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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.
| 401. |
Assertion (A): `0.1 M` solution of glucose has same increment in freezing point than has `0.1 M` solution of urea. Reason (R ): `K_(f)` for both has different value.A. If both assertion and reason are true and the reason is the correct explanation of the assertionB. If both the assertion and reason are true but reason is not the correct explanation of the assertionC. If assertion is true but reason is falseD. If assertion is false but reason is true. |
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Answer» Correct Answer - B The same increase in boiling point is due to same molality. |
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| 402. |
Assertion:Addition of solvent to a solution always lowers the V.P. Reason: The increase in relativer surface area given rise to an increases inV.P. for a given solution.A. If both assertion and reason are true and the reason is the correct explanation of the assertionB. If both the assertion and reason are true but reason is not the correct explanation of the assertionC. If assertion is true but reason is falseD. If assertion is false but reason is true. |
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Answer» Correct Answer - D Note that addition of solute to solvent (and not solvent to solute which will show reverse effect)shows a lowering in V.P.due to decrease in relative surface area. |
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| 403. |
The boiling point elevation contant for benzene is `2.57^(@)C//m`. The boiling water of benzene is `81.0^(@)C`. Determine the boiling point of a solution formed when `5 g` of `C_(14)H_(12)` is dissolved I `15 g` fo benzene. |
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Answer» Given weight of `(C_(14)H_(12)(W_(B)) = 5 g` Weight of benzene `(W_(A)) = 15 g` `DeltaT_(b) = K_(b) [(W_(B) /(Mw_(B)) xx 1000] /W_(A)]` ` = 2.57 [[5 /180 xx 1000] /15] = 4.76^(@)C` Now,`DeltaT_(b)=T_(b)-T_(b)^(@) [{:(T_(b)^(@)="boiling point of pure solvent"),(T_(b)="boiling point of solution"):}]` `:. T_(b) = DeltaT_(b) + T_(b)^(@)` `= 4.76 + 81.0` `=85.76^(@)C` |
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| 404. |
Assertion:A cook cries more in cutting onion rather than cutting an onion taken out from refrigerator . Reason:The cold onion has lower vapour pressure of its volatile content.A. If both assertion and reason are true and the reason is the correct explanation of the assertionB. If both the assertion and reason are true but reason is not the correct explanation of the assertionC. If assertion is true but reason is falseD. If assertion is false but reason is true. |
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Answer» Correct Answer - A Vapour pressure decreases with decrease in the temperature. |
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| 405. |
Benzene and toluene form two ideal solution A and B at 313 K, Solution A(total pressure `"P"_("A")`) contains equal mole fo toluene and benzene. Solution B contains equal masse of both(total pressure `"P"_("B")`). The vapour pressure of benzene and toluene are 160 and 60 mm of Hg respectively at 313 K. Calculate the value of `"P"_("A")//"P"_(B")`. |
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Answer» Correct Answer - 0.964 Benzene-B, Toluene-T `"P"_("A")="X"_("B")"P"_("B")^(0)+X_("T")"P"_("T")^(0)` `"n"_("B")="n"_("T")="n"` `"P"_("solution")=("n")/(2"n")xx160+("n")/(2"n")xx60` `"P"_("solution")=110"mm Hg"` Now for solution B `"w"_("B")="w"_("T")="w"` `"X"_("b")=(("w")/(78))/(("w")/(78)+("w")/(92)),"X"_("t")=1-(46)/(85)` `=(92)/(170)" "=(39)/(85)` `=(46)/(86)` `"P"_("solution")^(1)="X"_("b")"P"_("b")^(0)+"X"_("t")"P"_("t")^(0)` `46/85xx160+39/85xx60=1940/17` |
