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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.
| 1801. |
During the evaporation of liquidA. The Temperature of the liquid will riseB. The temperature of the liquid will fallC. May rise or fall depending on the natureD. The temperature remains unaffectd |
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Answer» Correct Answer - B In the process of evaporation , high energy molecules leave the surface of liquid ,hence average kinetic energy and consequently the temperature of liqiuid falls. |
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| 1802. |
A `5.8%` solution of `NaCI` has vapour pressure closets to:A. `5.8%` solution of ureaB. 2m solution of glucoseC. 1m solution of ureaD. `5.8%` solution of glucose |
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Answer» Correct Answer - B `m = ((5.8)/(58.5))/(94.2) xx 1000 = 1.06` So `2m` solution of glucose will have same vapour pressure of this `NaCI` solution. |
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| 1803. |
Boiling point of water is defined as the temperature at which:A. vapour pressure of water equal to that of atmospheric pressureB. bubbles are formedC. steam comes outD. none of the above |
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Answer» Correct Answer - A By definition |
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| 1804. |
The partial pressure of ethane over a solution containing `6.56xx10^(-2) g` of ethane is 1 bar. If the solution contains `5.00xx10^(-2) g` of ethane, then the partial pressure of the ethane gas will be |
| Answer» Correct Answer - `7.62` bar | |
| 1805. |
The degree of dissociation of `Ca(NO_(3))_(2)` in a dilute aqueous solution, containing `7.0 g` of the salt per `100 g` of water at `100^(@)C` is `70%`. If the vapour pressure of water at `100^(@)C` is `760 mm`, calculate the vapour pressure of the solution. |
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Answer» Correct Answer - 746.24 mm/Hg `("P"_("solvent")^(0)-"P"_("solution"))/("P"_("solution"))=[1+("n"-1)alpha]("w"_("solute")xx"m"_("solvent"))/("m"_("solute")xx"w"_("solvent"))" "(("P"_("s")^(0)-"P"_("s"))/("P"_("s"))=("in"_("B"))/("n"_("A")))` `(760-"P"_("solution"))/("P"_("solution"))=[1+(3-1)0.70]xx(7xx18)/(164xx100)` `"P"_("solution")=746.2"mm of Hg"` |
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| 1806. |
The boiling point elevation and freezing point depression of solutions have a number of partical applications. Ethylene glycol `(CH_(2)OH-CH_(2)OH)` is used in automobile radiatiors as an antifreeze because it lowers the freezing point of the coolant. The same substance also helps to prevent the radiator coolant from boiling away by elevating the boiling point. Ethylene glycol has low vapour pressure. We can also use glycerol as an antifreeze. In order for the boiling point elevation to occur, the solute must be non-volatile, but no such restriction applies to freezing point depression. For example, methanol `(CH_(3)OH)`, a fairly volatile liquid that boils only at `65^(@)C`, is sometimes used as an antifreeze in automobile radiators. `124 g` each of the two reagents glycol and glycerol are added in `5 kg` of water of the radiators in two cars. Which of the following statements is wrong?A. Both will act as antifreeze.B. Glycol will be better.C. Glycerol is better because its molar mass is greater than glycol.D. Glycol is more volatile than glycerol. |
| Answer» Correct Answer - C | |
| 1807. |
The boiling point of `C_(6)H_(6), CH_(3)OH, C_(6)H_(5)NH_(2)` and `C_(6)H_(5)NO_(2)` are `80^(@)C, 65^@ C ,184^@ C` and `212^(@)C` respectively. Which will show highest vapour pressure at room temerature:A. `C_(6)H_(6)`B. `CH_(3)OH`C. `C_(6)H_(5)NH_(2)`D. `C_(6)H_(5)NO_(@)` |
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Answer» Correct Answer - B Boiling point `darr` vapour pressure `uarr` |
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| 1808. |
The boiling points of `C_(6)H_(6),CH_(3)OH,C_(6)H_(5)NH_(2)` and `C_(6)H_(5)NO_(2)` are `80^@C,65^@C,184^@C,` and `212^@C` respectively. Which of the following will have highest vapour pressure at room temperature ?A. `C_(6)H_(6)`B. `CH_(3)OH`C. `C_(6)H_(5)NH_(2)`D. `C_(6)H_(5)NO_(2)` |
