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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.
| 51. |
If the temperature increases from `0^(@)C` to `50^(@)C` at atmospheric pressure, which of the following processes is expected to take place more in case of liquidsA. fusionB. vaporisationC. solubilizationD. none |
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Answer» Correct Answer - B An increase in temperature favoures evaporation due to increase in average kinetic energy of molecules. |
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| 52. |
The lubricating action of an oil is more if it possess:A. High vapour pressureB. low vapour pressureC. high surface tensionD. high density |
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Answer» Correct Answer - B Substances of high vapour pressure (e.g. gasoline ) evaporates more quickly than substances of low vapour pressure (e.g. motor oil) |
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| 53. |
A solution is 0.25% by mass. The weight of solvent containing 1.25g. Of solutes would beA. 506gB. 498.75 gC. 580.25 gD. 581.25 g |
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Answer» Correct Answer - B Mass of solution =100 g. Mass of solute = 0.25 g. :. Mass of solvent =100-0.25 = 99.75 g. 0.25 of solute is present in 99.75g of solvent :. 1.25 g. of solute will be present in `(99.75 xx 1.25)/0.25 =498.75` g of solvent. |
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| 54. |
In a solution if the amount of solvent is doubled, keeping the amount of solutes same, the share of solute in the solutionA. become halfB. would decrease but not a halfC. remain unchangedD. change unpredictably |
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Answer» Correct Answer - B No. of moles of solvent are increacsing but total no. of moles of solute solvent are not becoming double, However the no. of moles of solute are same. Hence the share of solute in the solution would decrease but not to half. |
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| 55. |
When 10g of a non-volatile solute is dissolved in 100 g of benzene, it raises boiling point by `1^(@)C` then molecular mass of the solute is (`K_(b)` for benzene `= 2.53 k-m^(-1)`)A. 223 gB. 233 gC. 243 gD. 253 g |
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Answer» Correct Answer - D `Delta T_(b)=(K_(b)xx w xx1000)/(m xx W)` `m=(K_(b)xx wxx 1000)/(Delta T_(b)xx W)=(2.53xx10xx1000)/(1xx100)=253 g` |
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| 56. |
If for a sucrose solution elevation in boiling point is `0.1^(@)C` then what will be the boiling point of NaCl solution for same molal concentrationA. `0.1^(@)C`B. `0.2^(@)C`C. `0.08^(@)C`D. `0.01^(@)C` |
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Answer» Correct Answer - B Elevation in a boiling point is a colligative property as it depends upon the number of particles. `Delta T_(b) ptop n` For sucrose, `n = 1, Delta T_(b) = 0.1^(@)C` For NaCl, n = 2, `Delta T_(b)=0.2^(@)C` |
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| 57. |
If 0.15 g of a solute dissolved in 15 g of solvent is boiled at a temperature higher by `0.216^(@)C` than that of the pure solvent. The molecular weight of the substance (molal elevation constant for the solvent is `2.16^(@)C` kg/mol) isA. `1.01`B. 10C. `10.1`D. 100 |
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Answer» Correct Answer - D `m=(K_(b)xxw xx 1000)/(Delta T_(b)xx W)=(2.16xx0.15xx1000)/(0.216xx15)=100`. |
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| 58. |
Assuming the degree of ionization to be equal, the ratio of osmatic pressures of equimolar solution of`A1_(2)(SO_(4))_(3),Na_(3)PO_(4) and K_(4)[Fe(CN)_(6)]is`A. `5:04:05`B. `4:05:06`C. `1:0.8:1`D. `0.8:1:1` |
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Answer» Correct Answer - C `Al_(2) (SO_(4))_(3)hArr 2 A1^(3+) + 3 SO_(4)^(2-),n=5` `Na_(3) PO_(4) hArr 3 Na^(+) + PO_(4)^(3-),n=4` `K_(4)[Fe(CN)_(6)] hArr 4K^(+)+[Fe(CN)_(6)]^(4-),n=5`:. Ratio is 5:4:5 i.e. 1:0.8:1 |
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| 59. |
Two solutions have different osmotic pressures. The solution of higher osmotic pressure is called:A. Isotonic solutionB. Hypertonic solutionC. Hypotonic solutionD. None |
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Answer» Correct Answer - C In a pair of two solution ,the one having higher osmotic pressure is called hypertonic and the othar having lower osmotic pressure is called hypotonic. |