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| 406. |
A very dilute saturated solution of a sparingly soluble salt `A_(3)B_(4)` has a vapour pressure of `20mm` of `Hg` at temperature T, while pure water exerts a pressure of `20.0126mm Hg` at the same temperature. Calculate the solubility product constant of `A_(3)B_(4)` at the same temperature. |
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Answer» Correct Answer - `5.4 xx10^(-13)` `P_(s) = 20 P^(@) = 20.0126` `(P^(@)-P_(s))/(P^(@)) = (0.0126)/(20) = (n)/(n+N) =(n)/(N)` `("mole of solute")/("moles of" H_(2)O) = 0.0063` 1 mole `H_(2)O = 18g = 18 mL` `18 mL` solution `= 0.0063` mole `1L` solution `= (0.00063)/(18) xx 1000 = 0.35` mole/L Let solubility of salt `A_(3)B_(4)` is s then `7s = 0.035` `s = 0.005` mole/L `k_(sp) = 3^(3).4^(4) (s)^(7) = 27 xx 256 xx (0.005)^(7)` `k_(sp) = 5.4 xx 10^(-13)` |
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| 407. |
For a binary ideal liquid solution, the variation total vapour pressure versus composition of solution is given by which of the curves?A. B. C. D. |
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Answer» Correct Answer - A::D Depending on the vapour pressure of the pure component 1 and 2 total vapour pressure over the solution decreases or increases with the increase of the mole fraction of component 1. |
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| 408. |
The vapour pressure of a saturated solution of sparingly soluble salt `("XCl"_(3))` was 17.20mm Hg at `27^(@)"C"`. If the vapour pressure of pure `"H"_(2)"O"` is 17.25 mm Hg at 300K, what is the solubility of sparingly soluble salt `"XCl"_(3)" in mole"//"Litre".`A. `4.04xx10^(-2)`B. `8.08xx10^(-2)`C. `2.032xx10^(-2)`D. `4.04xx10^(-2)` |
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Answer» Correct Answer - A Let solubility of `"XCl"_(3)="S mol lit"^(-1)` For `"XCl"_(3),"n"=4,alpha=1` `therefore"i"=4` As S is very small `therefore` Molality=Molarity `("P"^(0)-"P"_("solution"))/("P"^(0))xx"i"xx("n"_("solute"))/("n"_("solvent")` `(17.25-17.20)/17.25=(4"S")/55.55` `"S"=4.04xx10^(-2)"mol lit"^(-1)` |
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| 409. |
The value of van’t Hoff factor will be minimum for :(a) 0.05 M AlCl3(b) 0.2 M NaNO3 (c) 5.0 M glucose (d) 0.1 M H2SO4 |
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Answer» Option : (c) 5.0 M glucose |
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| 410. |
van’t Haff factor for K4[FeC(N)6] dissociated 10% is :(a) 1.1 (b) 1.4 (c) 0.86 (d) 1.6 |
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Answer» Option : (b) 1.4 |
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| 411. |
`1.22 g` of benzoic acid is dissolved in acetone and benzene separately. Boiling point of mixture with acetone increase by `0.17^(@)C` and boiling point of mixture with benzene increases by `0.13^(@)C`. `K_(b) ("acetone")=1.7 K kg "mol"^(-1)`, Mass of acetone `= 100 g`, `K_(b) ("benzene")=2.6 k Kg mol^(-1)`, Mass of benzene `= 100 g`, Find molecular weight of benzoic acid in acetone and in benzene solution. Justify your answer with structure. |
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Answer» Correct Answer - `122,224` (i) In first case, `DeltaT_(b)=K_(b)xxm=K_(b)xx("Wt. of solute")/("Mol.wt. of solute")xx(1000)/(" wt. of solvent")` or `0.17=1.7xx(1.22)/(Mxx100xx10^(-3))` `M=122gm//"mole"` Thus the benzoic acid exists as a monomer in acetone (ii) In second case, `DeltaT_(b)=K_(b)xx("Wt. of solute")/("Mol. wt. of solute")xx(1000)/("wt. of solvent")` or `0.13=2.6xx(1.22)./(Mxx100xx10^(-3))` `M=224` Double molecular weight of benzoic acid `(244)` in acetone solution indicates that benzoic acid exists as dimer in acetone. |
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| 412. |