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Answer» Correct Answer - B Lesser b.pt.More vapour pressure . |
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| 1809. |
The boiling point elevation and freezing point depression of solutions have a number of partical applications. Ethylene glycol `(CH_(2)OH-CH_(2)OH)` is used in automobile radiatiors as an antifreeze because it lowers the freezing point of the coolant. The same substance also helps to prevent the radiator coolant from boiling away by elevating the boiling point. Ethylene glycol has low vapour pressure. We can also use glycerol as an antifreeze. In order for the boiling point elevation to occur, the solute must be non-volatile, but no such restriction applies to freezing point depression. For example, methanol `(CH_(3)OH)`, a fairly volatile liquid that boils only at `65^(@)C`, is sometimes used as an antifreeze in automobile radiators. Which of the following is a better reagent for depression in freezing point but not for elevation in boiling point?A. `CH_(3)OH`B. C. D. `C_(6)H_(12)O_(6)` |
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Answer» Correct Answer - A `DeltaT_(f)=K_(f).m` and `K_(f) prop` molecular weight. Therefore, larger the molecular weight lesser will be depression in freezing point . |
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| 1810. |
Compartment `A` and `B` have the following combinations of solution: `{:(,A,B),(1,0.1 M KCl,0.2 M KCl),(2,0.1%(m//V) NaCl,10% (m//V) NaCl),(3,18 gL^(-1) "glucose",34.2 gL^(-1) "sucrose"),(3,20% (m//V) "glucose",10% (m//V) "glucose"):}` Indicate the solution(s) in which compartment `A` will show an increases in volume.A. `4`B. `2`C. `3`D. `5` |
| Answer» Correct Answer - A | |
| 1811. |
Compartment `A` and `B` have the following combinations of solution: `{:(,A,B),(1,0.1 M KCl,0.2 M KCl),(2,0.1%(m//V) NaCl,10% (m//V) NaCl),(3,18 gL^(-1) "glucose",34.2 gL^(-1) "sucrose"),(3,20% (m//V) "glucose",10% (m//V) "glucose"):}` The solutions in which compartment `B` is hypertonic.A. `1,2`B. `2,3`C. `3,4`D. `1,4` |
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Answer» Correct Answer - A For solution 1 and 2 the concentration in compartment `B` is higher than in `A`. |
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| 1812. |
Compartment `A` and `B` have the following combinations of solution: `{:(,A,B),(1,0.1 M KCl,0.2 M KCl),(2,0.1%(m//V) NaCl,10% (m//V) NaCl),(3,18 gL^(-1) "glucose",34.2 gL^(-1) "sucrose"),(3,20% (m//V) "glucose",10% (m//V) "glucose"):}` Answer the following question: Indicate the number of solutions which is/are isotonic.A. `1 only`B. `3 only`C. `4 only`D. `2 only` |
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Answer» Correct Answer - B Isontonic solution have same molarity. Molarity of `18 g` glucose =`(18/276)//(1000/1000)=0.065` Molarity of `34.2 g` sucrose = `(34.2/518)//(1000/1000) = 0.066` The concentration of `18 g L^(-1)` glucose is same to `34.2 g L^(-1)` sucrose. |
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| 1813. |
Compartment `A` and `B` have the following combinations of solution: `{:(,A,B),(1,0.1 M KCl,0.2 M KCl),(2,0.1%(m//V) NaCl,10% (m//V) NaCl),(3,18 gL^(-1) "glucose",34.2 gL^(-1) "sucrose"),(3,20% (m//V) "glucose",10% (m//V) "glucose"):}` The solution in which there will be no change in the level of the solution in the compartment `A` and `B` is.A. `1`B. `2`C. `4`D. `3` |
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Answer» Correct Answer - D Isotonic solutins have same concentration and therefore same rise in osmotic pressure. |
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| 1814. |
0.067 molar aqueous solution of a binary electrolyte `A^(+)B^(-)` shows 2.46 atm osmotic pressure at `27^(@)C` . What fraction of `A^(+)B^(-)` remains unionised ?A. 0.1B. 0.15C. 0.5D. Zero |
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Answer» Correct Answer - 0.03 |
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| 1815. |
The osmotic pressure of a solution contaiining 1.0 g protenin in 200 mL of aqueous solution at `27^(@)C` has been found to be `2.0xx10^(-1)`atm. Calculate the molecular mass of protein. (R=0.82 L atm `mol^(-1)`). |
