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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.
| 1601. |
Solve the following :The solubility of dissolved oxygen to 27 °C is 2.6 × 10-3 mol dm-3 at 2 atm. Find its solubility at 8.4 atm and 27 °C. |
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Answer» Given : Solubility of O2 = S1 = 2.6 × 10-3 mol dm-3 Initial pressure of O2 = P1 = 2 atm Final pressure of O2 = P2 = 8.4 atm Solubility of O2 = S2 = ? (i) By Henry’s law, S1 = KH x P1 ∴ Henry’s law constant KH is, KH = \(\frac{S_1}{P_1}\) = \(\frac{2.6 \times 10^{-3}}{2}\) = 1.3 × 10-3 mol dm-3 atm-1 (ii) Now, S2 = KH x P2 = P2 = 1.3 x 10-3 x 8.4 = 10.92 × 10-3 = 1.092 × 10-2 mol dm-3 ∴ Solubility of O2 = 1.092 × 10-2 mol dm-3 |
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| 1602. |
To a `25 mL` of `H_(2)O_(2)` solution, excess of acidified solution of `KI` was added. The iodine liberated required `20 mL` of 0.3 N `Na_(2)S_(2)O_(3)` solution. Calculate the volume strength of `H_(2)O_2` solution. Strategy : Volume strength of `H_(2)O_(2)` solution is related to its normality by the following relation Volume strength `(V)=5.6xx"Normality" (N)` where, Normality`=((meq)H_(2)O_(2))/V_(mL)` According to the law of equivalence `(meq)_(Na_(2)S_(2)O_(3))=(meq)_(I_(2))=(meq)_(H_(2)O_(2))` |
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Answer» `(meq)_(Na_(2)S_(2)O_(3))=V_(mL)xxN` `=(20)(0.3)=6` `=(meq)_(H_(2)O_(2))` Thus, normality of `H_(2)O_(2)=(meq)_(H_(2)O_(2))//V_(mL)` `6/25eq L^(-1)` Volume strength `(V)=5.6xx "Normality"` `=(5.6)(6//25)` `=1.344` |
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| 1603. |
Calculate the mole fraction of solute in a dilute aqueous solution from which ice begins to separate out at `-46^(@)C`. (`K_(f)` of `H_(2)O` =`1.86 K m^(-1)`) |
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Answer» `DeltaT_(f)=K_(f) xx m rArr m=(DeltaT_(f))/K_(f)=(0-(0-0.46))/1.86 = 0.46/1.86=0.247 m` Use the relation, `m=(n_(2)xx1000)/(n_(1)xxMw_(1))` `:.n_(2)/n_(1)=chi_(2)=(m xx Mw_(1))/1000` =`(0.247 xx 18)/1000` `=0.004` |
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| 1604. |
At freezing point of a solution there is alwaysA. An equilibrium solution (I)` hArr `solution (I)B. Vapour pressure of solution (I) = solvent (s)C. An equilibrium solvent (I)` hArr` solvent (s)D. Both (2) & (3) |
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Answer» Correct Answer - 4 |
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| 1605. |
Two liquids A and B form ideal solution. At `300 K` , the vapour pressure of a solution containing `1 mol` of A and 3 mol of B is 500 mm of Hg. At the same temperature, if one more mole of B is added to this solution , the vapour pressure of the solution increases by 10 mm of Hg. Determine vapour pressures of A and B in their pure states. |
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Answer» Let the vapour pressure of pure `A be = P_(A)^@`, and the vapour pressure of pure `B be = P_(B)^@`. Total vapour pressure of solution (1 mol A + 3 mol B ) `= chi_(A)P_(A)^@ + chi_(B)P_(B)^@` [`chi_(A)` is mole fraction of A and `chi_(B)` is mole fraction of B] `500 = 1/4p_(A)^@ + 500 = 3/4p_(B)^@` or `2000 =p_(A)^@ + 3p_(B)^@` Total vapour pressure of solution (1 mol A + 4 mol B) `=1/5 p_(A)^@ +4/5 p_(B)^@` `510 = 1/5 p_(A)^@ + 4/5 p_(B)^@` `2250=p_(A)^@ + 4p_(B)^@ ` Solving Eqs. (i) and (ii), we get `p_(B)^@ =550` mm Hg = Vapour pressure of pure B `p_(A)^@ =350` mm Hg = Vapour pressure of pure A |
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| 1606. |
Liquids A`(p_(A)^(@)=360" mm Hg")` and `B(p_(B)^(@)=320" mm Hg")` are mixed. If solution has vapour has vapour pressure 340 mm Hg, then number of mole fraction of B/mole solution will beA. `(33)/(34)`B. `(1)/(34)`C. `(1)/(2)`D. `(1)/(4)` |
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Answer» Correct Answer - 3 |