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| 60. |
The solution which has higher osmotic pressure than some other solution is known as………A. HypotonicB. HyperonicC. IsotonicD. Normal. |
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Answer» Correct Answer - B A solution with higher osmotic pressure than the other is called hypertonic. |
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| 61. |
At low concentrations, the statement that equimolal solutions under a given set of experimental conditions have equal osmotic pressure is true forA. all solutionsB. solutions of non-electrolytes onlyC. solutions of electrolytes onlyD. none of these. |
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Answer» Correct Answer - B Non electrolyets do not undergo dissociation. |
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| 62. |
The osmotic pressure of a `5% (weight// volume)` solution of cane sugar at `150^(@)C` isA. 4 atmB. 3.4 atmC. 5.078 atmD. 2.45 atm |
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Answer» Correct Answer - C `pi V = "Mass"/"Molar Mass" xx R xxT ` `pi xx 100/1000=5/342 xx0.0821 xx 423 :. pi = 5.078 "atm"` |
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| 63. |
If O.P. of 1 M of the following in water can be measured, which one will show the maximum O.P.A. `AgNO_(3)`B. `MgCl_(2)`C. `(NH_(4))_(3)PO_(4)`D. `Na_(2)SO_(4)` |
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Answer» Correct Answer - C `(NH_(4))_(3)PO_(4)` gives maximum ion. Hence, its osmotic pressure is maximum. |
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| 64. |
The phenomenon in which cells are shrinked down if placed in hypertonic solution is calledA. plasmolysisB. HaemolysisC. EndosmosisD. none |
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Answer» Correct Answer - A The exosmosis occurs from cell to solution (hypertonic or high osmotic pressure or high concentration), & cells are shrinked down called plasmolysis. |
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| 65. |
The osmotic pressure of a `5% (weight// volume)` solution of cane sugar at `150^(@)C` isA. 2.45 atmB. 5.078 atmC. 3.4 atmD. 4 atm |
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Answer» Correct Answer - B `C=(5)/(342)xx(1)/(100)xx1000=(50)/(342)` mol/1 `pi=(5)/(342)xx0.082xx423=5.07` atm |
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| 66. |
The osmotic pressure of a solution isA. `P=(RT)/(C )`B. `P=(CT)/(R )`C. `P=(RC)/(T)`D. `(P)/(C )=RT` |
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Answer» Correct Answer - D P = CRT or `(P)/(C )=RT` |
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| 67. |
As a result of osmosis the volume of the concentrated solution:A. remains constantB. increasesC. decreasesD. increases or decreases |
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Answer» Correct Answer - D Volume of dilute solution decreases while of concentrate solution increases. |
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| 68. |
An aqueous solution of sucrose,`C_(12) H_(2) O_(11)` containing 34.2 g/L has an osmotic pressure of 2.38 atmospheres at `70^(o) C. ` for an aquesous solution of glucose, `C_(6)H_(12)O_(6)` to be istonic with this solution ,it would have :A. 34.2g/LB. 17.1 g/LC. 18.0 g/LD. 36.0 g/L of glucose |
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Answer» Correct Answer - C `34.2 g L ^(-1) "sucrose" = 0.1 M,` `18 g L^(-1) "glucose" = 0.1 M` |
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| 69. |
Which salt shows maximum osmotic pressure in its `1 m` solution.A. `AgNO_(3)`B. `Na_(2)SO_(4)`C. `(NH_(4))_(3)PO_(4)`D. `MgCl_(2)` |
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Answer» Correct Answer - C `O.P. prop "mole",` i.e. `(NH_(4))_(3) PO_(4)`given 4 ions in solution. `(NH_(4))_(3) PO_(4) hArr 3 NH_(4)^(+) + PO_(4)^(-3)` |
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| 70. |
Two solutions of `KNO_(3)` and `CH_(3)COOH` are prepared separately. Molarity of both is `0.1M` and osmotic pressure are `P_(1) and P_(2)` respectively. The correct relationship between the osmotic pressure is :A. `P_(2)gt P_(1)`B. `P_(1)=P_(2)`C. `P_(1)gt P_(2)`D. `(P_(1))/(P_(1)+P_(2))=(P_(2))/(P_(1)+P_(2))` |
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Answer» Correct Answer - C `KNO_(3)` dissociates completely while `CH_(3)COOH` dissociates to a small extent. Hence, `P_(1)gt P_(2)`. |
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| 71. |
