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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.
| 551. |
Which of the following solution will have highest freezing point?A. `2MNaCl` SolutionB. `1.5M AlCl_(3)` SolutionC. `1M Al_(2)(SO_(3))_(4)` SolutionD. `3M Urea` Solution |
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Answer» Correct Answer - D Depression in freezing point`prop molality xx n` For `NaCl,DeltaT_(f)prop2xx2` For`AlCl_(3),DeltaT_(f)prop4xx1.5` For`Al_(2)(SO_(4))_(3)DeltaT_(f)prop5xx1` For Urea `DeltaT_(f)prop3xx1` Since, lowest depression is observed for solution containing urea, hence it will have highest freezing point. |
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| 552. |
A certain substance `A` tetramerizes in water to the extent of `80%`. A solution of `2.5 g` of `A` in `100 g` of water lowers the freezing point by `0.3^(@)C`. The molar mass of `A` is a.`120` , b.`61` ,c.`60` ,d.`62` |
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Answer» d. `alpha=(1-i)/(1-(1/n))` `0.8=(1-i)/(1-(1/4)) rArr i=0.4` Given that `4A rarr A_(4)` Alternate method to calculate (i) `i=("Number of ions" xxalpha)+(1-alpha)` `=(1/4xx0.8)+(1-0.8) ` [`alpha=80%` or `0.8`] `=0.2+0.2=0.4` `DeltaT=iK_(f)xxm` `0.3=0.4xx1.86xx(W_(B)xx1000)/(Mw_(B)xxW_(A))` `0.3=0.4xx1.86xx(2.5xx1000)/(Mw_(B)xx100) rArr Mw_(B)=62` |
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| 553. |
Addition of `0.643`g of compound to `50`ml of benzene (density `=0.879mol^(-1)` ) lower the freezing point from`50.51^(@)C`. If `k_(f)` for benzene is `5.12` the molecule of the compound isA. `156.00`B. `312.00`C. `78.00`D. `468.000` |
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Answer» Correct Answer - A Weight of bezene =`Vxxd=50xx0.879g` `W=43.95g` Weight of compound =`0.643gm` `DeltaT_(f)=50.51-50.03=0.48` `:.DeltaT_(f)=K_(f)xx(w)/(m)xx(1000)/(w)` `0.48=5.12xx(0.643)/(m)xx(1000)/(43.95)` ` :.m=156.05` |
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| 554. |
A certain substance `A` tetramerizes in water to the extent of `80%`. A solution of `2.5 g` of `A` in `100 g` of water lowers the freezing point by `0.3^(@)C`. The molar mass of `A` is a.`120` , b.`61` ,c.`60` ,d.`62`A. `122`B. `31`C. `244`D. `62` |
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Answer» Correct Answer - D `alpha =(1-i)/(1-(1)/(n))` `(1-i)/(1-(1)/(4)) ,i=0.4` `DeltaT=iK_(f)xxm` `0.3=0.4xx1.86xx(w_(b)xx1000)/(m_(B)xxw_(A))` `0.3=0.4xx1.86xx(2.5xx1000)/(m_(B)xxw_(A))` `m_(B)=62` |
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| 555. |
The freezing point of a solution containing `50 cm^(3)` of ethylene glycol in `50 g` of water is found to be `-34^(@)C`. Assuming ideal behaviour, Calculate the density of ethylene glycol `(K_(f)` for water = `1.86 K kg mol^(-1)`).A. `1.13g//cm_(3)`B. `2.00g//cm_(3)`C. `1.8g//cm_(3)`D. `2.25g//cm_(3)` |
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Answer» Correct Answer - A `DeltaT=K_(f)xx(wxx1000)/(mxxW)` `34=1.86xx(wxx1000)/(62xx50)` `w=56.66g` `V=(w)/(d)` `50=(56.66)/(d)` `d=1.13g//cm^(3)` |
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| 556. |
The freezing point of a solution containing `50 cm^(3)` of ethylene glycol in `50 g` of water is found to be `-34^(@)C`. Assuming ideal behaviour, Calculate the density of ethylene glycol `(K_(f)` for water = `1.86 K kg mol^(-1)`). |
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Answer» `DeltaT = (K_(f) xx W_(f) xx 1000)/(Mw_(2) xx W_(1))` `34 = 1.86xx(W_(2) xx 1000)/(62 xx50)` `W_(2) = 56.66 g` `(V) = (W_(2))/(d)` `(50) = (56.66)/(d)` ` d = 1.13 g cm^(-3)`. |