The elevation in boiling point of a solution of `13.44`g of `CuCl_(2)`(molecular weight =`134.4,k_(b)=0.52K "molality"^(-1))` in `1` kg water using the following information will be:A. 0.16B. 0.05C. 0.1D. 0.2 |
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Answer» Correct Answer - A Molality =`13.44/134.4=0.1` For `CuCl_(2)`,`i=3` `DeltaT_(b)=i xx K_(b) xx m` =`3 xx 0.52 xx 0.1` =`0.156^(@)` |
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| 413. |
The elevation in boiling point of a solution of `13.44`g of `CuCl_(2)`(molecular weight =`134.4,k_(b)=0.52K "molality"^(-1))` in `1` kg water using the following information will be:A. `0.16`B. `0.05`C. `0.1`D. `0.2` |
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Answer» Correct Answer - A `DeltaT_(b)=(1000xx0.52xx13.44)/(134.4xx1000)xx(1+2alpha)` `:.alpha =1` `0.052xx3=0.156` |
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| 414. |
Elevation in boiling point of an aqueous solution of urea is `0.52 ( k_(b) "for water"=0.52K"molality"^(-1))`. The mole fraction of urea in this solution is :A. `0.982`B. `0.018`C. `0.0567`D. `0.943` |
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Answer» Correct Answer - B `0.52 = 0.52 xx m` `m = 1` i.e., solution is 1m Thus, the solution contains 1 mol of urea in `1000g (55.5mol)` of water `:. x = (1)/(1+55.5) + (1)/(56.5) = 0.0177` |
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| 415. |
Calculate the molarity and mole fraction of the solute in aqueous solution containing `3.0 g` of urea per `250gm` of water (Mol. Wt. of urea = 60)A. ` 0.2m ,0.00357 `B. ` 0.4m, 0.00357 `C. ` 0.5m, 0.00357 `D. ` 0.7m, 0.00357 ` |
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Answer» Correct Answer - A Wt.of solute(urea)dissolved=`3.0g` wt. of the solvent (water)=`250g` Mol.wt.of the solute=`60` `3.0gm`of the solute =`(3.0)/(60)mol es ` `=0.05Mol es` thus `250g`of the solvent contain =`0.05`Moles of solute `:.1000g`of the solvent contain =`(0.05xx1000)/(250)=0.2`moles Hence molality of the solution =`0.2m` In short, Molality = No . of mol es of solute /1000g of solvent `:.` Molality `=(3//60)/(250)xx1000=0.2m` calculation of the mole fraction`3.0 g` of solute `=3//60 "moles" =0.05 "mole"` `250g`of water `=(250)/(18)`moles =`13.94`moles `:.` Mole fraction of the solute `=(0.05)/(0.05+13.94)=(0.05)/(13.99)=0.00357` |
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| 416. |
Elevation in `b.p.` of an aqueous urea solution is `0.52^(@), (K_(b)=0.622^(@) "mol"^(-1)kg)` Hence, mole-fraction of urea in this solution is :A. `0.982`B. `0.567`C. `0.943`D. `0.018` |
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Answer» Correct Answer - D As `DeltaT_(b)="molality"xxK_(b)` `0.52=mxx0.52` molality`=1 "mol" kg^(-1)` `therefore` urea `=1` mol ` moles of water `=(1000)/(18)=55.55` mole fraction of urea `=(1)/(56.55)=0.018` |
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| 417. |
The preservation of meat by salting and of fruits by adding sugar protects them from bacterial action becauseA. bacteria die of eating sugar or saltB. due to osmosis bacteria lose water on salted meat or candid fruit and dieC. due to osmosis bacteria gain water on salted meat or candid fruit and dieD. bacteria get stuck to the salt and sugar layers and die |
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Answer» Correct Answer - B The bacteria on salted meat or candid fruit loss water through osmosis, shrink and die |
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| 418. |
A solution containing `2.675g` of `COCl_(3).6NH_(3)` (molar mass `=267.5gmol^(-1)`) is passed through a cation exchanger, The chloride ions obtined in solution were treated with excess of `AgNO_(3)` to give`4.78g` of `AgCl ("molar mass" =143.5gmol^(-1)`.The formula of the complex is (Atomic mass of `Ag=108 u`)A. `[Co(NH_(3))_(6)]Cl_(3)`B. `[CoCl_(2)(NH_(3))_(4)]Cl`C. `[CoCl_(3)(NH_(3))_(3)]`D. `[CoCl(NH_(3))_(5)]Cl_(2)` |
