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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.
| 1251. |
Molarity of `K^(+)` ions in `0.33 M` potassium sulphate aqueous solution isA. less than 0.33 MB. equal to 0.33 MC. more than 0.33 MD. `0.66 M` |
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Answer» Correct Answer - 4 A `0.33 M` aqueous solution of potassium sulphate `(K_(2)SO_(4))` contains `0.33 mol` of the salt in 1 litre of solution. The molarities of the potassium and sulphateions in the solution are `0.66 M` and `0.33 M` respectively because `K_(2)SO_(4)` is a strong electroly to i.e., `1 mol K_(2)SO_(4)` gives `2 mol of K^(+)` ions and 1 mol of `SO_(4)^(2-)` ions on dissociation. |
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| 1252. |
In the laboratory we often measure the volume of one solution that is required to react with a given volume of another solution of known concentration. Then we calculate the concentration of the first solution. The process is calledA. neutralizationB. precipitationC. titrationD. combination |
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Answer» Correct Answer - 3 Titration is the process in which a solution of one reactant (the titrant) is carefully added to a solution of another reactant and the volume of titrant required for complete reaction is measured. To know when to stop a titration (i.e., when is the chemical reaction just complete), a few drops of an indicator solution are added to the solution to be titrated. An indicator is a substance that can exist in different forms, with different colours that depend on the concentration of `H^(+)` in the solution. At least one of these forms must be very intensely coloured so that even very small amounts of it can be seen. We can titrate an acid solution of unknown concentration by adding a standardized solution of sodium hydroxide dropwise from a buret. A common buret is graduated in large intervals of 1 mL and in smaller intervals of `0.1 mL`, so that it is possible to estimate the volume of a solution dispensed to within at least `+- 0.02 mL`. (Experienced individuals can often read a burst to `+- 0.01 mL`.) The solution `(HCl)` to be titrated is placed in an Erlenmeyer flask (volumetric flask), and a few drops of indictor are added. The analyst tries to choose an indictor that changes color clearly at which stoichiometrically equivalent amounts of acid and base have reacted, the equivalence point. The point at which the indicator changes colour and the titration is stopped is called the end point. Ideally, the end point should coincide with the equivalence point. Phenolphthalein is colourless in acidic solution and reddish violet in basic solution. In a titration in which a base id added to an acid, phenolphthalein is often used as an indicator. The end point is signaled by the first appearance of a faint pink coloration that persists for at least `15` seconds as the solution is swirled. The buret is usually filled with a standard solution (or the solution to be standardized). The volume of solution in the buret is read carefully. The meniscus decribes the surface of the liquid in the buret. Aqueous solutions wet glass so the meniscus of an aqueous solution is always concave. The position of the bottom of the meniscus is read and recorded. The volume of the solution is read again after the end point. The difference between the final and initial buret readings is the volume of the solution used. |
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| 1253. |
Reverse Osmosis |
| Answer» When the hydrostatic pressure applied on the solution side is more than the osmotic pressure of solution, then solvent molecule from solution side enter in to the solvent side through semi-permeable membrane. This process is called reverse osmosis. | |
| 1254. |
Consider the following solutions: I.1 M sucrose , II. 1 M KCl III.1 M benzoic acid in benzene IV.`1 M (NH_(3))_(3)PO_(4)` Which of the following is/are true?A. All solutions are isotonic.B. III is hypotonic of I,II, and IV.C. I,II, and III are hypertonic of IV.D. IV is hypertonic of I,II, and III. |