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| 1607. |
The vapour pressure of a `5%` aqueous solution of a non-volatile organic substance at `373 K`. Is `745 mm`. Calculate the molecular mass of the solute. |
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Answer» `5%` aqueous solution of the solute implies that `5 g` of the solute are present in `100 g` of the solution, i.e., Weight of solute `(W)_(2)=5 g` Weight of solution=100 g `:.` Weight of solvent `(W_(1))=100-5 =95 g` Further as the solution is aqueous, it means that the solvent is water and we know that vapour pressure of pure water at `393 K =760mm` Vapour pressure of solvent at `373 K (P_(s))=745 mm` (given) Molecular mass of solvent (water), `Mw_(1)=18` Molecular massof solute, `Mw_(2)`= To be calculated Using the formula for dilute solutions, viz `(P^(@)-P_(s))/P^(@)=n_(2)/n_(1)=(W_(2)//Mw_(2))/(W_(1)//Mw_(1))` We get (760-745)/760=(5//Mw_(2))/(95//18) ` or `Mw_(2)=(5 xx 18 xx 760)/(15 xx 95)=48` |
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| 1608. |
Calculate the mass of urea (NH2CONH2) required in making 2.5 kg of 0.25 molal aqueous solution. |
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Answer» 0.25 molal aqueous solution means that Moles of urea = 0.25 mole Mass of solvent (water) = 1 kg = 1000 g Molar mass of urea = 14 + 2 + 12 + 16 + 14 + 2 = 6O g mol-1 0.25 mole of urea = 60 x 0.25 mole = 15 g Total mass of the solution = 1000 + 15 g = 1015 g = 1.015 g Thus, 1.015 kg of solution contain urea = 15 g 2.5 kg of solution will require urea = \(\frac{15}{100}\) x 2.5 kg = 37g |
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| 1609. |
Two liquids `A` and `B` form an idea solution. What will be the vapour pressure at `27^(@)C` of a solution having `1.5 mol` of `A` and `4.5 mol` of `B`? The vapour pressure of `A` and `B` at `27^(@)c` is `0.116 atm` and `0.140 atm`, respectively. |
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Answer» `n_(A)=1.5 n_(B)=4.5` `chi_(A)=1.5/(1.5 + 4.5)=1.5/6.0 =1/4` `chi_(B)=1-(1/4)=(3/4)` `P=P_(A)^(@)chi_(A) + P_(B)^(@)chi_(B)=0.116xx(1/4) + 0.14 xx(3/4)` `P=0.029 + 0.105` `=0.134` |
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| 1610. |
Calculate the vapour pressure of a 0.1 M urea solution at 298 K. The vapour pressure of pure water at 298 K is 20 mm hg. Assume density of solution to be 1g `mL^(-1)` |
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Answer» Correct Answer - 199.6 mm Hg 0.1 M urea solution contains 6.1 mole of urea in 1000 mL of solution. `"Mass of 1000 mL urea solution"=Vxxd=(1000mL)xx(1g mL^(-1))=1000 g` `"Mass of 0.1 mole of urea"=(0.1 mol)xx(60 g mol^(-1))=6g` `"Mass of water"(n_(A))=((994g))/((18 g mol^(-1)))=55.22 mol` `"Mole fraction os water"(x_(A)) =((55.22 mol))/((55.22 mol+0.1 mol))=0.998` `Vapour pressure of solution (P)=P_(A)^(@)X_(A)=(200mm)xx(0.998)-199.6 mm Hg`. |
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| 1611. |
At `298 K` , the vapour pressure of water is `23.75 mm Hg`. Calculate the vapour pressure at the same temperature over `5%` aqueous solution of urea. `[CO(NH_(2))_(2)]`. |
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Answer» Appling Raoult,s law `(P^(@)-P_(s))/P^(@)=n_(2)/n_(1)=(W_(2)//Mw_(2))/(W_(1)//Mw_(1))=(W_(2)xxMw_(2))/(Mw_(1)xxW_(1))` Substituting the values, we get `(23.75-P_(s))/23.75 =5/60xx18/95` `23.75-P_(s)=5/60 xx18/95xx23.75 ` or `23.75-P_(s)=0.375 rArr P_(s)=23.375 mm` |
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| 1612. |
What will be the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0 g of polymer of molar mass 150,000 in 500 mL of water at `37^@C` ?A. 30.96 PaB. 34.36 PaC. 68.72 PaD. 48.25 Pa |
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Answer» Correct Answer - B No. of moles of polymer =`1/(150,000)``pi=CRT = n/VRT` `pi=1/(150,000)xx(8.314xx10^3xx310)/0.5=34.36` Pa (R=`8.314xx10^3` Pa L `K^(-1) mol^(-1)`) |
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| 1613. |