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Answer» Correct Answer - 6`15.7 g 1mol^(-1)` `W_(B)=1.0g, V=200mL=0.2 L, T=273^(@)C=27+273=300K, pi=2.0xx10^(-1)atm` `R=0.082" L atm mol"^(-1)K^(-1),M_(B)=?` `M_(B)=(W_(B)xxRxxT)/(pixxV)=((1g)xx(0.082" L atm K"^(-1)mol^(-1))xx(300K))/((2xx10^(-1)atm)xx(0.2L))=615.7" g mol"^(-1)`. |
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| 1816. |
Using molarity as a conversion factor: An experiment calls for the addition to a reaction vessel of `0.200 g` of sodium hydroxide `(NaOH)` in aqueous solution. How many milliliters of `0.200 M NaOH` should be added? Strategy: According to equating `2.13` `M=n_(solute)/V_(L)` First we need to convert grams `NaOH` to moles `NaOH`, because molarity relates moles of solute to volume of solution. Then, we convert moles `NaOH` to litres of solution, using the molarity as a conversion factor. Here, `0.200 M` means that `1 L` of solution contains `0.200` moles of solute `(NaOH)`, so the conversion factor is: `(1L soln)/(0.200 mol NaOH)` (Converts mol NaOH to L soln) |
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Answer» The formula mass of `NaOH` is 40.0 u formula unit. Thus, the molar mass of `NaOH` is `40.0 g mol^(-1)`. The required calculation is `Mass rarr "Mole" rarr Litre` `0.200 g NaOHxx(1 mol NaOH)/(40.0 g NaOH)xx(1 L soln)/(0.200 mol NaOH)` `=0.025 L soln` `=25 mL` We thus need to add `25 mL` of `0.200 M NaOH` solution to the reaction vessel. Alternatively, according to equation `2.14` `M=n_(solute)/V_(mL)xx(1000 mL)/L` `=mass_(solute)/((molar mass)_(solute) V_(mL))xx(1000 mL)/L` `0.200 M=(0.200 g NaOH)/((40.0 g NaOH)/(mol NaOH)xxV_(mL))xx(1000 mL)/L` `or V_(mL)=25 mL` |
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| 1817. |
Calculate molarity from mass and volume: A solution is prepared from `22.5 g` of glucose `(C_(6)H_(12)O_(6))` and enough water for a volume of `250 cm^(3)`. Find the molarity of the glucose. Strategy: We first convert mass glucose to moles, because the molarity equals moles of solute (giucose) divided by volume of solution in litres. To calculate moles, we must divide the given mass of glucose by its molar mass. Molar is gram atomic mass (if solute consists of atoms), gram molecular mass (if solute consists of molecules) or gram formula mass (if solute consista of formula units) Gram atomic mass is the mass in grams, numerically equal to the atomic mass expressed in amu. Gram molecular mass is the mass in grams, which is numerically equal to the molecular mass expressed in amu. Gram formula mass is the mass in grams, numerically equal to formula mass expressed in amu. |
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Answer» Glucose is a polar organic compound consisting of `C_(6)H_(12)O_(6)` molecules (180 u molecule -1). Thus, its molar mass will be expressed as gram molecular= `"Mass"/("Molar mass")` `(n_("glucose"))=(22.5 g "glucose")/(180 g "glucose"//"mole glucose")` =0.125 mol glucose Volume of the solution in liters `250 cm^(3)xx(1 L)/(1000 cm^(3))=0.25 L` solution Using equation `2.13` `M=n_("glucose")/V_(L)=(0.125 "mol glucose")/(0.25 L "solution")` `=0.5 mol L^(-1)` `=0.5 M` |
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| 1818. |
What type of colloidal sols are formed in the following ? (i) Through cooled water, vapours of sulphur are passed. (ii) White of an egg is mixed with water. |
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Answer» (i) sulphur molecules associate together to form molecular sols. (ii) Macromolecular sol because protein molecules present in the white of the egg are macromolecules soluble in water. |
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| 1819. |
What happens when persistent dialysis of colloidal solution is carried out? |
| Answer» The stability of a colloidal sol is due to the presence of a small amount of the electrolyte. On persistent dialysis, the electrolyte is completely removed. As a result, the colloidal sol becomes unstable unstable and gets coagulated. | |
| 1820. |
Why is adsorption always exothermic ? |