| Answer» Correct Answer - B::C::D | |
| 1255. |
The osmotic pressure of a solution depends onA. Nature of soluteB. Nature of solventC. TemperatureD. Molar concentration of solute |
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Answer» Correct Answer - C::D Osmotic pressure is a colligative property, therefore, it depends upon the number of constituents of solution. It also depends upon temperature as `pi=CRT`. |
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| 1256. |
Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to homegenous solution. These are called colligative properties. Applications of colligative properties are very useful in day to day life. One of its exaples is the use of ethylene glycol and water mixtue as anti-freezing liquid in the radiators of automobiles. A solution M is prepared by mixing ehanol and water. The mole fraction of ethanol in the mixture is 0.9 Given. `K_(f)"of water"=1.86 "K kg mol"^(-1)` `K_(f)"of ethanol" = 2.0 "K kg mol"^(-1)` `K_(b)"for water "=0.52 "K kg mol"^(-1)` `K_(b)" of ethanol"=1.2 "K kg mol"^(-1)` `T_(f)("water")=273 K` `T_(f)("ethanol")=155.7 K` `T_(b)^(@) ("water")=373 K` `T_(b)^(@)("ethanol")=351.5 K` `P^(@)("water")32.8 mm Hg.` `P^(@)("ethanol")=40 mmHg` `M("ethanol")==46 "g mol"^(-1)` `M("water")=18"g mol"^(-1)` In answering the following quesion, consider the solutions to be ideal dilute solutions ans solute to be non-voltile and non-dissoviative. (6) The vapour pressure of solution M is :A. `39.3 mm Hg`B. `36.0 mm Hg`C. `25.2 mm Hg`D. `28.8 mm Hg` |
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Answer» Correct Answer - b V.P. of solution (p) =`P_(A)^(@)X_(A)` =(40 mm Hg)xx(0.9)=36 mm Hg |
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| 1257. |
Which of the following combinations are correct for a binary solution, in which the solute as well as solvent are liquid?A. `C_(6)H_(6) and C_(6)H_(5)CH(3)`,`Delta_(sol)Hgt0`,`Delta_(sol)V=0`B. `CH_(3)-overset(O)overset(||)(C)-CH_(3)` and `CHCl_(3)`,`Delta_(sol)H lt O`,`Delta_(sol)V lt O`C. `H_(2)O` and `HCl`,`Delta_(sol)H lt O`,`Delta_(sol)V lt O`D. `H_(2)O` and `CH_(2)OH`,`Delta_(sol)H lt O`,`Delta_(sol)V lt O` |
| Answer» Correct Answer - B::D | |
| 1258. |
Binary solution can be of nine different types depending upon the nature of the solute and solvent waether solid, liquid or gas. They may be further classified as solid, liquid and gaseous solutions based on the component which act as the solvent. Howerver, the liquid solutions are the most important. Both solids and gases dissolve in liquids resulting in homegeneous mixtures i.e., solutions. The solubility is governed by number of factors such as nature of solute and solutions can be expressure etc. The concentrations of the solutions can be expressed in different ways such as normality, molarty, molality, mole fration etc. Out of these, molality and mole fraction are better as they donot change with the change in temperature. as they donot change with the change in temperature. (3) The molality of a sulphuric acid solution in which in mole fraction of water is 0.85 is :A. `9.80`B. `10.50`C. `10.58`D. `11.25` |
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Answer» Correct Answer - a `X_(H_(2)O)=0.85,X_(H_(2)SO_(4))=0.15` `n_(1)/(n_(1)+n_(2))=0.85,n_(2)/(n_(1)+n_(2))=0.15` `n_(2)/n_(1)=(0.15)/(0.85)` Taking `n_(1)=1000/18` moles `n_(2)=(0.15)/(0.85)xx1000/18=9.80` |
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| 1259. |
Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to homegenous solution. These are called colligative properties. Applications of colligative properties are very useful in day to day life. One of its exaples is the use of ethylene glycol and water mixtue as anti-freezing liquid in the radiators of automobiles. A solution M is prepared by mixing ehanol and water. The mole fraction of ethanol in the mixture is 0.9 Given. `K_(f)"of water"=1.86 "K kg mol"^(-1)` `K_(f)"of ethanol" = 2.0 "K kg mol"^(-1)` `K_(b)"for water "=0.52 "K kg mol"^(-1)` `K_(b)" of ethanol"=1.2 "K kg mol"^(-1)` `T_(f)("water")=273 K` `T_(f)("ethanol")=155.7 K` `T_(b)^(@) ("water")=373 K` `T_(b)^(@)("ethanol")=351.5 K` `P^(@)("water")32.8 mm Hg.` `P^(@)("ethanol")=40 mmHg` `M("ethanol")==46 "g mol"^(-1)` `M("water")=18"g mol"^(-1)` In answering the following quesion, consider the solutions to be ideal dilute solutions ans solute to be non-voltile and non-dissoviative. (5) The freezing point of solution M is:A. `268.7 K`B. `268.5 K`C. `234.5 K`D. `150.9 K` |