The osmotic pressure of solution increases if :A. Temperature is decreasedB. Solution concentration is increasedC. Number of solute molecules is increasedD. Volume is increased |
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Answer» Correct Answer - B::C As soon as the solute molecules increases the osmotic pressure of solution increase. |
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| 72. |
The osmotic pressure (pi) of a solution is given by reationA. `pi = (RT)/C`B. `pi = (CT)/R`C. `pi = (RC)/T`D. `pi/C =RT` |
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Answer» Correct Answer - D PV = n RT `:. P = n/v RT = CRT.` |
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| 73. |
The natural semipermeable membrane is:A. Gelatinous `Cu_(2)Fe (CN)_(6)`B. Gelatious `Ca_(3) (PO)_(4)_(2)`C. plant cellD. Phenol laver |
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Answer» Correct Answer - C A natural semipermeable membrane is one which exist in nature . |
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| 74. |
Which inorganic precipitate acts as semipermeable membrane ?A. Calcium sulphateB. Barium oxalateC. Nickel phosphateD. Copper ferrocyanide |
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Answer» Correct Answer - D Copper ferrocyanide ppt. acts as a semipermeable membrane. |
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| 75. |
Two solutions A and B are separated by semipermeable membrane. If liquid flows from A to B thenA. A is more concentrated than BB. A is less concentrated than BC. both solutions have same concentrationD. none |
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Answer» Correct Answer - B Osmosis occurs from dilute solution to concentrate solution. |
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| 76. |
The natural semipermeable membrane is:A. Calcium phosphate (gelatinous)B. Phenol layerC. Copper ferrocyanide (gelatinous)D. All |
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Answer» Correct Answer - D All are semipermeable membrane. |
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| 77. |
Which inorganic precipitate acts as a semipermeable membrane ?A. Calcium phosphateB. Nickel phosphateC. Calcium sulphateD. Copper ferrocyanide. |
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Answer» Correct Answer - D Copper ferrocyanide ppt acts as a semipermeable membrane. |
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| 78. |
Which one of the following in the ratio of the lowering of V.P. of 0.1 M aqueous solution of `BaCl_(2)`,NaCl and `Al_(2)(SO_(4)_(4)` respectively?A. `2 : 3 : 5 `B. `3 : 2 : 5 `C. `5 : 2 : 3 `D. `5 : 1 : 2 ` |
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Answer» Correct Answer - B `BaCl_(2) rarr Ba^(2+) + 2CI^(-)(3 "ions")` `NaCl rarr Na^(+) + CI^(-) (2 "ions")` `Al_(2)(SO_(4))_(3) rarr 2Al^(3+) + 3SO_(4)^(2-) (5"ions")` lowering of V.P. ratio = 3 : 2 : 5 respectively as lowering of V.P. |
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| 79. |
In equimolar solution of glucose, NaCl and `BaCl_(2)`, the order of osmotic pressure is as followA. Glucose `gt NaCl gt BaCl_(2)`B. `NaCl gt BaCl_(2)gt` GlucoseC. `BaCl_(2)gt NaCl gt` GlucoseD. Glucose `gt BaCl_(2)gt NaCl` |
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Answer» Correct Answer - C `BaCl_(2)rArr Ba^(2+)+2Cl^(-)=3` ion `NaCl rArr Na^(+)+Cl^(-)=2` ion Glucose `rArr` No ionisation `therefore BaCl_(2)gt NaCl gt` Glucose |
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| 80. |
150 mL of `C_(2)H_(5)OH`(density = 0.78 g `mL^(-1)`is diluted to one litre by adding water, molality of the solution isA. 2.54B. 11.7C. 2.99D. 29.9 |
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Answer» Correct Answer - C `m=(W_(2) xx 1000)/(M_(2) xx W_(1)(g))` 150 mL `C_(2)H_(5) OH=150 xx 0.78 g =117.0 g=W_(2)` Water =850 g = `W_(2)`=46 g `"mol"^(-1) `"Molality"= `(117 xx 1000)/ (46 xx 850)=2.99` |
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| 81. |
The number of moles of hydroxide `(HO^(-))`ion in 0.3 litre of 0.005 M solution of `Ba(OH)_(2)` isA. 0.0075B. 0.0015C. 0.003D. 0.005 |
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Answer» Correct Answer - C 0.3L of 0.005 M `Ba(OH)_(2)` sol as `M=n/V(L)` `:. N = 0.005 M Ba(OH)_(2)` sol as `M=n/(V(L))` `:.n = 0.005 xx 0.3` mol = 0.0015 mol `Ba(OH)_(2)` `(Ba^(2+)2OH^(-))` mol of `OH^(-)` = `2 xx 0.0015 = 0.0030` moles |
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| 82. |