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Answer» Correct Answer - Adsorption is accompanied by decrease of randomness, i.e. this factor opposes the process, i.e. `DeltaS` is -ve. For the process to be spontaneous, `DeltaG` must be -ve. Hence, according to eqn, `DeltaG=DeltaH-TDeltaS, DeltaG` can be -ve only if `DeltaH` is -ve. (a) `3I^(-)+S_(2)O_(8)^(2-) rarr I_(3)^(-)+2SO_(4)^(2-)` `(-d[S_(2)O_(8)^(2-)])/(dt)=1.5xx10^(-3)" "(-1)/(3)(Delta[I^(-)])/(Deltat)=(-Delta[S_(2)O_(8)^(2-)])/(Deltat)` `-1/3(Delta[I^(-)])/(Deltat)=1.5xx10^(-3)" "-(Delta[I^(-)])/(Deltat)=4.5xx10^(-3)` (b) `-(d[S_(2)O_(8)^(2-)])/(dt)=1/2(d[SO_(4)^(2-)])/(dt)=1.5xx10^(-3)=1/2(d[SO_(4)^(2-)])/(dt)` `(d[SO_(4)^(2-)])/(dt)=3xx10^(-3)` |
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| 1821. |
A solution contains 20 g sodium chloride in `95cm^(3)` solution. The density of solution is 1.25 g `cm^(-3) g cm^(-3)`. What is the mass present of NaCI ? |
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Answer» Correct Answer - 0.1684 Density of NaCI solution `=1.25 g cm^(-3)`: Volume of solution = 95`cm^(3)` `"Mass of solution" ="Volume"xx"density"=(95cm^(3))xx(1.25 g cm^(-3))=118.75 g` `"Mass percent of NaCI" =("Mass of NaCI")/("Mass of solution")xx100=((20g))/((118.75 g))xx100=16.84%`. |
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| 1822. |
Which of the following ions is most effective in the coagulation of an arsenious suphide solution?A. `K^(+)`B. `Mg^(2+)`C. `Al^(3+)`D. C |
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Answer» Correct Answer - C Effectiveness of ion in coagulation `prop` charge on coagulating ion. |
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| 1823. |
Which of the following ions is most effective in the coagulation of ferric hydroxide solution?A. `Cl^(-)`B. `Br^(-)`C. `NO_(2)^(-)`D. `SO_(4)^(2-)` |
| Answer» Correct Answer - D | |
| 1824. |
small liquid droplets dispersed in another liquid is called :A. SuspensionB. EmulsionC. GelD. True solution |
| Answer» Correct Answer - B | |
| 1825. |
Some common salt is added to ice taken in a flask kept at `0^(@)C`. With timeA. there is no change in the mass of the iceB. mass of ice decreasesC. vapour pressure in the flask increasesD. none of these |
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Answer» Correct Answer - B Ice mixed with salt has a melting point of `-32^(@)C`. As such it will melt at `0^(@)C`. |
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| 1826. |
A substance is efflorescent at a particular level if its vapour pressure isA. more than that of water vapour in the atmosphereB. equal to that of water vapour in the atmosphereC. less than that of water vapour in the atmosphereD. None of these |
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Answer» Correct Answer - A A level information |
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| 1827. |
A Beckmann thermometer is used to measureA. low temperaturesB. high temperaturesC. normal temperaturesD. None of these |
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Answer» Correct Answer - D A Beckmann thermometer is used to measure a small change in temperature only. |
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| 1828. |
The beakers A and B containing pure water (A) and an aqueous solution of sugar (B) are placed in a closed glass container. The temperature of the container is kept constant. It is observed thatA. level of water in beaker A decreases and that in beaker B increases with timeB. level of water in beakers A and B remains constant with timeC. level of water in beaker A increases and that in beaker B decreases with timeD. level of both the beakers decreases with time |
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Answer» Correct Answer - A Rate of condensation remains the same for both the beakers. While rate of evaporation is more for beaker A (containing pure water) than for beaker B (containing sugar solution). As such equilibrium cannot be attained. |
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| 1829. |
In which case depression freezing point is equal to cryscopic for water:A. `6%`by mass of urea in aqueous solutionB. `100`gof sucrose in `100`mL solutionC. `9`g of urea in `59`g aqueous solutionD. `1`M KCl solution |