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| 557. |
`1.4 g` of acetone dissolved in `100 g` of benzene gave a solution which freezes at `277.12 K`. Pure benzene freezes at `278.4 K`.`2.8` of solid `(A)` dissolved in `100 g` of benzene gave a solution which froze at `277.76 K`. Calculate the molecular mass of `(A)`. |
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Answer» We know that `DeltaT = (K_(f)) xx (W_(2) xx 1000)/(Mw_(2) xx W_(1))` where`DeltaT` = Depression in freezing point `K_(f)` = Molal depression constant of benzene` `W_(2)` = Mass of solute `Mw_(2)` = Molecular mass of solute `W_(1)` = Mass of solvent Case i: `(278.4 - 277.12) = K_(f)xx(1.4 xx 1000)/(58 xx 100)` `1.28 = K_(f) xx (14)/(58)` Case ii: `(278.4 - 277.76) = K_(f)xx(2.8 xx 1000)/(Mw_(A) xx 100)` `0.64 = K_(f) xx (28)/(Mw_(A))` Dividing Eq. (i) by (ii), we get `Mw_(A) = 232` |
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| 558. |
Pure benzene freezes at `5.45^(@)C` at a certain place but a 0.374m solution of tetrachloroethane in benzene freezes at `3.55^(@)C`. The `K_(f)` for benzene isA. `5.08 K kg mol^(-1)`B. `508 K kg mol^(-1)`C. `0.508 K kg mol^(-1)`D. `50.8^(@)C kg mol^(-1)` |
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Answer» Correct Answer - A `DeltaT_(f) = (5.45 - 3.55) = K_(f) xx m` or `1.90 = k_(f) xx 0.374` or `k_(f) = (1.90)/(0.374) = 5.08` |
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| 559. |
The osmotic pressure of a solution is `1.3 atm`. The density of solution is `1.3 g cm^(-3)`. Calculate the osmotic pressure rise. (`1 atm =76 cm Hg`,`d_(Hg)=13.6 g cm^(-3)`) |
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Answer» `pi=hdg` `1.3xx76xx13.6xxg=hxx1.1xxg` `h=(1.3xx76xx13.6)/(1.1)` cm `=1221` cm. |
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| 560. |
`pi_(1),pi_(2),pi_(3)andpi_(4)`are the osmotic pressure of `5%((W)/(V))`solution of ure ,frutose ,sucrose andKCl respectively at certain temperatures.The correct order of their magnitude isA. `pi_(1)gtpi_(4)gtpi_(2)gtpi_(3)`B. `pi_(1)gtpi_(4)gtpi_(2)gtpi_(3)`C. `pi_(4)gtpi_(1)gtpi_(2)gtpi_(3)`D. `pi_(4)gtpi_(1)gtpi_(3)gtpi_(2)` |
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Answer» Correct Answer - C Moles of urea`=(5)/(60)` , Moles of fructose `=(5)/(180)`, Moles of sucrose `=(5)/(342)` , Moles of KCl(effective) `=2xx(5)/(74.5)=(5)/(37.25)`. |
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| 561. |
Osmotic pressure of insulin solution at 298K is found to be `0.0072` atm. Hence height of water column due to this osmotic pressure is (density of `Hg = 13.6 g cm^(-3))`A. `0.76mm`B. `70.28mm`C. `74.42mm`D. `760mm` |
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Answer» Correct Answer - C Pressure = hdg `P = hdg` (for Hg) 1 atm = 76 mm of Hg `:. 0.0072` atm `-= 0.0072 xx 760mm` of Hg `P = 0.0072 xx 760 xx 13.6 xx 9.8` (for Hg) `P = h xx 1 xx 9.8` (for water) (density of water `= 1g//cm^(3))` `:. 9.8 xx h = 0.0072 xx 60 xx 13.6 xx 9.8` `h = 0.0072 xx 760 xx 13.6mm` `= 74.42 mm` |
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| 562. |
At a given temperature total vapour pressure (in Torr) of a mixture of volatile components A and B is given by `P = 12 - 75 X_(B)` hence vapour pressure of pure A and B respectively (in Torr) areA. 120,75B. 120195C. 120,45D. 75,45 |
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Answer» Correct Answer - C `P = P_(A)^(@) X_(A) + P_(B)^(@) X_(B)` `P = P_(A)^(@) (1-X_(B)) +P_(B)^(@) X_(B)` `P = P_(A)^(@) - (P_(A)^(@)-P_(B)^(@)) X_(B)` But `P = 120 - 75 X_(B)` `:. P_(A)^(@) = 120` Torr `P_(A)^(@) - P_(B)^(@) = 75` `P_(B)^(@) = P_(A)^(@) - 75 = 120 - 75 = 45` Torr |