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Answer» Correct Answer - A Mole of `CoCl_(3).6NH_(3)=(2.675)/(267.5)=0.01mol` `AgNO_(3)(aq)+Cl^(-)(aq)rarrAgCldownarrow(white)+NO_(3)^(-)(aq)` mol e of AgCl=`(4.78)/(143.5)=0.03mol` `0.01 mol CoCl_(3).6NH_(3)gives=0.03mol AgCl` `:.1 mol CoCl_(3).6NH_(3)`ionised to give=`3 molCl^(-)` Hence the formula of compound is `[CO(NH_(3))_(6)]Cl_(3)` |
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| 419. |
The molarity of 648 g of pure water isA. 36 mB. 55.5 mC. 3.6 mD. 5.55 m |
| Answer» Correct Answer - B | |
| 420. |
How many grams of glucose be dissolved to make one litre solution of `10%1` glucose:A. `10g`B. `180g`C. `100g`D. `1.8g` |
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Answer» Correct Answer - C `100ml`solution contains `= 10 gm` of glucose So `1000ml` wil contain `= (10)/(10^(3)) xx 1000 = 100gm`. |
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| 421. |
How will you prepare 0.25 m `CaVI_(2)` solution? |
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Answer» Correct Answer - 27.75 g `"Molality of solution"=("No. of moles of"CaCI_(2))/("Mass of solvent in kg ")` `"No. of moles of "CaCI_(2)="Molality"xx"Mass of solvent in kg"` `=(0.25" mol lg"^(-1))xx(1 kg )= 0.25 mol` `"Molar mass of "CaCI_(2)=40+2xx35.5=111 " g mol"^(-1)` `"Mass of "CaCI_(2)=(0.25 mol)xx(111 g mol^(-1))=27.75 g.` |
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| 422. |
Cancerntrated `HNO_(3)` used in the laboratory is usually 69% by mass of `HNO_(3)`. The density of `HNO_(3)`. The density of `HNO_(3)` solution is 1`.41 g `cm^(3)`. Find The molarity. |
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Answer» Correct Answer - 15.44 M `"Mass of solution"= 100 g, "Mass os "HNO_(3)=69 g` `"Volume of solution"=(100 g)//(1.41" g cm"^(-3))=70.92 cm^(3)=0.07092 dm^(3)` `"Molar mass of "HNO_(3)=1 +14 +3xx16=63" g mol"^(-1)` `"Molarity of solution (M)"=("Mass of"HON_(3)//"Molar mass")/("Volume of solution in dm"^(3))=((69g)//(63" g mol"^(-1)))/((0.07092" dm"^(3)))` `=15.44" mol dm"^(-3)=15.44 M`. |
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| 423. |
What is “Tyndall effect” |
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Answer» The colloidal particles can scatter light and so we can see them. This is Tyndall effect. |
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| 424. |
Which liquid pair represents positive deviation from Raoult’s law?(a) Water + HCl(b) Water + HNO3(c) Benzene + Methanol(d) Acetone + Chloroform |
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Answer» The answer is (c) Benzene + Methanol |
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| 425. |
Choose solvent solute in the following cases (i) Brine solution (ii) Crystalline copper sulphate (iii) Aq. Solution of copper sulphate |
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Answer» (i) is aq. Solution of NaCl here water is solvent as it is larger in % (w/w) and also final solution is in the liquid state that is the physical state of water. (ii) Crystalline copper is same as of solvent (iii) In an aqueous solution solvent is always water. (iv) 1 mole ethanol=46 g, 1 mole water =18 g `rArr` Solvent is ethanol |
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| 426. |
This is a property of an ideal solution(a) It obeys Raoult’s law(b) ∆Hmixing(c) ∆Vmixing(d) All of these |
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Answer» The answer is (d) All of these |
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| 427. |
1.00 g of a non-electrolyte solute dissolved in 50g of benzene lowered the freezing point of benzene by 0.40 K. The freezing point depression constant of benzene is 5.12 K kg `mol^(-1)`. Find the molar mass of the solute. |