For an ideal solution with `P_(A)^(0)gt P_(B)^(0)`, which of the following is true?A. `(chi_(A))_("liquid")lt (chi_(A))_("vapor")`B. `(chi_(A))_("liquid") gt (chi_(A))_("vapor")`C. `(chi_(A))_("liquid")= (chi_(A))_("vapor")`D. `(chi_(A))_("liquid") and (chi_(A))_("vapor")` cannot be correlated witheach other |
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Answer» Correct Answer - 1 `P_(A)^(@) lt P_(B)^(@)`. This implies that component `A` is more volatile that component `B`. The vapor in equilibrium with a liquid solution of two or more volatile components always has a higher mole fraction of the more volatile component than does the liquid solution. |
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| 1614. |
What do you understand by solubility of gases? Explain the factors affecting solubility of gases in liquid. |
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Answer» Solubility of gases in liquids: All gases are less or more soluble in water or soluble in certain limit and form a true solution. The volume (in mL) of a gas which dissolved in 1 mL of water is called absorption coefficient.The solubility of gases in liquids is affected by the following factors: 1. Nature of gas: The gases which react with solvent or ionized in solution are soluble in water. e.g. NH3, HCl(g) and SO2 are more soluble in water and form NH4OH, HCl(l) and H2SO4 compounds respectively. 2. Nature of solvent: The polar gases are soluble in polar solvent while non-polar gases are soluble in non-polar solvents. It is like dissolves like rule e.g. HCl gas is more soluble in water than liquid benzene. 3. Effect of Temperature: Temperature has a marked effect on the solubility of a solid in liquid. The solubility may increase or decrease with the rise in temperature depending upon the value of ∆sol H < 0. Let us recall, that saturated solution represents equilibrium between undissolved solute and dissolved solute. (i) If the value of ∆sol H < 0 i.e. the solution process is exothermic, then according to Le-chatelier’s principle, the increase of temperature will push the solution equilibrium in the backward direction. In other words, the solubility of solutes decreases with rise in temperature. Some examples are Li2SO4, Na2SO4 etc. (ii) If ∆sol H > 0, i.e. solution process is endothermic, then increase of temperature will push solution equilibrium in the forward direction. In other words, the solubility of such solutes increases with the rise in temperature. Some examples of such solutes are KCI, KNO3, NaNO3 etc. 4. Effect of pressure: The solubility of gases is much affected by change in pressure. Henry proposed a relationship between pressure (P) and the composition of solution which is called Henry’s law. According to this law. |
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| 1615. |
On increasing temperature, the solubility of H2 gas in water(a) increases(b) decreases(c) remains unchanged(d) None of these |
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Answer» The answer is (a) increases |
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| 1616. |
`200 cm^(3)` of an aqueous solution of a protein contains `1.26 g `of the protein. The osmotic pressure of such a solution at `300 K` is found to be `2.57 xx10^(-3)` bar. Calculate the molar mass of the protein.A. `61038 g mol^(-1)`B. `51022 g mol^(-1)`C. `122044 g mol^(-1)`D. `31011 g mol^(-1)` |
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Answer» Correct Answer - 1 we have `pi=CRT=n_(2)/VRT=(W_(2)RT)/(M_(2)V)` or `M_(2)=(W_(2)RT)/(piV)` Substituting the given values, we get `M_(2)=((1.26 g)(0.083 L "bar" mol^(-1) K^(-1))(300 K))/((2.57xx10^(-3))(200xx10^(-3)L))` `=61038 g mol^(-1)` |
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| 1617. |
`200 cm^(3)` of an aqueous solution of a protein contains `1.26 g `of the protein. The osmotic pressure of such a solution at `300 K` is found to be `2.57 xx10^(-3)` bar. Calculate the molar mass of the protein. |