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| 563. |
Moles of `K_(2)SO_(4)`to be dissolved in `12`mol water to lower its vapour pressure by `10`mmHg at a temperature at which vapour pressure of pure water is `50`mm is:A. 3 molB. 2 molC. 1 molD. 0.5 mol |
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Answer» Correct Answer - C Lowering ofV.P.is colligative property thus,`i_(K_(2)SO_(4))=1+(y-1)x=1+2x=3` `:.` If `(cancel(O)p)/(P^(@))=(n_(1)i)/(n_(1)i+n_(2))` `(10)/(50)=(3n_(1))/(3n_(1)+12)=(n_(1))/(n_(1)+4)n_(1)=1` |
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| 564. |
Moles of `K_(2)SO_(4)` to be dissolved in 12 moles of water of lower its vapour pressure by 10mm Hg at a temperature at which vapour pressure of pure water is 50mm Hg isA. 3molB. 2molC. 1molD. 0.5mol |
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Answer» Correct Answer - C `(Deltap)/(p) = ("in"_(B))/( "in"_(B)+n_(A))` (Please note this step) Here `i=3` `(10)/(50) = (3n_(B))/(3n_(B) +12)` `3n_(B) +12 = 15 n_(B)` `12 n_(B) = 12` `n_(B) = 1` |
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| 565. |
In depression of freezing point method camphor is a suitable solvent as itsA. `K_(f)` is highB. Sublimation is easierC. Volatility is largeD. Density is low |
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Answer» Correct Answer - 1 |
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| 566. |
Molarity of pure water isA. 1B. 55.55C. 0.1D. 5.55 |
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Answer» Correct Answer - 2 |
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| 567. |
Vapour pressure of pure water is 40 mm. if a non-volatile solute is added to it, vapour pressure falls by 4 mm. Hence, molarity of solution isA. 6.17mB. 3.09mC. 1.54mD. 0.77m |
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Answer» Correct Answer - A `(Deltap)/(p^(@)) = x_(B)` `x_(B) = (4)/(40) = 0.1` Thus, solution contains 0.1 mol of solute in 0.9 mol of water Mass of water `= 0.9 xx 18 = 16.2 g` Molarity of solution `= (0.1)/(16.2) xx 1000` `=(100)/(16.2) = 6.17m` |
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| 568. |
`0.01` mole of a non-electrolyte is dissolved in 10 g of water. The molality of the solution is :A. 0.1mB. 0.5mC. 1.0mD. 0.18m |
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Answer» Correct Answer - C Molarity `= (0.01)/(10) xx 1000 = 1m` `[x_(B) ~~ (n_(B))/(n_(A))]` |
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| 569. |
Calculate molality of an aerated drink having 2.5 gm of carbonic acid dissolved in 150 gm of water. |
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Answer» Molality= moles of solute / Mass of solvent in Kg = [(2.5 / 62.)/ (150 / 1000)] = 0.268. |
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| 570. |
10 g of glucose is dissolved in 150 g of water. The mass percentage of glucose is :A. 0.05B. 0.0625C. 0.9375D. 0.15 |
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Answer» Correct Answer - B Mass `% = (10)/(10+150) xx 100 = 6.25%` |
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| 571. |
To a 4L of 0.2M solution of `NaOH, 2L` of `0.5M NaOH` are added. The molarity of resulting solution isA. 0.9MB. 0.3MC. 1.8MD. 0.18M |
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Answer» Correct Answer - B Total moles of `NaOH` `= 4 xx 0.2 +2 xx 0.5` `= 0.8 xx 1.0 = 1.8` Molarity `= (1.8)/(6) = 0.3M` |
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| 572. |
A solution containing 10.2 g of glycrine per litre is found to be isotonic with 2% solution of glucose `("molar mass"=180 g mol^(-1))`. Calculate the molar mass of glycrine.A. 91.8 gB. 1198 gC. 83.9 gD. 890.3 g |