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Answer» Substituting the values of various terms involved in equation `M_(2)=(K_(f)xx w_(2)xx1000)/(DeltaT_(f)xx w_(1))` we get, `M_(2)=(5.12" K kg mol"^(-1)xx1.00" g"xx1000" g kg"^(-1))/(0.40xx50" g")=256" g mol"^(-1)` Thus, molar mass of the solute `=256" g mol"^(-1)` |
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| 428. |
`1.0 g` of a non-electrolyte solute( mol. Mass `250.0 g mol^(-1)`) was dissolved in `5.12 g` benzene. If the freezing point depression constant, `K_(f)` of benzene is `51.2 K kg mol^(-1)`, the freezing point of benzene will be lowered by:A. `0.2 K`B. `0.4 K`C. `0.3 K`D. `0.5 K` |
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Answer» Correct Answer - 2 We need to calculate `DeltaT_(f)` (depression of freezing point of solvent) which is given as `DeltaT_(f)=iK_(f)m=iK_(f) W_(solute)/(mm_(solute)W_(solv))` we are given `i=1` as the solute is nonelectrolyte `K_(f)=5.12 K kg mol^(-1)` `W_(solute)=1.00 g` `W_(solvent)=51.2/1000kg` `mm_(solute)=250 g mol^(-1)` Substituting these results, we get `DeltaT_(f)=(1)(5.12 K kg mol^(-1))((1.00 g))/((250 g mol^(-1))(5.12/1000kg))` `=0.4 K` |
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| 429. |
`1.0 g`of a non-electrolyte solute (molar mass`250gmol^(-1)` was dissolved in 51.2 g of benzene. If the freezing point depression constant of benzene is `5.12K kgmol^(-1)` the lowering in freezing point will beA. 0.4 KB. 0.3 KC. 0.5 KD. 0.2 K |
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Answer» Correct Answer - A Molality of non-electrolyte solute `=(("Weight of solute in (g)")/("molecular weight of solute"))/("weight of solvent in (kg)")=((1)/(250))/(0.0512)` `=(1)/(250xx0.0512)=0.0781 m` `Delta T_(f)=K_(f)xx` molality of solution `=5.12xx0.0781 ~~ 0.4 K` |
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| 430. |
`1.0 g`of a non-electrolyte solute (molar mass`250gmol^(-1)` was dissolved in 51.2 g of benzene. If the freezing point depression constant of benzene is `5.12K kgmol^(-1)` the lowering in freezing point will beA. `0.5K`B. `0.4K`C. `0.2K`D. `0.3K` |
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Answer» Correct Answer - B `DeltaT=(1000xxK_(f)xxw)/(mxxW)=(1000xx5.12xx1)/(250xx51.2)=0.4K` |
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| 431. |
0.90g of a non-electolyte was dissolved in 87.9 g of benzene. This reised the boiling point of benzene by `0.25^(@)C`. If the molecular mass of the non-electrolyte is 103.0 g `mol^(-1)` , calculate the molal elecation constant for benzene. |
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Answer» `K_(B)=(DeltaT_(b)xxM_(B)xxW_(A))/W_(B)` `W_(B)=0.90g,W_(A)=87.9g=0.0879 kg, M_(B)=103.0 g mol^(-1),` `DeltaT_(b)=0.25 K` `K_(b)=((0.25K)xx(103.0gmol^(-1))xx(0.0879kg))/((0.90g))=2.52 K kg mol^(-1)` |
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| 432. |
1.0 g of a non-electolyte solute (molar mass 250 g `mol^(-1)`) was dissolved in 51.2 g of benzene. If the `K_(f)` for benzene is 5.12 K kg `mol^(-1)`, the freezing point of benzene will be lowerd by :A. 0.2 KB. 0.4 KC. 0.3 KD. 0.5 K. |
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Answer» Correct Answer - b `DeltaT_(f)=(K_(f)xxW_(B))/(M_(B)xxW_(A))` `=((5.12 kg mol^(-1))xx(1.0g))/((250gmol^(-1))xx(0.0512kg))=0.4 K` |
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| 433. |
The mole fraction of a solute in a solutions is `0.1`. At `298K` molarity of this solution is the same as its molality. Density of this solution at 298 K is `2.0 g cm^(-3)`. The ratio of the molecular weights of the solute and solvent, `(MW_("solute"))/(MW_("solvent"))` is |