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Answer» The given values are `pi=2.57xx10^(-3)` bar `V=200 cm^(3)=0.2 L` `T=300 K` `R=0.083 L` bar `mol^(-1) K^(-1)` Using the formula, `Mw_("solute")=(W_("solute")xxRxxT)/(pixxV)` `= (1.26 g xx 0.083 L` "bar" `"mol"^(-1) K^(-1)xx300 K)/(2.57xx10^(-3) "bar"xx0.2 L)` `= 61022 g mol^(-1)` Thus, the molecular weight of solute is `61022 g mol^(-1)`. |
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| 1618. |
If the observed and theoretical molecular mass of `NaCI` is found to be `31.80` and `58.50`, then the degree of dissociation of `NaCI` is:A. 0.8396B. 0.0839C. 0.9D. 1 |
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Answer» Correct Answer - A `i =(58.50)/(30.80) = 1.8396` `alpha = (i-1)/(m-1) = (1.8396 -1)/(2-1)` `= 0.8396` or `83.96%` |
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| 1619. |
Which of the following aqueous solutions will have minimum elevation in boiling point ?(a) 0.1 M KCl (b) 0.05 M NaCl (c) 1 M AIPO4 (d) 0.1 M MgSO4 |
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Answer» Option : (b) 0.05 M NaCl |
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| 1620. |
Which of the following solutions shows maximum depression in freezing point ?(a) 0.5 M Li2SO4 (b) 1 M NaCl (c) 0.5 M Al2(SO4)3(d) 0.5 M BaCl2 |
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Answer» Option : (c) 0.5 M Al2(SO4)3 |
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| 1621. |
To `10 mL` of `1 M BaCl_(2)` solution `5mL` of `0.5 M K_(2)SO_(4)` is added. `BaSO_(4)` is precipitated out. What will happen?A. Freezing point will increase.B. Boiling point will increase.C. Freezing point will lower down.D. Boiling point will lower down. |
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Answer» Correct Answer - B::C Adding `K_(2)SO_(4)` into `BaCl_(2)` solution will result into precipitation of `BaSO_(4)` which is non-volatile, and therefore boiling point of solution will be increased and its freezing point will decrease. |
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| 1622. |
If the freezing point of 0.1 m solution is 272.814 K, then the freezing point of 0.2 m solution will be :(a) 545.628 K (b) 265.64 K (c) 272.628 K (d) 0.482 K |
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Answer» Option : (c) 272.628 K |
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| 1623. |
A difference between diffusion and osmosis isA. A semi-permeable membrane is required for osmosis while diffusion requires no semi-permeable membrane.B. In osmosis movement of molecules is only in one direction whereas in diffusion movement is on both sides.C. In osmosis only the solvent moves while in diffusion both solute and solvent move.D. None of these |
| Answer» Correct Answer - A::B::C | |
| 1624. |
Which of the following statements is/are correct?A. The freezing point of water is depressed by the addition of glucose.B. The degree of dissociation of a weak electrolyte decrease as its concentration decreases.C. Energy is released when a substance dissolves in water provided that the hydration energy of the substance is more than its lattice energy.D. If two liquids that form an ideal solution are mixed, the change in entropy is positive. |
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Answer» Correct Answer - A::C::D There will be a depression in the freezing point by the addition of glucose in water. `alpha` increases as concentration decreases. `Delta_(sol)H= underset(-ve)(Delta_("latice")H )+ underset (-ve)(Delta_(hyd)H)` Mixing of solution is always accompanied by an inceases in entropy (randomness). |
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| 1625. |
