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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.
| 351. |
A plant cell shrinks when it is kept inA. hypotonic solutionB. a hypertonic solutionC. a solution is isotonic with cell sapD. water. |
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Answer» Correct Answer - B A hypertonic solution has higher concentration. |
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| 352. |
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? Here `n_(2)=` mass of solute, V = volume of solution, `pi =` osmotic pressure.A. `Mp=((m)/(pi))VRT`B. `Mp=((m)/(V))(RT)/(pi)`C. `Mp=((m)/(V))(pi)/(RT)`D. `Mp=((m)/(V))pi RT` |
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Answer» Correct Answer - B `pi=(n)/(V)RT rArr M_(P)=((m)/(V))(RT)/(pi)` |
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| 353. |
When a crystal of the solute is introduced into a super saturated solution of the soluteA. the solute dissolvesB. the excess solute crystallises outC. the solution becomes unsaturatedD. the solution remains super saturated |
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Answer» Correct Answer - B Super saturated state is metastable state. |
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| 354. |
When a cystal of the solute is introduced into a super saturated solution of the soluteA. The solute dissolvesB. The excess solutes crystallizes outC. The solution becomes unsaturatedD. The solution remains super saturated |
| Answer» Correct Answer - B | |
| 355. |
When the solute is present in trace quantities the following expression is usedA. Gram per millionB. Milligram percentC. Microgram percentD. Parts per million |
| Answer» Correct Answer - D | |
| 356. |
The distribution law is applied for the distribution of basic acid betweenA. Water and ethyl alcoholB. Water and amyl alcoholC. Water and sulphuric acidD. Water and liquor ammonia |
| Answer» Correct Answer - B | |
| 357. |
Molar solution means 1 mole of solute present inA. 1000 g of solventB. 1 litre of solventC. 1 litre of solutionD. 1000 g of solution |
| Answer» Correct Answer - C | |
| 358. |
Which of the following is an extensive property?A. Molar volumeB. MolarityC. Number of molesD. Mole fraction |
| Answer» Correct Answer - A | |
| 359. |
The solution in which the blood cells remain their normal shape, with regard to the blood, areA. IsotonicB. IsomoticC. HypertonicD. Equinormal |
| Answer» Correct Answer - A | |
| 360. |
Normal level of haemoglobin per 100 ml of blood in a woman isA. 8 mgB. 80 mgC. 200 mgD. 800 mg |
| Answer» Correct Answer - B | |
| 361. |
Density of `2.05 M` solution of acetic acid in water is `1.02g//mL`. The molality of same solution is:A. `1.14 mol kg^(-1)`B. `3.28 mol kg^(-1)`C. `2.28 mol kg^(-1)`D. `0.44 mol kg^(-1)` |
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Answer» Correct Answer - C Mass of 1 litre solution of 2.05 M acetic acid = 1020 g Mass of acetic acid present in it = `2.05xx60=123 g` Mass of solvent (water) in this solution `= 1020-123 =897 g` `therefore` Molality of the solution `=(2.05)/(897)xx1000=2.28 mol kg^(-1)` |
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| 362. |
0.5 molal aqueous solution of a weak acid (HX) is 20% ionised. If `K_(f)` for water is `1.86K kg mol^(-1)`, the lowering in freezing point of the solution isA. `-1.12 K`B. 0.56 KC. 1.12 KD. `0.56 K` |
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Answer» Correct Answer - C `alpha=(i-1)/(n-1)rArr 0.2=(i-1)/(2-1)rArr i=1.2` `rArr Delta T=ixxK_(f)xx m` |
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| 363. |
Calculate the mass of urea `(NH_(2)CONH_(2))` required in making `2.5 kg `of `0.25 `molal aqueous solution.A. 37 gB. 35 gC. 34 gD. 32 g |
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Answer» Correct Answer - a Mass of solvent = 1000 g Molar mass of urea `(NH_(2)CONH_(2)) = 60 g mol^(-1)` `0.25` mole of urea `= 0.25xx60 = 15g` Total mass of solution `= 100 + 15 = 1.015 kg` 1.015 kg of solution contain urea `= 15 g` 2.5 kg of solution `= (15)/(1.015)xx2.5 = 37 g` |
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| 364. |
The boiling point of water at `750mm Hg` is `99.63^(@)C`. How much sucrose is to be added to `500 g` of water such that it boils at `100^(@)C`. |