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Answer» Correct Answer - d Molality of solution (m) `=("No. of moles of water")/("Mass of ethanol in kg")` `n_("ethanol")=0.1 mol` `W_("ethnol")=((0.9"mol"))/((0.0414 kg))=2.415 " mol kg"^(-1)` =2.415 m. `DeltaT_(f)=K_(f)xxm=(2.0" K kg mol mol"^(-1))xx(2.415 "mol kg"^(-1))=4.83 K.` f.p. of solution =155.7-4.83=150.87 K. |
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| 1260. |
Binary solution can be of nine different types depending upon the nature of the solute and solvent waether solid, liquid or gas. They may be further classified as solid, liquid and gaseous solutions based on the component which act as the solvent. Howerver, the liquid solutions are the most important. Both solids and gases dissolve in liquids resulting in homegeneous mixtures i.e., solutions. The solubility is governed by number of factors such as nature of solute and solutions can be expressure etc. The concentrations of the solutions can be expressed in different ways such as normality, molarty, molality, mole fration etc. Out of these, molality and mole fraction are better as they donot change with the change in temperature. as they donot change with the change in temperature. (4) 1000 g of a 4% solution of salt contains :A. 4 g saltB. 4 moles of saltC. 40 g of solventD. 30 g of salt. |
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Answer» Correct Answer - d 4% of salt solution means 4 g of salt in 100 g of solution 1000 g of solution will contain 40 g of the salt. |
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| 1261. |
At equlibirium the rate of dissolutiono of a solid solute in a valatile liquid slvent is ……A. less then the rate of crystallisationB. greater than the rate of crystallisationC. equal to the rate of crystallisationD. zero |
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Answer» Correct Answer - C At equilibrium the rate of dissolution of solid in a volatile liquid solvent is equal to the rate of carystallisation. |
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| 1262. |
All of the water in a `0.20M` solution of `NaCl` was evaporated and a `0.150` mol of `NaCl` was obtained. What the original volume of the sample?A. `30mL`B. `333mL`C. `750mL`D. `1000mL` |
| Answer» Correct Answer - C | |
| 1263. |
How many grams of methanol would have to be added to water to prepare 150mL of solution which is `2M CH_(3)OH`?A. `9.6`B. `2.4`C. `9.6 xx 10^(3)`D. `4.3 xx 10^(2)` |
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Answer» Correct Answer - A `1000 mL` of `1M CH_(3)OH` requires methanol = 32 g `150mL` of `2M CH_(3)OH` requires mathanol `= (32)/(1000) xx 150 xx 2 = 9.6g` |
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| 1264. |
At`25^(@)C`the highest osmotic pressure is exhibited by `0.1M` solution ofA. `CaCI_(2)`B. `KCI`C. GlucoseD. Urea |
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Answer» Correct Answer - A Highest osmotic pressure is given by solution which produces maximum no of ions i.e., `CaCI_(2)`. |
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| 1265. |
At`25^(@)C`the highest osmotic pressure is exhibited by `0.1M` solution ofA. `CaCl_(2)`B. `KCl`C. GlucoseD. Urea |
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Answer» Correct Answer - A Highest osmotic pressure is given by solution which produces maximum number of ions i.e,`CaCl_(2)` |
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| 1266. |