Consider the two solutions: I: `0.5 M NaCl` aqueous solution at `25^(@)C`, `NaCl` is complete ionized. II:`2.0 M C_(6)H_(5)COOH` in benzene at `25^(@)C`, `C_(6)H_(5)COOH` dimerizes to the full extent. Which of the following statements(s) is (are) correct?A. Both the solutions display equal osmotic pressure.B. Both have equal vapour pressure.C. Solution II is hypertonic.D. Solution II has greater depression in freezing point than solution I. |
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Answer» Correct Answer - A::D `NaCl`:0.5 M (i=2) `rArr C_(eff) =0.5 xx 2 =1 M` `C_(6)H_(5)COOH:2.0 M (i=1/2) rArr C_(eff) = 2 xx 1/2=1 M` `pi=C_(eff) RT rArr pi` is same as `C_(eff)` is same. Since both the solutions have different solvents, vapour pressure will be different. Solutions are isotonic. `DeltaT_(f) = iK_(f).m rArr DeltaT_(f)` for solution II will be more as `(K_(f))_( "benzene") gt (K_(f))_( "water")` |
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| 1626. |
The freezing point of a pure solvent is 315 K. On addition of 0.5 mole of urea in 1 kg of the solvent, the freezing point decreases by 3 K. Calculate the molal depression constant for the solvent. |
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Answer» Given : m = Molality of urea = 0.5 m ΔTf = Depression in the freezing point = 3 K K = Molal depression constant for the solvent = ? ΔTf = Kf × m ∴ Kf = \(\frac{ΔT_f}{m}\) = \(\frac{3}{0.5}\) = 6 K kg mol-1 ∴ Molal depression constant = 6 K kg mol-1 |
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| 1627. |
Which pair(s) of liquids on mixing is/are expected to show no net volume change and no heat effect?A. Acetone and ethanolB. Chlorobenzene and bromobenzeneC. Chloroform and benzeneD. n-Butyl chloride and n-butyl bromide |
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Answer» Correct Answer - B::D `Delta_("mixing") V= 0`,`Delta_("mixing") H= 0`,`rArr` Look for ideal solutions. |
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| 1628. |
Solute when dissolved in water:A. increases the vapour pressure of waterB. decreases the boiling point of waterC. decreases the freezing point of waterD. all of the above |
| Answer» Correct Answer - C | |
| 1629. |
Glucose, an organic compound, dissolves in water because of theA. dipole-dipole interactionsB. dispersion forcesC. hydrogen bondingD. ion-dipole interactions |
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Answer» Correct Answer - 3 Glucose, a polar organic solid, has five OH groups per molecule. It dissolves in water because of the hydrogen bonds that form between the water molecules and the OH groups. Glucose does not dissolve in octane as there no favorable interactions between glucose molecules and octane molecules to compensate for the energy required to break the hydrogen bonds between glucose molecules in the solid. |
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| 1630. |
The process in which dissolved solute comes out of solution and forms crystals is calledA. crystallizationB. precipitationC. peptisationD. any of these |
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Answer» Correct Answer - 1 Both precipitation and crystallization describe the separation of excess solid substance from a supersaturated solution. However, solid formed in the two processes differ in appearance. We normally think of precipitates as being made up of small particle, whereas crystals may be large and well formed. |
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| 1631. |
(a) A solution containing `0.5g` of naphithalene in `50g C Cl_(4)` yield a boiling point elevation of `0.4K`, while a solution of `0.6g` of an unknown solute in the same mass of the solvent gives a boiling point elevation of `0.65K`. Find the molar mass of the unknown solute. (b) The boiling point of a solution of `0.1g` of a substance in `16g` of ether was found to be `0.100^(@)C` higher that of pure ether. What is the molecular mass of the substance. `K_(b)` (ether) `=2.16K kg "mol"^(-1)` |