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Answer» Correct Answer - A `pi_"glycerine"=pi_"glucose"` `n_1/V_1RT=n_2/V_2RT` `10.2/Mxx1/1=2/180xx1000/100 ` `rArr M=(10.2xx18)/2=91.8 g ` (Density of water = `1 g//cm^3`) |
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| 573. |
State Henry's law about partial pressure of a gas in a mixture. |
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Answer» Henry's law states that the mass of a gas dissolved in given volume of the liquid at constant temperature depends upon the pressure applied. |
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| 574. |
Define mole fraction. |
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Answer» Mole fraction is the ratio of the number of moles of one component to the total number of moles in a mixture. |
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| 575. |
Which of the following chemical entities can act as semipermeable membrane ?A. `Cu_(2)[Fe(CN)_(6)]`B. `Cu(SCN_(2)`C. `BaC_(2)O_(4)`D. `BaSO_(4)` |
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Answer» Correct Answer - 1 |
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| 576. |
Correct statement among the following regarding osmosis isA. Solvent flows from high concentration to low concentrationB. Solvent flows from low concentration to high concetrationC. Solute flows from high concentration to low concetrationD. Solute flows from low concentration to high concentration |
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Answer» Correct Answer - 2 |
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| 577. |
1 mole glucose is added to 1 L of water `K_(b)(H_(2)O)=0.512" K kg mole"^(-1)` boiling point of solution will beA. `373.512^(@)C`B. `100.512^(@)C`C. `99.488^(@)C`D. `372.488^(@)C` |
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Answer» Correct Answer - 2 |
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| 578. |
Calculate boiling point of a solution pepared by dissoving 15.0 g of NaCI in 250.0 g of water (`K_(b)` for water = 0.512 K kg `mol^(-1)`), Molar mass of NaCI=58.44 g `mol^(-1)`). |
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Answer» Correct Answer - 374.05 K Step I. Calculation of elevantion in b.p. temperature `(DeltaT_(b))` NaCI dissociates in aqueous solution as : `NaCI(s)overset(("aq"))toNa^(+)("aq")+CI^(-)("aq")` `i=2, W_(B)=15.0 g, M_(B)=58.44" g mol"^(-1), W_(A)=250 g=0.25 kg` `K_(b)=0.152" K kg mol"^(-1)` `DeltaT_(b)=ixxK_(b)xxm=(ixxK_(b)xxW_(B))/(M_(B)xxW_(A))` `DeltaT_(b)=(2xx(0.512" K kg mol"^(-1))xx(15.0g))/((58.44"g mol"^(-1))xx(0.25 Kg))=1.05 K` Step II. Calculation of boilling point of solution `(T_(b))` `T_(b)=T_(b)^(@)+DeltaT_(b)=(373+1.05)K=374.05 K` |
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| 579. |
Calculate the mass of urea `(NH_(2)CONH_(2))` required in making 2.5 kg of 0.25 molal aqueous solution. |
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Answer» `"Molality of solution"=("Mass of urea/Molar mass of urea")/("Mass of water in kg")` `"Molality of solution"=0.25 m=(0.25 mol kg^(-1))` Molar mass of urea `(NH_(2)CONH_(2))=2xx14+1xx12+1xx16+4xx1= 60 g mol^(-1)`. Mass of solvent (water)=2.5 kg `(0.25 mol kg^(-1)),=("Mass of urea")/((60g mol^(-1))xx(2.5 kg))` `"Mass of urea"= (0.25 mol kg^(-1))xx(60 g mol^(-1))xx(2.5 kg)=37.5` g |
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| 580. |