Osmotic pressure of `0.1M` solution of `NaCl`and `NaSO_(4)`will beA. sameB. osmotic pressure of`NaCl`solution will be more than`Na_(2)SO_(4)`solutionC. osmotic pressure of`Na_(2)SO_(4)`solution will be more than`NaCl`D. osmotic pressure of`Na_(2)SO_(4)`will be less than that of `NaCl` solution |
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Answer» Correct Answer - C `N_(2)SO_(4)`have more osmotic pressure than`NaCl`solution because `Na_(2)SO_(4)`gives `3`ions. |
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| 1267. |
Which of the following aqueous solutions has the highest boiling point?A. `0.1M KNO_(3)`B. `0.1M Na_(3)PO_(4)`C. `0.1M BaCI_(2)`D. `0.1M K_(2)SO_(4)` |
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Answer» Correct Answer - B Greater the number of ions in the electrolyte higher is the elevation in boiling point. |
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| 1268. |
The molarity of 98% by wt. `H_(2)SO_(4)` (d = 1.8 g/ml) isA. 6MB. 18MC. 10MD. 4M |
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Answer» Correct Answer - B `98% H_(2)SO_(4)` by weight `rArr 98 g` of `H_(2)SO_(4)` in 100 g of solution volume of solution `=(100g)/(1.8 g mL^(-1))=55.55 mL` `=0.056 L` Moles of `H_(2)SO_(4)=98//98=1 mol` Molarity `=(1 mol)/(0.056L) = 17.86 M` |
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| 1269. |
200.0 mL of calcium chloride solution sotains `3.10xx10^(22)CI^(-)` inoc. Calculate the molarity of the solution. Assume that calcium cholride is complately ionised. |
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Answer» Correct Answer - 0.125 M Setp I. Calculation of mass of `CaCI_(2)` present in solution. lonisation of `CaCI_(2)` and no. of different species in the soluion may be represented as follows : `CaCI(s)overset((aq))toca^(2+)(aq)+2CI^(-)(aq)` `(1.505xx10^(22)"molecules")(1.505xx10^(22))(3.01xx10^(22))` `6.022xx10^(22)"molecules of"CaCI_(2) " correspond to mass"=(1.505xx10^(22))/(6.022xx10^(23))xx(111g)=2.774 g` Step II. Calculation of molarity of solution. `"Molarity of soluion (M)"=("Mass of "CaCI_(2)//"Molar mass")/("Volume of solution in Litress")` `((2.774g)//(111" g mol"^(-1)))/(200//1000L)=0.125" mol L"^(-1)=0.125 M` |
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| 1270. |
If the density of a 1-ltre solution of `98% H_(2)SO_(4) (wt.//vol.)` is `1.88 g mol^(-1)`, the molality of the solution will beA. `13.13`B. `10.10`C. `11.11D. `12.12` |
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Answer» Correct Answer - 3 `98% H_(2)SO_(4) (wt.//vol.)` implies that every `100 mL` of solution contains `98 g` of `H_(2)SO_(4)`. Thus `n_(H_(2)SO_(4))=(mass_(H_(2)SO_(4)))/(molar mass_(H_(2)SO_(4)))` `=(98 g)/(98 g//mol)=1 mol` Mass of solution =`(Volume)_("soln")xx("density")_("soln")` `=(100 mL)(1.88 g mL^(-1))` `=188 g` Mass of solvent =(Mass of solution) - (Mass of solute) `=(188 g) - (98 g)` `=90 g` Molality (m)`= n_(H_(2)SO_(4))/g_("solvent")xx(1000 g)/(kg)` `=(1 mol)/(90 g)xx(1000 g)/(kg)` `=11.11 m` |
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| 1271. |
Magnitude of crystal lattice enthalpies generallyA. increase with increasing charge and size of ionsB. increase with decreasing charge and size of ionsC. increase with increasing charge and decreasing size of ionsD. increase with decreasing charge anf increasing size of ions |
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Answer» Correct Answer - 3 The sizes of the lattice enthalphy increase as the ionic charge densities increase and therefore as the strength of electrostatic attractions within the crystal increase. The charge//radius ratio is the ionic charge divided by the ionic radius in angstroms. This is a measure of the charge density around the ion. Hydration enthalpies vary in the same order. Crystal lattice enthalpies and hydration enthalpies are generally much smaller in magnitude for molecular solids than for ionic solids. Hydration and the effects of attractions in a crystal oppose each other in the dissolution process. Hydration enthalpies and lattice anthalpies are usually of about the same magniyude for low-charge species, so they often nearly cancel each