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Answer» Correct Answer - (a) `M=94.52,` (b) `m=135` (a) Let the molar mass of the unknown solute is `m` Now, for nephthalene solution `DeltaT_(b)=K_(b)="molality"` `0.4=K_(b)xx(0.5//128)/(50//1000)` .... (i) for unknown solution `0.65=K_(b)xx(0.6//m(50//1000)` ....(ii) Dividing eq. (i) by (ii), we get (b) `DeltaT_(b)=K_(b)` molality `0.100=2.16xx(0.1//m)/(1000)` |
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| 1632. |
What is molality of pure water?A. 1B. 18C. 55.5D. None of these |
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Answer» Correct Answer - C Molality of water means no of moles of water in 1000 g of water which is `=(1000)/(18)=55.55` mol |
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| 1633. |
Dry air was passed through bulbs containing a solution of `40` grams of non electrolytic solute in `360` grams of water, then through bulbs containing pure water at the same temperature and finally through a tube in which pumice moistened with strong `H_(2)SO_(4)` was kept. The water bulbs lost `0.0870` grams and the sulphuric acid tube gained `2.175` grams. Calculate the molecular weight of solute. |
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Answer» Correct Answer - `M=48` `(P-P_(s))/(P)=(0.087)/(2.175)= ((40)/(M))/((40)/(M)+(360)/(18))` `rArr (2.175)/(0.087)=1+(360)/(18)xx(M)/(40)` `rArr25=1+(M)/(2)` |
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| 1634. |
The increase in boiling point of a solution containing 0.6g urea in 200 gram water is `0.50^(@)C`. Find the molal elevation constantA. `10 K "gram" mol^(-1)`B. `1.0 K kg mol^(-1)`C. `10 K kg mol^(-1)`D. `10 K kg` mol |
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Answer» Correct Answer - C Molality, `m=(0.6)/(60)xx(1000)/(200)=0.05m` `Delta T_(b)=0.5 K` now `Delta T_(b)=K_(b)xxm` or `K_(b)=(Delta T_(b))/(m)=(0.5)/(0.05)=10K kg mol^(-1)` |
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| 1635. |
Dry air was drawn through bulbs containing a solution of 40 grams of urea in 300 gram of water, then through bulbs containing pure water at the same temperature and finally through a tube in which pumice moistened with strong.`H_(2)SO_(4)` was kept. The water bulbs lost 0.0870 grams and the sulphide acid tube gained 2.036 grams. Calculate the molecular weight of urea. |
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Answer» Correct Answer - `M=53.8` Weight loss in first bulb `("w"_(1))prop"P"_("solution")` weight loss in second bulb `("w"_(2))prop"P"^(0)-"P"_("solution")` weight gain in tube `("w"_(1)+"w"_(2))prop"P"^(0)` `("P"^(0)-"P"_("s"))/("P"_("s"))=("n"_("urea"))/("n"_("H"_(2)"O"))=("w"_(1))/("w"_(2))` `"w"_(2)=0.0870` `"w"_(1)=1.949` `(40xx18)/("m"_("urea")xx300)=(0.0870)/(1.949)` `"m"_("urea")=(40xx18xx1.949)/(300xx0.0870)=53.8"gm mol"^(-1)` |
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| 1636. |
Blood cells do not shrink in blood because blood isA. hypotonicB. isotonicC. equimolarD. hypertonic |
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Answer» Correct Answer - B Blood cells will not shink in an isotonic solution. |
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| 1637. |
Solubility curve of a hydrated salt in water with temperature is given. The curve indicates that the solution process is A. exothermic temperatureB. exotehrmic till `60^(@)C` and endothermic after `60^(@)C`C. endothermic till `60^(@)C` and exothermic there-afterD. endothermic |
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Answer» Correct Answer - C Since solubility increases with increase in temperature upto `60^(@)C`, the process is endothermic upto `60^(@)C`. After `60^(@)C` solubility decreases with increase in temperature, thus after `60^(@)C`, the process is exothermic. |
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| 1638. |