Which of the following is (are) correct statements? i. `0.1 M CaCl_(2)` has higher boiling point than `0.1 M NaCl`. ii. `0.05 M Al_(2)(SO_(4))_(3)` has higher freezing point than `0.1 M K_(3)[Fe(CN)_(6)]`. iii.`0.1 M` glucose exerts higher osmotic pressure than `0.08 M CH_(3)COOH` (25% dissociated). iv. Vapour pressure of `0.05 M` urea solution is greater than that of `0.05 M KCl` solution. a.i,ii ,b.ii,iv , c. i,ii,iii , d. i,ii,iv |
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Answer» d. `T_(b)` is higher if im is higher. `T_(f)~` is higher if im is lower. `pi` is higher if im is higher. `VP` is higher if im is lower. `CaCl_(2):Im=3xx0.1`,`Al_(2)(SO_(4))_(3)`:` Im=5xx0.05`, `"Glucose":im=1xx0.1`,`NaCl:im=2xx0.1`, `K_(3)[Fe(CN)_(6)]:im=4xx0.1`, `CH_(3)COOH: im=(1+alpha)m=1.25xx0.08=0.1` Urea:`im=1xx0.05`,`KCl: im=2xx0.05=0.1` |
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| 581. |
Which of the following is not a characteristic of chemisorption?A. it is irreversibleB. it is specificC. it is multilayer phenomenonD. heat of adsorption of about `-400 kJ` |
| Answer» Correct Answer - A::B::D | |
| 582. |
The rate of chemisorption :A. decreases with increase of pressureB. increases with increase of pressureC. is independent of pressureD. is independent of temperature |
| Answer» Correct Answer - B | |
| 583. |
If all symbols have their usual meaning, then for a non-volatile non-electrolyte solute `Lt_(m rarr 0) ((DeltaT_(f))/(m))` is equal toA. zeroB. InfinityC. `DeltaT_(f)`D. `K_(f)` |
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Answer» Correct Answer - D For a non-volatile non-electrolyte solute `DeltaT_(f) = K_(f)m` `(DeltaT_(f))/(m) = K_(f)` `Lt_(m rarr 0) ((DeltaT_(f))/(m)) = Lt_(m rarr 0) K_(f) = K_(f)` |
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| 584. |
`1.0 g` of non-electrolyte solute dissolved in `50.0 g` of benzene lowered the freezing point of benzene by `0.40 K`. The freezing point depression constant of benzene is `5.12 kg mol^(-1)`. Find the molecular mass of the solute. |
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Answer» `M_(B)-(K_(f)xxW_(B))/(DeltaT_(f)xxW_(A))` Mass of non-electrolyte`(W_(B))=10 g` Mass of benzene `(W_(A))=50 g=0.005 Kg`. Depression in freezing point `(DeltaT_(P))=0.40 K` Molal depresion constant`(K_(f))=5.12 K kg mo^(-1)` `T_(f)=((5.12 K kg mol^(-1))xx(1.0 g))/((0.40 K)xx(0.005 kg))=256 g mol^(-1)` |
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| 585. |
In context with the industrial preparation of hydrogen from water gas `(CO+H_(2))`. Which of the following is the correct statement?A. CO is removed by adsorption in aqueous `Cu_(2)Cl_(2)` SolutionB. `H_(2)` is removed through occusion with PdC. CO is oxidized to `CO_(2)` with steam in the presence of a catalyst followed by adsorption of `CO_(2)` in alkaliD. CO and `H_(2)` are fractionally separated using differences in their densities |
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Answer» Correct Answer - C CO is oxidised to `CO_(2)` with steam in the presence of a catalyst followed by absorption of `CO_(2)` in alkali. `CO+H_(2) underset("catalyst")overset("Steam")(rarr)CO_(2)+2H_(2) overset(KOH)(rarr)K_(2)CO_(3)` |
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| 586. |
Which of the following represents correcty the changes in thermodynamic properties during the formation of 1 mole of an ideal binary solution :A. B. C. D. |
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Answer» Correct Answer - C For ideal solution `Delta "H"=0` `Delta"S for mixing is positive"implies"T"Delta"S"_("mix")=+"v"e` `Delta"G"_("mix")=Delta"H"_("mix")-"T"Delta"S"_("mix")=-"ve"` |
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| 587. |
