetic theory of gases can be liquefied ifA. Its temperature is more than its critical temperature `(T_(c))`B. Its pressure is more than its critical pressure `(P_(c))`C. Its pressure is more than `P_(c)` at a temperature less than `T_(c)`D. It cannot be liquefied at any value of `P` and `T` |
| Answer» Ideal gas has no force of attraction and has negligible volume. Hence, it cannot be liquefied at any `T` and `P`. | |
| 187. |
50 mL of hydrogen diffuse through a small hole from a vessel in 20 mintues time. Time taken for 40 ml of oxygen to diffuse out under similar conditions will be :A. 12 min.B. 64 min.C. 8 min.D. 32 min. |
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Answer» Correct Answer - b `(r_(O_(2)))/(r_(H_(2)))= sqrt((d_(H_(2)))/(d_(O_(2))))` `(40)/(t) xx 20/50 = sqrt((1)/(16)) = 1/4` or `t = (40 xx 20 xx 4)/(50) = 64 mm`. |
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| 188. |
An ideal gas obeying the kinetic theory of gases can be liquefied ifA. can be liquefied if its temperature is more than critical temperatureB. can be liquefied at any value of T and P.C. cannot be liquefied under any value of T and P.D. can be liquefied if its pressure is more than critical pressure. |
| Answer» Correct Answer - c | |
| 189. |
Which of the following is a network solid ?A. `SO_2` (Solid )B. `I_2`C. DiamondD. `H_2O` (Ice) |
| Answer» Correct Answer - c | |
| 190. |
There is a colllection of crystalline substances in a hexagonal close packing. If the density of matter is `2.6 g//cm^3`, what would be the average density of matter in collection ? What fraction of the space is actually unoccupied ? |
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Answer» In hexagonal close packing , packing efficiency is 74 % `therefore` Density of matter = Packing fraction x Total density =`74/100xx2.6=1.924 "g/cm"^3` % empty space (space unoccupied ) =100-74=26% |
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| 191. |
A real gas obeying van der Waal equation will resemble ideal gas if theA. constants a and b both are smallB. a is large and b is smallC. a is small and b is largerD. constants a and b both are large . |
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Answer» Correct Answer - A According to van der Waals equation for 1 mole of a real gas `(P+(a)/(V^(2)))(V-b)=RT` If a and b are small, `a//V^(2)` and b can neglected as compared to P and V and the equation reduces to PV = RT. Hence, a real gas will resemble an ideal gas when constants a and b are small. |
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| 192. |
The van der Waal equation of gas is `(P + (n^(2)a)/(V^(2))) (V - nb) = nRT`A. (i) `CO_(2)` , (ii) `H_(2)`B. (i) `CH_(4)` , (ii) `CO_(2)`C. (i) `H_(2)` , (ii) `CO_(2)`D. `O_(2)`, (ii) `H_(2)` |
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Answer» Correct Answer - A (i) Order of intermolecular attraction is `CO_(2)gt CH_(4)gt O_(2)gt H_(2)` In `CO_(2)`, intermolecular forces increases with number of electrons in a molecule. (ii) Size increases in order of `H_(2)lt O_(2)CH_(4)lt CO_(2)` Hence, `H_(2)` will have lowest value of b. |
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| 193. |
The van der Waal equation of gas is `(P + (n^(2)a)/(V^(2))) (V - nb) = nRT` |
| Answer» Correct Answer - True | |
| 194. |
Which of the following is the most volatile liquid?A. Diethyl ethylB. Methyl alcoholC. BenzeneD. Water |