`18`g of glucose `(C_(6)H_(12)O_(6))`is added to `178.2` g of water. The vapour pressure of water for this aqueous solution at `100^@C`is :A. `76.0`B. `752.4`C. `759.0`D. `7.6` |
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Answer» Correct Answer - B Moles of glucose `=(18)/(180)=0.1` Moles of water `=(178.2)/(18)=9.9` `rArr n_("Total")=10 rArr (Delta P)/(P^(@))=(0.1)/(10)rArr Delta P=0.01 P^(@)` `=0.01xx760=7.6` torr `P_(S)=760-7.6=752.4` torr |
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| 83. |
The vapour pressure of acetone at `20^(@)C` is 185 torr. When `1.2 g` of non-volatile substance was dissolved in `100g` of acetone at `20^(@)C` its vapour pressure was 183 torr. The moalr mass `(g mol^(-1))` of the substance is:A. 32B. 64C. 128D. 488 |
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Answer» Correct Answer - B `(P_(@)-P_(S))/(P_(S))=(n)/(N)` `(185-183)/(183)=(1.2//M)/(100//58)rArr M ~~ 64 g//mol` |
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| 84. |
If the density of some lake water is `1.25 g mL^(-1)` and contains `92g` of `Na^(o+)` ions per `kg` of water, calculate the molality of `Na^(o+)` ions in the lake. |
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Answer» Molar mass of `Na=23 " g mol"^(-1)` `therefore ` No. of moles of `Na^(+)` ions present `=(92)/(23)=4 ` moles `therefore "Molality"=(4xx1000)/(1000)=4 m` |
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| 85. |
Which one of them is more volatile component?A. `CH_(2)Cl_(2)`B. `CHCl_(3)`C. Both a and bD. Not able to determine |
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Answer» Correct Answer - a Since, `CH_(2)Cl_(2)` is a more volatile component than `CHCl_(3).[p_(CH_(2)Cl_(2))^(@) = 415 mm Hg` and `p_(CHCl_(3))^(@) = 200 mm Hg]` and the vapour phase is also richer in `[y_(CH_(2)Cl_(2)) = 0.82]` and `[y_(CHCl_(3)) = 0.18]` |
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| 86. |
Among the following compounds, identify which are insoluble, partially soluble, and highly soluble in water ? `a.` Phenol `b.` Toluene `c.` Formic acid `d.` Ethylene glycol `e.` Chloroform `f.` Pentanol |
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Answer» (i) Phenol (having polar-OH group)-Partially soluble. (ii) Toluene (non-polar)-Insoluble. (iii) Formic acid (form hydrogen bonds with water molecules )-Highly soluble. (iv) Ethylene glycol (form hydrogen bonds with water molecules) Highly soluble. (v) Chloroform (non-polar)-Insoluble. (vi) Pentanol (having polar-OH)-Partially soluble. |
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| 87. |
Based on solute `-` solvent interactions, arrange the following in order of increasing solubility in `n-`octane and explain the result. Cyclohexane, `KCl,CH_(3)OH,CH_(3)CN`. |
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Answer» (a) Cyclohexane and n-octane both are non-polar. They mix completely in all proportions. (b) KCl is an ionic compound, KCl will not dissovle in n-octane. (c ) `CH_(3)OH` is polar. `CH_(3)OH` will dissolve in n-octane. (d) `CH_(3)CN` is polar but lesser than `CH_(3)OH`. Therefore, it will dissolve in n-octane but to a greater extent as compared to `CH_(3)OH`. Hence, the oreder is `KCl lt CH_(3)OH lt CH_(3)CN lt "Cyclohexane"`. |
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| 88. |
Arrange the following in the increasing order of their solubility in n-octane based on solute-solvent interaction:A. `KCl lt CH_(3)CN lt CH_(3)OH lt` CyclohexaneB. `KCl lt` Cyclohexane `lt CH_(3)OH lt CH_(3)CN`C. `KCl lt CH_(3)OH lt CH_(3)CN lt` CyclohexaneD. `KCl lt` Cyclohexane `lt CH_(3)CN lt CH_(3)CN` |
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Answer» Correct Answer - c `KCl lt CH_(3)OH lt CH_(3)CN lt` Cyclohexane |
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| 89. |
To observe an elevation of boiling point of `0.05^(@)C`, the amount of a solute (molecular weight = 100) to be added to 100 g of water `(K_(b) = 0.5)` isA. 2 gB. 0.05 gC. 1 gD. 0.75 g |
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Answer» Correct Answer - c Elevation of boiling point, `DeltaT_(b) = (wxxK_(b)xx1000)/(MxxW_("gram"))` On substituting values, we get `0.05 = (wxx0.5xx1000)/(100xxx100)` or `w = (0.05xx100xx100)/(0.5xx1000) = 1g` |
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| 90. |
For converting a solution if 100 ml KCl of 0.4 M concentration into a solution of KCl 0.05 M concentration. The quantity of water added isA. 900 mlB. 700 mlC. 500 mlD. 300 ml |