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Answer» Correct Answer - C `because DeltaT=K_(f)xx "molarity"` `59g`solution has 9g glucose and 50g water `:.DeltaT=K_(f)xx(9)/((180xx50)/(1000))=K_(f)` |
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| 1830. |
Density of 2M NaCI solution is 1.25 g/cc.The molality of the solution is :A. 2.79 molalB. 0.279 molalC. 1.279 molalD. 3,85 molal |
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Answer» Correct Answer - a `"Mass of 3 M NaCI solution" = "density" x "volume" =(1.25g/ccxx1000cc)=1250g ` Molar mass of NaCI =58.5 g/mol Mass of 3 moles of NaCI=`(58.5 g mol)xx(3 mol)=1.074 kg` `"Molality of solution"=("No.of moles of NaCI")/("Mass of solvent in kg")` `=((3mol))/((1.074 kg))=2.79 mol kg^(-1)=2.79 molal` |
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| 1831. |
The freezing point of the dilute solution of acetamide in glacial acetic acid is `298K`. This is the value when crystals ofA. Acetamide first appearsB. Acetic acid from appearsC. Both appear togetherD. Ice first appears |
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Answer» Correct Answer - B `CH_(3)COOH` has intermolecular hydrogen bonding |
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| 1832. |
15 g of an unknow molecular substance is dissolved in 450 g of water. The reulting solution freezen at `-0.34^(@)C`. What is the molar mass of the substance (`K_(f)` for water=1.86 K kg `mol^(-1)`) |
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Answer» Correct Answer - 182.35 g `mol^(-1)` `W_(B)=15.0g, W_(A)=540g, 0.45kg DeltaT_(f)=0-(-0.34)^(@)C` `=0.34^(@)C=0.34 K, K_(f)=1.86" K kg mol"^(-1)` `M_(B)=(K_(f)xxW_(B))/(DeltaT_(f)xxW_(A))((1.86" K kg mol"^(-1))xx(15.0g))/((0.34K)xx(0.45 kg))=182.35" g mol"^(-1)` |
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| 1833. |
Calculate the molality of a solution that contains 51.2 g of naphthanlene `(C_(10)H_(18))` in 500 mL of carbon tetrechloride. Density of `C CL_(4)` is 1.60 g/mlA. 0.250 mB. 0.500 mC. 0.840 mD. 1.69 m |
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Answer» Correct Answer - b `"Mass of" C CI_(4)(W_(A))="Volume "xx"density "` `=(500mL)xx(1.60 g mL^(-1))=800g =0.8 kg` `"Molality of "C CI_(4)(m)=("No.ofmoles of napthalene")/("Mass of "C CI_(4)"in kg")` ` =((51.2g)//(128 g mol^(-1)))/((0.8 kg))` 0.5 `mol kg^(-1)=0.5 m` |
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| 1834. |
A solution containing 25.6 g of sulphur dissolved in 1000 g of naphthanlene whose melting point is `80.1^(@)C`, gave a freezing point lowering of `0.68^(@)C`. Calculate the formula of sulphur. `K_(f)` for naphthanlene is 6.8 K/m. |
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Answer» Correct Answer - `S_(8)` `W_(B)=25.6g, W_(A)=1 kg, DeltaT_(f)=0.68K, K_(f)=6.8" k kg mol"^(-1)` `M_(B)=(K_(f)xxW_(B))/(DeltaT_(f)xxW_(A))=((6.8" K kg mol"^(-1))xx(25.6 g))/((0.68K)xx(1kg))=256" g mol"^(-1)` `"Atomicity of sulphur"=((256" g mol"^(-1)))/((32" g mol"^(-1)))=8` `"Therefore, molecular fomula of sulphur"=S_(8)`. |
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| 1835. |
Calculate the tmperature at which a solution containing 54 g glucose `(C_(6)H_(12)O_(6))` in 80 g of water will freeze (`K_(f)`for water=1.86 K`mol^(-1)`kg). |
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Answer» Correct Answer - 270.77 K `W_(B)=54 g, M_(B)=180" g mol"^(-1), W_(A)=250 g=0.25 kg` `"No. of moles of glucose"(n_(B))=W_(B)/M_(B)=((54g))/((180"g mol"^(-1)))=0.3mol` `"Molarity of glucose solution (m)"=n_(B)/w_(B)=(0.3mol)/(0.25kg)=1.2" mol kg"^(-1).` `DeltaT_(f)=K_(f)xxm=(1.86" K kg mol"^(-1))=2.23 K`. `"Freezing point of solution"=(273.0-2.23)K=270.77 K` |
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| 1836. |
Calculate the freezing point of a solution conataining 54 g glucose `(C_(6)H_(12)O_(6)` in 250 gof water will freeze `K_(f)` for waer is 1.86 `Km^(-1))` |
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Answer» Correct Answer - 272.93K `W_(B)=0.52g, M_(B)=180" g mol"^(-1), W_(A)=80 g=0.08 kg` `DeltaTK_(f)=1.86" K m"^(-1)=1.86" K kg mol"^(-1)`. `DeltaT_(f)=(W_(B)xxK_(f))/(M_(B)xxW_(A))=((0.52g)xx(1.86" K kg mol"^(-1)))/((180" g mol"^(-1))xx(0.08" kg"))=0.067 K` `"Freezing point of solution" = 273 K-0.067 K=272.93 K` |