Why is reverse osmosis of great practical utility to meet potable water requirements these days? Can it make sea water drinkable? Do we use ordinary semi permeable membranes for this? |
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Answer» Reverse osmosis is of great practical utility since it can result in absolute purification of any water sample because semi permeable membrane allows only solvent molecules to pass through. It can make sea water drinkable though ordinary semi permeable membrane may not be able to sustain high pressure observed in desalination of sea water i.e. reverse osmosis of sea water. Semi permeable membrane of cellulose acetate is generally used for this. |
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| 588. |
Explain the terms ideal and non-ideal solutions in the light of forces of interactions operating between molecules in liquid solutions. |
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Answer» △mixH = 0, △mix V= 0 Non-ideal solutions: When a solution does not obey Raoult's law over the entire range of concentration, it is called non-ideal solution. Positive deviations: Vapour pressure of such solutions shows higher value than the predicted value. △mixH = +ve △mixV = +ve A—B interactions <A—A and B—B interactions. Negative deviation: Vapour pressure of such solutions shows lower value than expected. △mixH = —ve △mixV = —ve A—B interactions > A—A and B—B interactions. Refer to page 45 of NCERT textbook for Class XII. |
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| 589. |
List three general applications of phenomenon of osmosis? |
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Answer» 1. Shrinking and swelling of blood cells when put into aqueous NaCl having concentration more or less than 0.9%. 2. Water movement from soil into plant body. 3. Preservation of meat and fruits by salting against bacterial infection |
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| 590. |
Why is it not possible to obtain pure ethanol by fractional distillation? What general name is given to binary mixtures which show deviation from Raoult’s law and whose components cannot be separated by fractional distillation. How many types of such mixtures are there? |
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Answer» The solution or mixture having same composition in liquid as well as in vapour phase and boils at a constant temperature is known as azeotropes. Due to constant composition it cannot be separated by fractional distillation. There are two types of azeotropes |
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| 591. |
Calculate the freezing point of an aquteous containing 10.50 g of `MgBr_(2)` in 200 g of water (molar mass of `MgBr_(2)=184 mol^(-1),K_(f) for water=1.86 K kg mol^(-1)`) |
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Answer» Calculation of depression in freezing point temperature `(DeltaT_(f))` `MgBr_(2)` dissociates in aqueous solution as: `MgBr_(2)overset((aq))toMg^(2)(aq)+2Br^(-)(aq)` `i=3, W_(B)=10.50g, m_(B)-184 g mol^(-1)`, `DeltaT_(f)=ixxK_(f)xxm=(ixxK_(f)xxW_(B))/(M_(B)xxW_(A))` `=(3xx(1.86 K kg mol^(-1))xx(10.50g))/((184 g mol^(-1))xx(0.2 kg))=1.59 K` Calculation of freezing point of solution `T_(f)=T_(f)^(@)-DeltaT_(f)(273-1.59)K=271.41 K` |
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| 592. |
Which of the following solutions has the minimum freezing pointA. 1 molal NaCl solutionB. 1 molal KCl solutionC. `1 molal CaCl_(2)` solutionD. 1 molal urea solution |
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Answer» Correct Answer - C More is `DeltaT_(f)` lesser is freezing point. |