other. As a result, the dissolution process is slightly endothermic for many ionic substances. Ammonium nitrate `(NH_(4)NO_(3))` is an example of a solid salt that dissolves in water in a very endothermic process, absorbing heat from its surroundings. This property is used in "instant cold packs" for early treatment of injuries, such as sprains and bruises to minimize swelling. For this purpose `NH_(4)NO_(3)` and `H_(2)O` are packaged in a plastic bag in which they are kept separate by a partition that is easily broken when squeezed. As the `NH_(4)NO_(3)` dissolves in the `H_(2)O`, the mixture absorbs heat from its surroundings and the bag becomes cold to the touch. Some ionic solids dissolve with the release of heat. Examples are anhydrous sodium sulphate `(Na_(2)SO_(4))`, calcium acetate, `Ca(CH_(3)COO)_(2)`, calcium chloride, `CaCl_(2)`, and lithium sulphate hydrate, `Li_(2)SO_(4). H_(2)O`. As the charge-to-size ratio (charge density) increases for ions in ionic solids, the magnitude of the crystal lattice anthalpy usually increases more than the hydration enthalphy. This makes dissolution of solids that contain highly charged ions (such as `AlF_(3), MgO` and `Cr_(2)O_(3))` very endothermic. As a result, these compounds are not very soluble in water. |
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| 1272. |
180 ml of pure water at `4^(@)C` is saturated with `NH_(3)` gas, yielding a solution of density 0.8 gram/ml and containing `NH_(3)` (40% by weight ). (a) Volume of `NH_(3)` solution. (b) Volume of `NH_(3)` gas present in saturated solution. (c ) Volume of `NH_(3)` gas at 115^(@)C` and 950 mm of Hg in saturated solution. |
| Answer» 375 ml, 195 ml, 162.19 ml | |
| 1273. |
An aqueous solution of `H_(2)SO_(4)` has density `1.84" g"//"ml"`. Solution contains 98% `H_(2)SO_(4)` by mass.Calculate (i) Molarity of solution (ii) Molar volume of solution (iii) Relative lowering of vapour pressure w.r.t. water, assuming `H_(2)SO_(4)` as non-electrolyte at this high concentration. |
| Answer» (i) 18.4 M, (ii) `48.914" ml"//"mol"`, (iii) 0.9 | |
| 1274. |
Which of the following collingative properties can peovide molar mass of proteins, polymers of colloids with greater precision ?A. Relative lowering in vapour pressureB. Elevation of boiling pointC. Depression in freezing pointD. Osmotic pressue |
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Answer» Correct Answer - d is the correct answer. |
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| 1275. |
A 0.5% (by weight) solution of `A_(2)B` in solvent C was found to freeze at `-3.25^(@)C`. Calculate the degree of dissociation of `A_(2)B` in solvent C into `A_(2)^(-)` and `B_(2)^(-)`. (Given freezing point of pure C is `-3^(@)C`, molar weight of `A_(2)B` is 60 and `K_(f)` of C is 2 `K^(-1)` "molality"^(-1)`). |
| Answer» Correct Answer - 0.5 | |
| 1276. |
Protective sols are:A. Lyophilic colloids - reversible solsB. lyophobicC. both (A) and (B)D. none of (A) and (B) |
| Answer» Correct Answer - A | |
| 1277. |
A `500gm` liquid consist of `15gm` ethane at any temp. `T`, at a pressure `=2` atm. Find Pressure of gas required to dissolve `30gm` gas in `300gm` liquid. |
| Answer» Correct Answer - `p_(2)=6.66` atm. | |
| 1278. |
Depression in freezing point of 1.10-molal solution of HF is `0.201^(@)C`. Calculate percentage degree of dissoviation of HF (`K_(f)`=1.856 K kg `mol^(-1)`). |
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Answer» Correct Answer - 8.06%18.67 atm `DeltaT_(f)=ixxK_(f)xxmori=(DeltaT_(f))/(K_(f)xxm)` `i=((0.201K))/((1.86" K kg mol"^(-1))(0.10"mol kg"^(-1)))=1.0806.` `"Degree of dissociation of HF "(alpha)=(i-1)/(n-1)=(1.0806-1)/(2-1)` =0.0806 = 8.06 %. |
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| 1279. |
Define the term ‘abnormal molar mass’. |