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Answer» Correct Answer - B We know that, `M_(1)V_(1)= M_(2)V_(2)` `0.05xx V_(1)=0.4xx100` `therefore V_(1)=(0.4xx100)/(0.05)=800` `V_(2)-V_(1)=800-100 rArr 700 ml`. |
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| 91. |
35.4 mL of HCl is required for the neutralization of a solution containing 0.275 g of sodium hydroxide. The normality of hydrochloric acid isA. 0.97 NB. 0.142 NC. 0.194 ND. 0.244 N |
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Answer» Correct Answer - C We know that 1 g equivalent weight of NaOH = 40 g `therefore` 40 g of NaOH = 1 g eq. of NaOH `therefore 0.275 g` of `NaOH =(1)/(40xx0.275 eq.` `=(1)/(40)xx0.275xx1000=6.88` meq. `therefore underset((HCl))(N_(1)V_(1))=underset((NaOH))(N_(2)V_(2))` `N_(1)xx35.4=6.88 " " (because meq = NV)` `N_(1)=0.194`. |
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| 92. |
When `HgI_(2)` is added to aqueous solution of KI, which of the following are not correct statementsA. Freezing point is raisedB. Freezing point does not changeC. Freezing point is loweredD. Boiling point does not change |
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Answer» Correct Answer - B::C::D Explanation : (b) is correct choice as the statement is not correct because freezing point will change. (c ) Is correct choice as the statement is not correct because freezing point is raised in this case as explained in (a). (d) Is correct choice as the statemetn given is not correct because boiling point will change. (a) Is not correct choice as the statement is correct because number of particles will decrease in number due to formation of complex, `2KI+HgI_(2)rarr K_(2)HgI_(4)`, therefore freezing point will be raised. |
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| 93. |
`K_(2)HgI_(4)` is 50% ionised in aqueous solution. Which of following are correct ?A. n = 7B. n = 3C. i = 2D. i = 4 |
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Answer» Correct Answer - B::C Explanation : `K_(2)HgI_(4)rarr 2K^(+)+HgI_(4)^(2-)rArr n=3` `alpha=(i-1)/(n-1),0.5=(i-1)/(3-1)rArr i=2` `therefore` (b) and (c ) are correct while (a) and (d) are not correct. |
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| 94. |
Identify the correct statementA. 1 M NaCl has higher freezing point than 1 M glucose solutionB. 1 M NaCl solution has same boiling point as 1 M KCl solutionC. Molecular weight NaCl will be less than 58.5 in water because it undergoes diffociationD. `i gt 1` when solute undergoes association |
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Answer» Correct Answer - B::C Explanation : (b) is correct both have i=2 and isotonic (same) molar conc. (c ) Is correct because `NaCl rarr Na^(+)+Cl^(-)`, molecular weight will be less than 58.5. (a) Is wrong because non-volatile solute lowers down the freezing point. (d) Is not correct because ilt1, when association takes place. |
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| 95. |
Value of gas constant R isA. 0.082 litre atmB. 0.987 cal `mol^(-1)K^(-1)`C. `8.3 J mol^(-1)K^(-1)`D. 83 erg `mol^(-1)K^(-1)` |
| Answer» Correct Answer - C | |
| 96. |
Which of the following form/s an ideal solution ?A. Ethyl bromide and ethyl iodideB. `CHCl_(3)` and acetoneC. Benzene and tolueneD. Ethanol and water |
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Answer» Correct Answer - A::C Explanation : (a) and (c ) are ideal solution because they are compounds belonging to same homologous series and forces of attraction between A-B is equal to A-A and B-B. (b) Forms non-ideal solution showing -ve deviation. (d) Forms non-ideal solution showing +ve deviation. |
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| 97. |
The temperature , at which the vapour pressure of a liquid becomes equal to the atmospheric pressure is known asA. Freezing pointB. Boiling pointC. Absolute temperatureD. None of these |
| Answer» Correct Answer - B | |
| 98. |
At 300 K , when a solute is aded to a solvent ,its vapour pressure over mercury reduces from 50 mm to 45 mm . The value of mole fraction of solute wil be _____.A. `0.005`B. `0.010`C. `0.100`D. `0.900` |
| Answer» Correct Answer - C | |
| 99. |
Azeotropic mixtures areA. Constant temperature boiling mixturesB. Those which boils at different temperaturesC. Mixture of two solidsD. None of the above |
| Answer» Correct Answer - A | |
| 100. |
A mixture of benzene and toluene formsA. An ideal solutionB. Non-ideal solutionC. SyspensionD. Emulsion |
| Answer» Correct Answer - A | |