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| 1837. |
Two beakers of capacity 500mL were taken. One of these beakers, labelled as “A”, was filled with 400mL water whereas the beaker labelled “B” was filled with 400mL of 2M solution of NaCl. At the same temperature both the beakers were placed in closed containers of same material and same capacity as shown in Fig. 2.2. At a given temperature, which of the following statement is correct about the vapour pressure of pure water and that of NaCl solution. (i) vapour pressure in container (A) is more than that in container (B). (ii) vapour pressure in container (A) is less than that in container (B). (iii) vapour pressure is equal in both the containers. (iv) vapour pressure in container (B) is twice the vapour pressure in container (A). |
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Answer» (i) vapour pressure in container (A) is more than that in container (B). |
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| 1838. |
On the basis of information given below mark the correct option. Information: (A) In bromoethane and chloroethane mixture intermolecular interactions of A–A and B–B type are nearly same as A–B type interactions. (B) In ethanol and acetone mixture A–A or B–B type intermolecular interactions are stronger than A–B type interactions. (C) In chloroform and acetone mixture A–A or B–B type intermolecular interactions are weaker than A–B type interactions. (i) Solution (B) and (C) will follow Raoult’s law. (ii) Solution (A) will follow Raoult’s law. (iii) Solution (B) will show negative deviation from Raoult’s law. (iv) Solution (C) will show positive deviation from Raoult’s law. |
| Answer» (ii) Solution (A) will follow Raoult’s law. | |
| 1839. |
`K_(H)" value for Ar(g)," CO_(2)(g)`,HCHO(g) and `CH_(4)(g)` are 40.39, 1.67, `1.83xx10^(-5)` and 0.413 respectively.A. `HCHOltCH_(4)ltCO_(2)ltAr`B. `HCHOltCO_(2)ltCH_(4)ltAr`C. `ArltCO_(2)ltCH_(4)ltHCHO`D. `ArltCH_(4)ltCO_(2)ltHCHO` |
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Answer» Correct Answer - c Higher th KH value, lesser is the solubility of a gas in a given solvent. Therefore, it Is the correct increasing order of relative solubilities. |
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| 1840. |
On the basis of information given below mark the correct option.Information : On adding acetone to methanol some of the hydrogen bonds between methanol molecules break.(i) At specific composition methanol-acetone mixture will form minimum boiling azeotrope and will show positive deviation from Raoult’s law.(ii) At specific composition methanol-acetone mixture forms maximum boiling azeotrope and will show positive deviation from Raoult’s law.(iii) At specific composition methanol-acetone mixture will form minimum boiling azeotrope and will show negative deviation from Raoult’s law.(iv) At specific composition methanol-acetone mixture will form maximum boiling azeotrope and will show negative deviation from Raoult’s law. |
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Answer» (ii) At specific composition methanol-acetone mixture forms maximum boiling azeotrope and will show positive deviation from Raoult’s law. |
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| 1841. |
KH value for Ar(g), CO2(g), HCHO(g) and CH4(g) are 40.39, 1.67, 1.83×10–5 and 0.413 respectively.Arrange these gases in the order of their increasing solubility.(i) HCHO < CH4 < CO2 < Ar(ii) HCHO < CO2 < CH4 < Ar(iii) Ar < CO2 < CH4 < HCHO(iv) Ar < CH4 < CO2 < HCHO |
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Answer» (iii) Ar < CO2 < CH4 < HCHO |
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| 1842. |
Intermolecular forces between two benzene molecules are nearly of same strength as those between twop toluene molecules. For a mixture of benzene and toluene, which of the following are not true ?A. `Delta_(mix)H-zero`B. `Delta_(mix)V=zero`C. These will form minimum boiling azeotrope.D. These will not form ideal solution. |