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Answer» Molecular masses can be calculated by measuring any of the colligative properties. The relation between colligative properties and molecular mass of the solute is based on following assumptions. (1) The solution is dilute, so that Raoult‘s law is obeyed. (2) The solute neither undergoes dissociation nor association in solution. In case of solutions where above assumptions are not valid we find abnormal molar masses. (i) Association of solute molecules: Certain solutes in solution are found to associate. This eventually leads to a decrease in the number of molecular particles in the solutions. Thus, it results in a decrease in the values of colligative properties therefore higher values are obtained for molecular masses than normal values for unassociated molecules. (ii) Dissociation of solute molecules: A number of electrolytes dissociate in solution to give two or more particles (ions). Therefore, the number of solute particles, in solutions of such substances, is more than the expected value therefore molecular masses of such substances as calculated from colligative properties will be less than their normal values. |
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| 1280. |
Give two difination of solubility of a substance. Show that the total vapour pressure over that solution of two liquids A and B at a p[articular temprature varies linerluy wioth the mole fraction of the solute. |
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Answer» It may be defined as the maximum amount of the slute that can desso=ilve in 100 grams of the solvent to from a saturated solution at a given temperature. For answer consult section 6. (Ideal and non-ideal solutions) |
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| 1281. |
Give resons for the following : Measurementy of osmotic pressure method is preferred for the determination of moar masses of macromolecules. Such as proteins and polymers. Aquatic animals are more comfortable in cold water than in warm water. Elevation in boiling point of a 1 M KCI solution is nearly doble than that of 1M sugar solution. |
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Answer» For answer consult section 12. For answer consult section 3. For answer consult answer to 14. |
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| 1282. |
A sample of alcohol at 200 torr and `70^(@)C` is cooled at constant volume to `25^(@)C`. What would exist at `25^(@)C`? Vapour pressure of alcohol at `25^(@)C` is 100 torrA. Alcohol liquid and vapour at 100 torrB. Alcohol liquid at 100 torrC. Alcohol liquid at 200 torrD. Alcohol vapour at 200 torr |
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Answer» Correct Answer - A At `25^(@)C` both alcohol liquid as well as vapour will exist. |
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| 1283. |
Equal volume of `0.1 M` urea and `0.1M` glucose are mixed. The mixture will haveA. same osmotic pressureB. lower osmotic pressureC. higher osmotic pressureD. none of these |
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Answer» Correct Answer - A Let volume of each solution mixed `= VL` `M_("Total") = (1M xx VL)/(2VL) +(1M xx VL)/(2VL) = 1M` As molarity of solution is not changed, there is no change in osmotic pressure. |
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| 1284. |
A solution has `1:4` mole ratio of pentane to hexane . The vapour pressure of pure hydrocarbons at `20^@C`are `440` mmHgfor pentane and `120`mmHg for hexane .The mole fraction of pentane at vapour phase would beA. `0.786`B. `0.549`C. `0.478`D. `0.200` |
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Answer» Correct Answer - C `P_(M)=P_(C_(5)H_(12))^@.X_(C_(5)H_(14))^@+P_(C_(6)H_(14))^@.X_(C_(5)H_(14))` Thus `P_(M)=440xx(1)/(40)+120xx(4)/(5)=184` lt brgt Now, `P_(C_(5)H_(12))=P_(C_(5)H_(12))^0. X_(C_(5)H_(14)(l))=P_(M).X_(C_(5)H_(12)(g))` `:.440xx(1)/(5)=184xxX_(C_(5)H_(12)(g))` `:. X_(C_(5)H_(12)(g))= 0.478` |
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| 1285. |
A solution contains non-volatile solute of molecular mass`M_(2)`which of the following can be used to calculate the molecular mass of solute in terms of osmotic pressure? (`m_(2)` =mass of solute,V=volume of solution,`pi` =osmotic pressure)A. `M_(2)=((m_(2))/(pi))VRT`B. `M_(2)=((m_(2))/(V))(RT)/(pi)`C. `M_(2)=((m_(2))/(V))piRT`D. `M_(2)=((m_(2))/(V))(pi)/(RT)` |