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| 593. |
Which aqueous will have the highest boiling point?A. 1% glucose in waterB. 1% sucrose in waterC. 1% NaCl in waterD. `1% CaCl_(2)` in water |
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Answer» Correct Answer - C More is `DeltaT_(b)` more is boiling point. |
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| 594. |
The factor `(DeltaT_(f)//K_(f))` representsA. MolarityB. FormalityC. NormalityD. Molality |
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Answer» Correct Answer - D `DeltaT_(f)=K_(f) xx Molarity` |
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| 595. |
An ideal solution is formed when its components sameA. have no volume change on mixingB. have no enthalpy change on mixingC. have both the above characteristicsD. have high solublity |
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Answer» Correct Answer - C In ideal solution there is no change in enthalpy and volume after mixing, i.e. `Delta V_("mixing")=0, Delta H_("mixing")=0` and `T_("mix")=0` |
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| 596. |
Which aqueous solution has minimum freezing point?A. `0.01 M NaCl`B. `0.005 M C_(2)H_(5)OH`C. `0.005 M Mgl_(2)`D. `0.005 M MgSO_(4)` |
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Answer» Correct Answer - A Minimum freezing point means more number of ions `0.01 xx 2 =0.02` `0.005 xx 1 =0.005` `0.005 xx 3 =0.015` `0.005 xx 2 =0.01` |
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| 597. |
Which of the following aqueous solution has minium freezing point?A. `0.005 m MgSO_(4)`B. `0.005 m C_H_(5)OH`C. `0.00 m MgI_(2)`D. `0.01 m NaCl` |
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Answer» Correct Answer - 4 According to modified equation `DeltaT_(f)= i K_(f)m` Thus, `DeltaT_(f)` is directly proportional to i m. `0.005 m MgSO_(4) implies (2)(0.005 m)=0.01 m` `0.005 m C_(2)H_(5)OH implies (1)(0.005 m)=0.005 m` `0.005 m MgI_(2) implies (3)(0.005 m)=0.015 m` `0.005 m NaClimplies (2)(0.01m)=0.02 m` Thus, `0.01 NaCl` having the highest value of has maximum `DeltaT_(f)` and hence minimum freezing point. |
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| 598. |
All form ideal solution exceptA. `C_(6)H_(6)` and `C_(6)H_(5)CH_(3)`B. `C_(2)H_(5)Cl` and `C_(2)H_(5)l`C. `C_(6)H_(5)Cl` and `C_(6)H_(5)Br`D. `C_(2)H_(5)l` and `C_(2)H_(5)OH` |
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Answer» Correct Answer - D When we mix `C_(2)H_(5)I` and `C_(2)H_(5)OH`, there is change in enthalpy and volume, so it is an example of non-ideal solution. |
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| 599. |
Which aqueous solution has minimum freezing point?A. 0.01 m NaClB. `0.005 m Mhgl_(2)`C. `0.005 m C_(2)H_(5)OH`D. `0.005 m MgSO_(4)` |
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Answer» Correct Answer - A 0.01 molal NaCl solution have minimum freezing point because its molality is more and it given two ions after dissociation of one molecule. |
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| 600. |
Blood cells retain their normal shape in solution which areA. hypotonic to bloodB. isotonic to bloodC. hypertonic to bloodD. equinormal to blood |
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Answer» Correct Answer - B When blood cells are placed in a solution of similar concentration as that of blood, then they neither swell nor shrink it means the concentration of solution is same as that of inside the blood cells, i.e. they are isotonic to each other. |
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