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Answer» Correct Answer - `(c,d)` are not ture statemnents |
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| 1843. |
Which of the following factor (s) affect the solubility of a gaseous solute in the fixed volume of liquid solvent?(a) nature of solute (b) temperature (c) pressure(i) (a) and (c) at constant T(ii) (a) and (b) at constant P(iii) (b) and (c) only(iv) (c) only |
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Answer» (i) (a) and (c) at constant T (ii) (a) and (b) at constant P |
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| 1844. |
Intermolecular forces between two benzene molecules are nearly of same strength as those between two toluene molecules. For a mixture of benzene and toluene, which of the following are not true?(i) ΔmixH = zero(ii) ΔmixV = zero(iii) These will form minimum boiling azeotrope.(iv) These will not form ideal solution. |
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Answer» (iii) These will form minimum boiling azeotrope. (iv) These will not form ideal solution. |
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| 1845. |
Relative lowering of vapour pressure is a colligative property because _____________. (i) It depends on the concentration of a non electrolyte solute in solution and does not depend on the nature of the solute molecules. (ii) It depends on number of particles of electrolyte solute in solution and does not depend on the nature of the solute particles. (iii) It depends on the concentration of a non electrolyte solute in solution as well as on the nature of the solute molecules. (iv) It depends on the concentration of an electrolyte or nonelectrolyte solute in solution as well as on the nature of solute molecules. |
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Answer» (i) It depends on the concentration of a non electrolyte solute in solution and does not depend on the nature of the solute molecules. (ii) It depends on number of particles of electrolyte solute in solution and does not depend on the nature of the solute particles. |
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| 1846. |
Show that relative lowering of vapour pressure is a colligative property. |
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Answer» Consider a binary solution containing a nonvolatile solute. If P and P are vapour pressures of a pure solvent and the solution respectively then, Lowering of vapour pressure = ΔP = P0 – P Relative lowering of vapour pressure = \(\frac{P_0-P}{P_0}\) By Raoult’s law, \(\frac{P_0-P}{P_0}\) = x2 Where x2 is mole fraction of the solute. Therefore the relative lowering of vapour pressure is a colligative property. |
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| 1847. |
10g of cane sugar (molecular mass = 342) in `1 xx 10^(-3) m^(3)` of solution produces an osmotic pressure of `6.68 xx 10^(4) Nm^(-2)` at 273 K. Calculate the value of S in `SI` units. |
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Answer» `S=(PV)/(nxxT)` Given : `V=1xx10^(-3)m^(3),P=6.68xx10^(4)Nm^(-2)`, `n =(10)/(342),T=273K` `S=(6.68xx10^(4)xx1xx10^(-3)xx342)/(10xx273)=8.2684" J K"^(-1)"mol"^(-1)`] |
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| 1848. |
If a thin slice of sugar beet is placed in concentrated solution of `NaCl`, thenA. Sugar beet will lose water from its cells.B. Sugar beet will absorb water from solution.C. Sugar beet will neither absorb nor lose waterD. Sugar beet will dissolve in solution. |
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Answer» Correct Answer - A Osmosis occurs from dilute solution to concentrated solution, i.e., exosmosis. |
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| 1849. |
What is depression of freezing point ? |
| Answer» The freezing point of a solution containing non-volatile solute is always lower than the freezing point of pure solvent. The difference in freezing points between solution and pure solvent is called depression of freezing point. | |
| 1850. |
What is vapour pressure of a liquid ? |
| Answer» The pressure exerted by vapour over liquid when it is in equilbrium with the liquid is called vapour pressure. | |