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Answer» Correct Answer - A Easily vaporized liquids are called volatile liquids , and they have relatively high vapor pressures. The most volatile liquid is diehtyl ether while water is the least volatile : Diethyl ether gt Methyl alcohol gt Benzene gt Water We can understand the order of volatility of the four liquids by considering the strength of their intermolecular attractions. Water has the lowest vapor pressure (strongest cohesive forces) because each molecule has two `H` atoms to act as hydrogen -bond donors and each molecule can accept (through two lonw pairs on `O` atom) hydrogen bonds from two other molecules. Methyl alcohol has only one potential hydrogen-donor, so its average cohesive forces are weaker than those in water and its vapor pressure is higher. In benzene and diethyl ether, the `h` atoms are all bonded to `C`, so strong `H` bonds are not possible . However , `e^(-) s` can move easily throughout the delocalized `pi`-bonding orbitals of benzene. Thus, benzene is quite polarizable and exhibits significant dispersion forces. In addition , the `H` atoms that are bonded to `C`. The `H` atoms of `C_(6)H_(6)` are attracted to the electron -rich `pi` bonding regions of nearby molecules. The accumulation of these forces , resulting in a lower vapor pressure than we might except for a hydrocarbon. The diethyl ether molecule is only slightly polar , resulting in weak dipole-dipole forces and a high vapor pressure. Also note that dispersion forces generally increase with increasing molecular size , so substances composed of larger molecules have lower vapor pressures. |
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| 195. |
The vapor density of a gas depends upon theA. total number of electrons in the moleculesB. total number of neutrons in the moleculesC. total number of moleculesD. molecular mass of the gas |
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Answer» Correct Answer - D According to the ideal gas law, `d = pM//RT`. |
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| 196. |
Match the column I with column II and mark the appropriate choice . A. `(A) to (i) , (B) to (ii) , (C ) to (iv) , (D) to (iii)`B. `(A) to (iii) , (B) to (i) ,(C ) to (iv) , (D) to (ii)`C. `(A) to (ii) , (B) to (iii) , (C ) to (i) , (D) to (iv)`D. `(A) to (iv) , (B) to (ii) , (C ) to (iii) , (D) (i)` |
| Answer» Correct Answer - B | |
| 197. |
Match the columm I with column II and mark the appropriate choice . A. `(A) to (iii) , (B) to (ii) , (C ) to (i)`B. `(A) to (i) , (B) to (ii) , (C ) to (iii)`C. `(A) to (iii) , (B) to (i) , (C ) to (iii)`D. `(A) to (ii) , (B) to (iii) , (C ) to (i)` |
| Answer» Correct Answer - A | |
| 198. |
The correct expression of partial pressure in terms of mole fraction isA. `p_(1) = x_(1) p_("total"), p_(2) = x_(2)P_("total")`B. `P = x_(1) x_(2) x_(2) P_("total")`C. `P_("total") = P_(1) x_(1) , P_("total") = P_(2) x_(2)`D. `P_(1) + P_(2) = x_(1) + x_(2)` |
| Answer» Correct Answer - A | |
| 199. |
In the mixture of nonreacting ideal gases, the partial pressure of gas, mole fraction of gas, and total pressure of mixture are related asA. `p_(gas) = (p t o t a l)/(chi_(gas))`B. `p_(gas) = chi_(gas) p_(t o t a l)`C. `p_(gas) = p_(t o t a l)chi_(gas)^(2)`D. `p_(gas) = (chi_(gas))/(p_(t o t al))` |
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Answer» Correct Answer - B Consider a mixture of two nonreacting ideal gases `A` and `B` in a closed container of volume `V` at temperature `T`. `p_(A)V = n_(A)RT` `p_(B)V = n_(B)RT` `p_(t o t a l)V = n_(t o t al)RT` On dividing `p_(A)` by `p_(t o tal)`, we get `(p_(A))/(p_(t o tal)) = chi_(A)` where `chi_(A)` is called the mole fraction of gas `A`. Thus, `p_(A) = chi_(A) p_(t o tal)` In general, `p_(i) = chi_(i)p_(t o tal)` where `p_(i)` and `chi_(i)` , respectively are partial pressure and mole fraction of the `ith` gas. |
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| 200. |
What are the untis of coefficient of viscosity ? |
| Answer» Correct Answer - Poise. | |