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Answer» Given `DeltaT_(b)=100-96.63=3.37^(@)` Mass of water `w_(1)=500g` Molar mass of water `M_(1)=18" g "mol^(-1)` Molar mass of sucrose, `M_(2)=342" g "mol^(-1)` To find: Mass of sucrose `w_(2)=?` Solution we know, `DeltaT_(b)=K_(b)xxm` `=K_(b)xx(w_(2))/(M_(2))xx(1000)/(w_(1))` `impliesw_(2)=(M_(2)xxw_(1)xxDeltaT_(b))/(1000xxK_(b))=(342xx500xx3.37)/(1000xx0.52)` `w_(2)=1108.2g` `therefore` mass of solute `w_(2)=1.11` kg |
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| 365. |
Calculate the `(a)` molality, `(b)` molarity, and `(c)` mole fraction of `KI` if the density of `20% (` mass `//` mass `)` aqueous `KI` is `1.202 g m L^(-1)`. |
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Answer» 20% aq. KOH solution `implies20g` of KI in 100g solution `therefore` Mass of solvent `=100-20=80g` (i). Molality `=("no. of moles of KI")/("mass of solvent (kg)")` `=(0.120)/(0.080)=1.5" mol "kg^(-1)` (ii). Density of solution `=1.202" g "mL^(-1)` Volume of solution `=(100)/(1.202)=83.2mL` `=0.0832L` `therefore`Molarity `=(0.120)/(0.0832)=1.44M` (iii). No. of moles of KI `=0.120` `n_(H_(2)O)=(80)/(18)=4.44` `x_(KI)=(0.120)/(0.120+4.44)` `=(0.120)/(4.560)=0.0263`. |
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| 366. |
Two solutions A and B are separated by semi-permeable membrane. If liquid flows A to B thenA. A is less concentrated than BB. A is more concentrated than BC. Both have same concentrationD. None of these |
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Answer» Correct Answer - A Osmosis occur from dilute solution to concentrate solution. Therefore solution A is less concentrated than B. |
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| 367. |
A 5.25% solution of a substance is isotonic with a 1.5% solution of urea (molar mass = 60 g `mol^(-1)`) in the same solvent. If the densities of both solutions are assumed to be equal to `1.0 g cm^(-3)`, molar mass of the substance will beA. `90.0 g mol^(-1)`B. `115.0 g mol^(-1)`C. `1050.0 g mol^(-1)`D. `210.0 g mol^(-1)` |
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Answer» Correct Answer - D Solutions with the same osmotic pressure are isotonic Let the molar mass of the substances be M `pi = C_(1)RT=C_(2)RT=pi_(2)` So, `C_(1)=C_(2)` As density of the solutions are same So, `(5.25)/(M)=(15)/(60)rArr M =(5.25xx60)/(1.5)=210` |
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| 368. |
Osmotic pressure of a urea solution at `10^(@)C` is 500mm. Osmotic pressure of the solution become 105.3 mm. When it is diluted and temperature raised to `25^(@)C`. The extent of dilution isA. 6 TimesB. 5 TimesC. 7 TimesD. 4 Times |
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Answer» Correct Answer - B `pi V = nRT` `(500 V_(1))/(105.3V_(2))=(nRxx283)/(nR xx298), (V_(1))/(V_(2))=(1)/(5)` so `V_(2)=5V_(1)` |
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| 369. |
Molarity of `H_(2)SO_(4)` is `18 M`. Its density is `1.8 g//cm^(3)`, hence molality is :A. 36B. 200C. 500D. 18 |
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Answer» Correct Answer - C Molality (m) = `("Molarity")/("Density"-("Molarity " xx " Molecular mass")/(1000))` `=(18)/(1.8-(18xx98)/(1000))=500`. |
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| 370. |
On dissolving 1 mole of each of the following acids in 2 litre water, the which does not give a solution of strength 1 N isA. HClB. Perchloric acidC. `HNO_(3)`D. Phosphoric acid |
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Answer» Correct Answer - D `H_(3)PO_(4)to H^(+)+H_(2)PO_(4)^(-)` `H_(2)PO_(4)^(-)to H^(+)+HPO_(4)^(2-)` `HPO_(4)^(2-)to H^(+)+PO_(4)^(3-)` Phosphoric acid does not give 1N strength. |
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| 371. |
Find the molality fo `H_(2)SO_(4)` solution whose specific gravity is `1.98 g mL^(-1)` and 98% (Weight/volume) `H_(2)SO_(4)`.A. 2 NB. 19.8 NC. 39.6 ND. 98 N |
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Answer» Correct Answer - C Strength of `H_(2)SO_(4) = 98xx19.8` g/litre S = eq. wt. `xx N , N=(S)/(eq. wt.)=(98xx19.8)/(49)=39.6` |
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| 372. |
Calculate the mass percentage of aspirin `(C_(9)H_(8)O_(4))` in acetonitrile `(CH_(3)CN)` when `6.5g` of `C_(9)H_(8)O_(4)` is dissolved in `450g` of `CH_(3)CN`. |
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Answer» Mass percentage of aspirin `("Mass of aspirin")/("Mass of aspirin+ Mass of acetonitrile")xx100` `=(6.5)/(6.5+450)xx100=1.424%` |
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| 373. |