The molar volume of liquid benzene (density `0.877 g mL^(-1))` increases by a factor of `2750` as it vapourises at `20^(@)C` and that of liquid toluene(density `0.867 g mL)` increases by a factor of `7720`at `20^(@)C`. A solution of benzene and toluene at `20^(@)C` has a vapour pressure of 46.0 torr. Find the mole fraction of benzene in the vapour above the solution. |
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Answer» Correct Answer - `0.732` `V_(B) = (78)/(0.877) xx 2750 mL = 244.583 L` `V_(T) = (92)/(0.867) xx 7720 mL = 819.192 L` `P_(B) = (1 xx 0.0821 xx293 xx760)/(244.583) = 74.74` torr `P_(T) = (1 xx 0.0821 xx 293 xx 760)/(819.192) =22.317` `46 = 74.74 X_(B) +23.317 (1-X_(B))` `52.423 X_(B) = 23.683` `X_(B) = 0.451` `Y_(B) = (1 xx 0.0821 xx 293 xx 760)/(819.192) = 22.317` `46 = 74.74 X_(B) +23.317 (1-X_(B))` `54.423 X_(B) = 23.683` `X_(B) = 0.451` `Y_(B) = (P_(B)^(@)-X_(B))/(P_(T)) = (74.74xx 0.451)/(46) = 0.732` |
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| 1639. |
The system that forms maximum boiling azeotrope isA. acetone-chloroformB. ethanol-acetoneC. n-hexane-n-heptaneD. carbon disulphide - acetone |
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Answer» Correct Answer - A Due to stronger intermolecular interactions in acetone and chloroform lesser number of molecules vaporise resulting in low vapour pressure and high boiling point . |
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| 1640. |
Which will from maximum boiling azeotrope ?A. `HNO_(3)+ H_(2)O` solutionB. `C_(2)H_(5)OH +H_(2)O` solutionC. `C_(6)H_(6)+C_(6)H_(5)CH_(3)` solutionD. None of these |
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Answer» Correct Answer - A `HNO_(3)+H_(2)O` solution shows negative deviation from ideal behaviour. Hence it forms azeotrope with maximum boiling point. |
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| 1641. |
Consider the following aqueous solutions and assume 100% ionisation in electrolytes : I. `0.1 m` urea II. `0.04 m Al_(2)(SO_(4))_(3)` III. `0.0.5 m CaCl_(2)` IV. `0.005m NaCl` The correct statement regarding the above solutions is :A. Freezing point will be lowest for solution 1B. Freezing point will be highest for solution 4C. Boiling point will be highest for solution 4D. Vapour pressure will be highest for solution 2 |
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Answer» Correct Answer - B Particle concentration. (1) 0.1 m (2) `0.04 xx =0.2 m` (3) `0.05 xx 3 = 0.15 m` (4) `0.005 xx 2 = 0.01 m` Solution (4) has the lowest concentration, depression in freezing point will be minimum and hence actual freezing point will be highest. |
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| 1642. |
5 litre of a solution contains 25 mg of `CaCO_(3)`. What is its concentration in ppm? (mol.wt of `CaCO_(3)` is 100)A. 25B. 1C. 5D. 250 |
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Answer» Correct Answer - C 5L dilute solution = 5 kg = 5000 g Mass of solute `= 25 mg = 25xx10^(-3)g` ppm of `CaCO_(3)` in solution `= (25xx10^(-3)g)/(5000 g)xx10^(6) = 5` ppm |
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| 1643. |
At `100^(@)"C"`, benzene & toluene have vapour pressure of 1375 & 558 torr respectively. Assuming these two form an ideal binary solution, calculate the composition of the solution that boils at 1 atm & `100^(@)"C"`. What is the composition of vapour issuing at these conditions? |
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Answer» Correct Answer - `X_(b)=0.2472,Y_(b)=0.4473` `"P"_("total")="X"_("benzene")("P"_("benzene")^(0)-"P"_("benzene")^(0))+"P"_("toluene")^(0)` `760="X"_("benzene")(1375-558)+558` `"X"_("benzene")=0.2472` `"Y"_("benzene")=("X"_("benzene")xx"P"_("benzene")^(0))/("P"_("total"))=(0.2472xx1375)/(760)=0.4473` |
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| 1644. |
The vapour pressure of water at `80^(@)C` is `355` torr. A `100 ml` vessel contained water-saturated oxygen at `80^(@)C`, the total gas pressure being `760` torr. The contents of the vessel were pumped into a `50.0 ml`. Vessel at the same temperature. What were the partial pressures of oxygen and of water vapour, what was the total pressure in the final equilibrated state? Neglect the volume of any water which might condense. |