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Answer» Correct Answer - 9 When `M = m` `(w_(B)xx1000)/(m_(B)xxV)=(w_(B)xx1000)/(m_(B)xxw_(A))` i.e., `V=w_(A)` Volume of solution = Mass of solvent `("Mass of solution")/("Density")=` Mass of solvent `(w_(A)+w_(B))/(2)=w_(B)` `w_(A)=w_(B)` `x_(A)xxm_(A)=x_(B)xxm_(B)` `0.9xxm_(A)=0.1xxm_(B)` `(m_(B))/(m_(A))=9` `(MW_("solute"))/(MW_("solvent"))=9`] |
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| 434. |
Define osmotic pressure. How is the osmotic pressure related to the concentration of a solute in a solution ? |
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Answer» The external pressure applied on the solution side to stop the flow of solvent across the semipermeable membrane i.e., osmosis is known as osmotic pressure. The osmotic pressure is directly proportional to the concentration of the solution i.e.,π = CRT. |
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| 435. |
Explain why on addition of 1 mol glucose to 1 litre water the boiling point of water increases. |
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Answer» Vapour pressure of the solvent decreases in the presence of non-volatile solute (glucose) hence boiling point increases. Detailed Answer: Addition of non-volatile solute ie, glucose in water, lowers the vapour Pressure of water. As a result, boiling point of water increases. |
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| 436. |
Will the elevation in boiling point be same if 0.1 mol of sodium chloride or 0.1 mol of sugar is dissolved in 1L of water ? Explain. |
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Answer» No, the elevation in boiling point is not the same. Elevation in boiling point is a colligative property which depends on the number of particles. NaCl is an ionic compound which dissociates in solution to give more number of particles whereas sugar is made up of molecules and thus does not dissociate. |
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| 437. |
State Henry's law and mention two of its important applications. |
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Answer» According to Henry's law, the partial pressure of a gas in vapour phase (p) is directly proportional to the mole fraction (x) of the gas in the solution. Henry's law is applied in the production of carbonated beverages and for the deep-sea-divers. |
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| 438. |
State Raoult's law for a solution containing volatile components. How does Raoult's law becomes a special case of Henry's law ? |
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Answer» Partial vapour pressure of a liquid components is directly proportional to its mole fraction in its solution. The partial pressure of the volatile component or gas is directly proportional to its mole fraction in solution. Only the proportionality constant KH differs from PA°. Thus, Raoulfs law becomes a special case of Henry's law in which Ks become equal to PA°. |
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| 439. |
What is meant by + ve and - ve deviations from Raoult's law and how is the sign of △H solution related to + ve and - ve deviations from Raoult's law ? |
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Answer» Solutions showing positive deviations, the partial vapour pressure of each component of solution is greater than the vapour Pressure as expected according to Raoult's law. Solutions showing negative deviations, the partial vapour pressure of each component of solution is less than vapour pressure as expected according to Raoult's law. For positive deviation △mix H = + ve For negative deviation △mix H = - ve |
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| 440. |
Rubbing isopropyl alcohol often gives a cooling sensation to the skin. Why? |