Concentrated aqueous solution of sulphuric acid is `98 %` by mass and has density of `1.80 "g mL"^(-1)`. What is the volume of acid required to make one liter `0.1 M H_(2)SO_(4)` solution ?A. 11.10 mLB. 16.65 mLC. 22.20 mLD. 5.55 mL |
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Answer» Correct Answer - D `H_(2)SO_(4)` si 98% by weight wt. of `H_(2)SO_(4) = 98 gm` wt. of solution `= (100)/(180)mL=(100)/(180xx1000)` litre `therefore M_(H_(2)SO_(4))=(98)/(98xx(100)/(1.80xx1000))=18.014` Let V mL of this `H_(2)SO_(4)` are used to prepare 1 litre of 0.1 `M H_(2)SO_(4)` solution. `M_(1)V_(1)=M_(2)V_(2)rArr Vxx18.014=1000xx0.1 , V=5.55 mL` |
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| 374. |
Iodine is more soluble in alcohol than in carbon tetrachloride becauseA. iodine and alcohol both are non-polarB. randomness factor is greater in alcohol than in `C Cl_(4)`C. dissolution of iodine in alcohol is exothermic whereas it is endothermicD. dissolution of both is endothermic but heat of dissolution in alcohol is less than in `C CI_(4)` |
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Answer» Correct Answer - D in edothermic process of dissolution, energy factor opposes the dissolution.Less is the heat of dissolution, i.e.,less is the opposing factor,greater is the solubility. |
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| 375. |
Suger is soluble in water due toA. High solvation energyB. Lonic character of sugarC. High dipole moment of waterD. Hydrogen bond formation with water. |
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Answer» Correct Answer - D _OH groups present in sugar from H-bonds with `H_(2)O.` |
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| 376. |
Solubility of a gas in a liquid solvent increases withA. increase of pressure and increase of temperatureB. decrease of pressure and increase of temperatureC. increase of pressure and decrease of temperatureD. decrease of pressure and decrease of temperature |
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Answer» Correct Answer - C Solubility of a gas in a liquid increases with increase of pressure and decrease of temperature. |
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| 377. |
V.P. of pure solvent water is …. Than 2 M `CuSO_(4)` solution.A. lowerB. higherC. sameD. can not be said |
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Answer» Correct Answer - B Pure liquid water has maximum vapour pressure. |
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| 378. |
Solubility curve of `Na_(2)SO_(4)`. `10H_(2)O` in water with temperature is given as A. Solution process is exothermicB. Solution proscess is exothermic til `34^(@)` C and endothermic after `34^(@)`C. Solution process is endothermic till `34^(@)`C and exothermic thereafterD. Solution process is endothermic. |
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Answer» Correct Answer - C Solubility increases with temp. upto `34^(@)`C and then decreases.So the process is endothermic upto `34^(@)` C and exothermic thereafter. |
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| 379. |
`CuSo_(4)`.` 5 H_(2)O` is aA. solution of a solid in a liquid.B. solution of liquid in a solidC. salt only and cannot be called a solutionD. coordination compound of copper with water molecules as the ligands. |
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Answer» Correct Answer - B Liquid `H_(2)O` is uniformly mixed into solid `CuSo_(4)`. |
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| 380. |
Two solutions A and B have same mole fractions of the solute. If 1 `dm^(3)` of A is mixed with `2 dm^(3)` of B, the mole fraction of the solute in the mixture wouldA. decreaseB. increaseC. ramain unchangedD. change |
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Answer» Correct Answer - C As the two solutions have same mole fractions the ratio of moles of solute to total moles (i.e.mole fraction) would remain same even after mixing. |
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| 381. |
A solution containing 10 mole of ethylene glycol dissolved in 1000 g of water (`K_(f) = 1.86 K "molality^(-1)` will freeze at:A. 273 KB. 2.544 KC. 254.4 KD. 25.44 K |
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Answer» Correct Answer - C `Delta T_(f)= "molality" xx K_(f) = 10 xx 1.86 =18.6` `Delta T_(f) = T^(@) - T rArr T=0 - 18.6` `=-18.6^(@) + 273 K = 254.4 K` |