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Answer» Correct Answer - B We know ,`pi=CRT` `!= =(n)/(V)RT,piV=(m_(2))/(M_(2))RT` |
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| 1286. |
Assertion: Boiling point of rectified spirit is `78.15^(@)C`. Reason: Ethyl alcohol and water form non ideal solution with negative deviation.A. Both A and R are true and R is the correct explanation of AB. Both A and R are true but R is not a correct explanation of AC. A is true but R is falseD. A is false but R is true |
| Answer» Correct Answer - C | |
| 1287. |
Lowering of vapour pressure due to a solute in `1` molal aqueous solution at `100^(@)C` is a.`13.44 mm Hg` ,b. `14.14 mm Hg` ,c.`13.2 mm Hg` ,d. `35.2 mm Hg`A. 13.44 mm HgB. 14.12 mm HgC. 31.2 mm HgD. 35.2 mm Hg |
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Answer» Correct Answer - A `p = p_(A) + p_(B)` `p = p_(A)^(@) x_(A) + p_(B)^(@) x_(B)` `= 120 xx (2)/(4) + 80 xx (2)/(4)` `= 60 + 40 = 100` torr `y_(B) =` mole fraction of B in the vapour phase `= (p_(B))/(p_("total")) = (40)/(100) = (2)/(5)` |
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| 1288. |
At `120^@C`, the vapour pressure of pure cholorobenzene `(C_(6)H(5)Cl)`is `0.736`atm. What is the vapour pressure of a solution of `5.00`g of naphthalene `(C_(10)H_(8)`in `50.0`g of chlorobenzene? (assume that naphthalene is not volatile)A. `0.736`atmB. `0.091`atmC. `0.677`atmD. `1.00`atm |
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Answer» Correct Answer - C `(0.736-P_(S))/(0.736)=((5)/(152))/((5)/(152)+(50)/(112.5))` on solving `P_(S)=0.677` atm |
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| 1289. |
Lowering of vapour pressure due to a solute in `1` molal aqueous solution at `100^(@)C` is a.`13.44 mm Hg` ,b. `14.14 mm Hg` ,c.`13.2 mm Hg` ,d. `35.2 mm Hg`A. `13.44`mmHgB. `14.12`mmHgC. `31.2`mmHgD. `35.2`mmHg |
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Answer» Correct Answer - A `m=(x_(B)xx1000)/(1-x_(B)m_(A))x_(B)` =mole fraction of solute `1=(x_(B)xx1000)/((1-x_(B))xx18)m_(A)` =mole mass of solvent `x_(B)=0.0176` `x_(A)=1-0.0176=0.9824` `p=p^(@)x_(A)` `760xx0.9824=746.62` `Deltap=P^(@)-p=760-746.62=13.4` |
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| 1290. |
Lowering of vapour pressure due to a solute in `1` molal aqueous solution at `100^(@)C` is a.`13.44 mm Hg` ,b. `14.14 mm Hg` ,c.`13.2 mm Hg` ,d. `35.2 mm Hg` |
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Answer» a. `Mw_(B) = (chi_(B)xx1000)/((1 - chi_(B))Mw_(A))` `{:[(chi_(B)= "mole fraction of solute"),(Mw_(A)="molar mass of solvent")]:}` `1= (chi_(B)xx1000)/((1 - chi_(B))xx18)` `chi_(B) = 0.0176` `chi_(A) = 1 - 0.0176 = 0.9824` `P=P^(@)chi_(A) = 760 xx 0.9824 = 746.62` `DeltaP = P^(@) -P = 760 - 746.62 ~~13.41` |
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| 1291. |
A solution of an organic compound is prepared by dissolving `30 g` in `100 g` water. Calculate the molecular mass of compound and the osmotic pressure of solution at `300 K`, when the elevation in boiling point is `0.52` and `K_(b)` for water is `0.52 K m^(-1)`. |
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Answer» The given values are `W_(B)=30 g`,`W_(A)=100 g`,`T=300 K ` `DeltaT_(b)=0.52 K m^(-1)` Now, using formula `Mw_(B)=(K_(f)xx1000xxW_(B))/(W_(A)xxDeltaT_(b))` `=(0.52xx1000xx30)/(100xx0.52)=300 g` For osmotic pressure, `pi=(piRT)/V= W_(B)/Mw_(B)xx(RxxT)/V=30/300xx(0.082xx300)/(100//1000)` =2.45 atm |
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| 1292. |
Which of the following solutions `(1 molal)` will have the maximum freezing point, assuming equal ionization in each case?A. `[Fe(H_(2)O)_(6)]Cl_(3)`B. `[Fe(H_(2)O)_(5)Cl]Cl_(2).H_(2)O`C. `[Fe(H_(2)O)_(4)Cl_(2)]Cl.2H_(2)O`D. `[Fe(H_(2)O)_(3)Cl_(3)].3H_(2)O` |
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Answer» Correct Answer - D Larger the value of (i), Smaller the freezing point (d) is a case of non-electrolytes, `i=1` |