| Answer» Correct Answer - `P_(O_(2))=810 mm Hg, P_(H_(2)O)=355 mm Hg, P_("total")=1165 mm Hg` | |
| 1645. |
The vapor pressures of benzenen, toluene and `a` xylene are `75` Torr, `22` Torr and `10` Torr respectively at `20^(@)C` Which of the following is not a possible value of the vapor pressure of an equimolar binarytemary solution of these at `20^(@)C`? Assume all form ideal solution with each other?A. `48(1)/(2)`B. 16C. `35(2)/(3)`D. `53(1)/(2)` |
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Answer» Correct Answer - D Possible vapor pressures are `(75+22)/(2),(75+10)/(2),(22+10)/(2) "and" (75+22+10)/(2)=48(1)/(2),42(1)/(2),16,35(2)/(3)`. |
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| 1646. |
Calculate molarity of `CaCO_(3)`(aq.) solution which has concentration of `CaCO_(3)=200`ppm. |
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Answer» `200g` of `CaCO_(3)` in `10^(6)g` of water. `(200)/(1000)=2` moles of `CaCO_(3)` liters of water. (density`=1g//mL`) So molarity`=(2)/(10^(3))=2xx10^(-3)M`. |
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| 1647. |
The vapor pressure of water at `80^(@)C` is `355` torr. A `100 ml` vessel contained water-saturated oxygen `80^(@)C`, the total gas pressure being `760` torr. The contents of the vessel were pumped into a `50.0m` vessel at the same temperature. What were the partial pressure of oxygen and of water vapor, with was the total pressure in the final equilibrated state? Neglect the volume of any water which condense. |
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Answer» Correct Answer - `P_(O_(2))=810mmHg, P_(H_(2))O=355mm Hg, P_("total")=1165mm Hg` In `100ml` vessel which contained water-saturated oxygen, the pressure of `O_(2)` gas`=760-355=405` torr when the contents of this vessel were pumped into `50ml` at the same temperature, pressure of oxygen gets doubled i.e. `P_(O_(2)=810` torr. But pressure of water vapour will remain constant, as some vapour in this `50ml` vessel, gets condense. So `P |
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| 1648. |
Calculate the molality and molarity of a solution made by mixing equal volumes of `30%` by weight of `H_(2)SO_(4)` (density`=1.20g//ml`) and `70%` by weight of `H_(2)SO_(4)` (density`=1.60g//mL`) |
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Answer» Correct Answer - Molality `=11.44 m`, Molarity `=7.55M` Suppose that the solution contains `100 ml` of each variety of `H_(2)SO_(4)`. Total solution is, `200 ml` or `0.2` litre `wt.` of `100 ml` of `H_(2)SO_(4)` solution `(30%)=1060xx100=160` gram `wt.` of `H_(2)SO_(4)(30%)=120xx(30)/(100)=36` gram `wt.` of `H_(2)SO_(4)(70%)=160xx(70)/(100)=112` gram Total `wt.` of `H_(2)SO_(4)` (salute) `=36+112=148` gram `therefore wt.` of `H_(2)O`(solvent) `=wt.` of solution`-wt.` of solute `=(120+160)-148` `=280-148=132` gram Moles of `H_(2)SO_(4)=(148)/(98)=1.51` (mol `wt.` of `H_(2)SO_(4)=98`) Molality `=(1.51)/(132)xx1000=11.44m` Molarity `=(1.51)/(0.2)=7.55 M` |
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| 1649. |
Find the molality of` H_(2)SO_(4) `solution whose specific gravity is ` 1.98 g ml^(-1) `and ` 95% `by volume ` H_(2)SO_(4) `A. ` 7.412 `B. ` 8.412 `C. ` 9.412 `D. ` 10.412 ` |
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Answer» Correct Answer - C ` H_(2)SO_(4) ` is ` 95% `by volume wt. of ` H_(2)SO_(4) = 95 g` vol . Of solution =` 100 ml ` ` :. ` Moles of ` H_(2)SO_(4) =(9)/(98), ` and weight of solution ` 100xx 1.98 =198 g ` Weight of water =` 198-95 =103 g ` Weight of water =` (95xx1000)/(98xx103) =9.412 ` |
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| 1650. |
Calculate molality of `1.2M H_(2)SO_(4)` solution? If its `p=1.4g//mL` |
| Answer» Molarity`=(1.2xx1000)/(1000xx1.4-1.22xx98)=0.936` | |