| Answer» Isopropyl alcohol `(CH_(3)CHOHCH_(3))` being a volatile liquid absorbs the required latent heat of vaporisation from the skin thereby giving a cooling sensation. | |
| 441. |
Blood cells are isotonic with 0.9% sodium chloride solution. What happens if we place blood cells in a solution containing; (a) 7.2% sodium chloride solution ? (b) 0.4% sodium chloride solution ? |
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Answer» (a) If we place blood cells in solution containing 1.2% sodium chloride solution, they shrink. (b) If we place blood cells in solution containing 0.4% sodium chloride solution, they swell. |
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| 442. |
Explain the following : (a) Henry's law about dissolution of a gas in a liquid, (b) Boiling point elevation constant for a solvent. |
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Answer» (a) The partial pressure of the gas above the liquid (vapour phase) is directly proportional to the mole fraction of the gas dissolved in the liquid. (b) Boiling Point Elevation Constant: It is equal to elevation in boiling point of 1 molal solution, i.e.,mole of solute is dissolved in 1kg of solvent. |
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| 443. |
Does osmosis occur from hypertonic solution to hypotonic solution ? |
| Answer» No, it always occurs from hypotonic to hypotonic to hypertonic solution. | |
| 444. |
The bottle of liquid ammonia is generally cooled before opening the seal. Assign reson. |
| Answer» On cooling, the gas will tend to liqueft and its vapour pressure will decrese. Therefore, the gas will not come with force upon opening the seal. In fact, ammonia vapours are extremely poisonous and they affect eyes as well as respiratory system. | |
| 445. |
Why is liquid ammonia bottle first cooled in ice before opening it? |
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Answer» At room temperature, the vapour pressure of liquid ammonia is very high. On cooling, vapour pressure decreases. Hence the liquid ammonia will not splash out. |
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| 446. |
Why is liquid ammonia bottle first cooled in ice before opening it ? |
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Answer» The vapour pressure of liquid ammonia is very high at room temperature. On cooling, the vapour pressure decreases restricting ammonia from splashing out. |
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| 447. |
Define molal elevation constant or ebullioscopic constant. |
| Answer» Molal elevation constant is defined as the elevation in boiling point that takes place when the molality of solutions is unity. | |
| 448. |
What is molal elevation constant or ebullioscopic constant? |
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Answer» The elevation in boiling point which takes place when molality of the solution is unity, is known as ebullioscopic or molal elevation constant. |
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| 449. |
What is molal elevation constant or ebullioscopic constant? Write its units. |
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Answer» Molal elevation constant or ebullioscopic constant is the elevation in boiling point for one molal solution. Its unit is K Kg mol-1 |
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| 450. |
What are the units of molal elevation constant? |
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Answer» Boiling point elevation, ΔTb is given by, ΔTb = Kb × m where m is molality in mol kg-1 and Kb is molal elevation constant or abullioscopic constant. ∴ Kb = \(\frac{ΔTb}{m}\) ∴ K has units K kg mol-1 (or °C kg mol-1) Therefore, molal elevation constant is the elevation in boiling point produced by 1 molal solution of a nonvolatile solute. |
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