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| 1293. |
The lowering of vapour pressure due to a solute in a `1 m` aqueous solution at `100^(@)C` isA. 13.44 torrB. 14.12 torrC. 312 torrD. 352 torr |
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Answer» Correct Answer - A `(DeltaP)/P^(@) =(W_(2) xx Mw_(2))/(Mw_(2) xx W_(1))` `DeltaP=(1 xx 18)/(1000) xx 760 =13.68` (at `100^(@)C`,`VP=760 "torr"`) |
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| 1294. |
pH of `0.1` M monobasic acid is found to be 2 . Hence its osmotic pressure at a given temp. T K is :A. `0.1 RT`B. `0.11 RT`C. `1.1 RT`D. `0.01 RT` |
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Answer» Correct Answer - A `pH=2` `[H^(o+)]=0.01 M =Cx=0.1x` x=0.1 `i=1 + x=1.1` `pi=i(n/V)RT=iMRT = 1.1 xx 0.1RT =0.11 RT` |
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| 1295. |
A non-volatile electrolyte dissolved in an aqueous solution in same molal proportion as non-eletrolyte producesA. same coligative effectB. higher colligative effectC. lower colligative effectD. no colligative effect |
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Answer» Correct Answer - B Electrolyte generally dissociates in solution and thus, increases the concentration of solute particles. |
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| 1296. |
Lowering of vapour pressure of an aqueous solution of a non-volatile non-electrolyte 1 m aqueous solution at `100^(@)C` is,A. 14.12 TorrB. 312 TorrC. 13.45 TorrD. 352 Torr |
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Answer» Correct Answer - C `p^(@) = 760` Torr A 1m aqueous solution contains 1 mol of solute in 100g (55.5 mol) of water `:. X_(B) = (1)/(1+55.5) = (1)/(56.5)` `(Deltap)/(p) = x_(B)` `Deltap = px_(B)` `=(760 "Torr")/(56.5) = 13.45` Torr |
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| 1297. |
Osmotic pressure of blood is 7.40 atm at `27^(@)C`. Number of mol of glucose to be used per `L` for an intravenous injection that is to have the same osmotic pressure as blood is :A. `0.3`B. `0.2`C. `0.1`D. `0.4` |
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Answer» Correct Answer - A `pi=CRT 7.40=nxx0.0821xx300` `pi=(n)/(v)Rt n=(7.4)/(0.0821xx300)=0.3` |
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| 1298. |
A solution of urea in water has boiling point of `100.15^(@)C`. Calculate the freezing point of the same solution if `K_(f)` and `K_(b)` for water are `1.87 K kg mol^(-1)` and `0.52 K kg mol^(-1)`, respectively. |
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Answer» `DeltaT_(b)=(100.15-100)=0.15^(@)C` We know that, `DeltaT_(b)=` molality `xx K_(b)` Molality `= (DeltaT_(b))/(K_(b))=(0.15)/(0.52)=0.2884` `DeltaT_(f)=` molality `xxK_(f)` `=0.02884xx1.87=0.54^(@)C` Thus, the freezing point of the solution `=-0.54^(@)C`. |
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| 1299. |
Some of the following gases are soluble in water due to formation of their ions : `I:CO_(2), II:NH_(3), III:HCL, IV:CH_(4), V:H_(2)` Water insoluble gases can be:A. I,IV,VB. I,VC. I,II,IIID. IV,V |
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Answer» Correct Answer - D `CO_(2)+H_(2)OrarrH_(2)CO_(3)rarr2H^(+)+CO_(3)^(-2)` `NH_(3)+H_(2)OrarrNH_(4)OHrarrNH_(4)^(+)+OH^(-)` `HCl+H_(2)OrarrH_(3)O^(+)+Cl^(-)` But `CH_(4)` and `H_(2)` are insoluble gases in water. |
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| 1300. |
Which has maximum osmotic pressure at temperature `T`? a. `100 mL` of `1 M` urea solution b. `300 mL` of `1 M` glucose solution c. Misture of `100 mL` of `1 M` urea solution and `300 mL` of `1 M` glucose solution d. All are isotonic |
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Answer» d. All are isotonic Osmotic pressure `(pi)` depends upon the concentration of solution, i.e., `pi=C_("effective")RT` For (a), `C_("effective")= 1 M` (since it is non-electrolytic solution) For (b), `C_("effective")= 1 M` (since it is non-electrolytic solution) For c, `C_("effective")= [100/(100 +300) xx 1 +(300)/(100 +300) xx1] =1 M` |
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