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| 382. |
25.3 g of sodium carbonate, `Na_(2)CO_(3)` is dissolved in enough water to make 250 mL of solution. If sodium carbonate dissociates completely, molar concentration of sodium ions, `Na^(+)` and carbonate ions, `CO_(3)^(2-)` are respectively (Molar mass of `NaCO_(3)=106g mol^(-1)`)A. 0.477 M and 0.477 MB. 0.955 M and 1.910 MC. 1.910 M and 0.955 MD. 1.90 M and 1.910 M |
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Answer» Correct Answer - C Molarity `(M)=(wt)/(mol.wt.)(1000)/("vol(in ml)")` `=(25.3)/(106)xx(1000)/(250)=0.955` mol/L of `Na_(2)CO_(3)` and `Na_(2)CO_(3)rarr 2Na^(+)+CO_(3)^(-2)` therefore `[Na^(+)]=2xx0.955=1.910 M` `[CO_(3)^(-2)]=0.955 M` |
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| 383. |
The solutions containing 6 g of urea (molecular mass =60) per `dm_(3)` of water and another containing 9 g of solute A per `dm^(3)` of water freeze at the same temp. The molecular mass of A isA. 12B. 90C. 54D. 150 |
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Answer» Correct Answer - B Molality of urea solution =`6/(60 xx 1kg)=0.1M` "As freezing point of both the solution are same they must have same molalities". "Hence for solute A", `"Molecular mass"="mass"/("molality" xx 1 kg)=9/(0.1 xx 1 )=90` |
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| 384. |
Equal volumes of 10% solution (by wt) of the solute A and 15 % solution (by wt) of the solute B are mixed. The mass percent of A and B in the mixture would be respectivelyA. 5 and 7.5B. 10 and 25C. 5 and 10D. 20 and 30 |
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Answer» Correct Answer - A As equal volume of two solution are mixed the final volume i.e. the volume of mixture would be doubled. Hence each % will become half. |
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| 385. |
The molarity of a 0.2 N `Na_(2)CO_(3)` solution will be :A. 0.05 MB. 0.2 MC. 0.1 MD. 0.4 M |
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Answer» Correct Answer - C `M=(N)/(2)=(0.2)/(2)=0.1M` |
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| 386. |
Some properties as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogeneous solution. These are called colligative properties. Applications of colligative properties are useful in day-to day life. Its example is the use of ethylene glycol and water mixture as anti-freezing liquid in the radiator of automobiles. A solution M is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is 0.9. Given : Freezing point depression constant of water `(K_(f)^("water"))=1.86" K kg mol"^(-1)` Freezing point depression constant of ethanol `(K_(f)^("ethanol"))=2.0"K kg mol"^(-1)` Boiling point elevation of water `(K_(b)^("water"))=0.52" K kg mol"^(-1)` Boiling point elevation constant of ethanol `(K_(b)^("ethanol"))=1.2" K kg mol"^(-1)` Standard freezing point of water = 273 K Standard freezing point of ethanol = 155.7 K Standard boiling point of water = 373 K Standard boiling point of ethanol = 251.5 K Vapour pressure of pure water = 32.8 mm Hg Vapour pressure of pure ethanol = 40 mm Hg Molecular weight of water `= 18 g mol^(-1)` Molecular weight of ethanol `= 46 g mol^(-1)` In answering the following questions, consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative. Water is added to the solution M such that the mole fraction of water in the solution becomes 0.9. The boiling point of this solution isA. 380.4 KB. 376.2 KC. 375.5 KD. 354.7 K |
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Answer» Correct Answer - B `{:("Solute ethanol","Solvent " H_(2)O),(0.1,0.9),(4.6 gm,16.2 gm):}` `Delta T_(b)=0.52((4.6xx1000)/(46xx16.2))` `Delta T_(b)=3.20` `T_(b)=3.20+373=376.20 K` |
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| 387. |
An aqueous solution of urea containing 18 g urea in 1500 `cm^(3)` of solution has a density of 1.5 `g//cm^(3)` . If the molecular weight of urea is 60. Then the molality of solution is:A. `0.200`B. `0.192`C. `0.100`D. `1.200` |
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Answer» Correct Answer - B `M=V xx D=1500 xx 1.052g =1578g.` Solvent =1560 g, solute = 18g `"Molality"=("mass" xx 1000)/"Molar mass" xx "Mass of solvent is g"` `18 xx 1000/(60 xx (1578-18))=(19 xx 1000)/(60 xx 1560)=0.192` |
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| 388. |
The wt. percent of `NO_(3)^(-)` in the solute in a solution is 20. The volume of solvent containing 60 g of solute (d = 1.2 g/cc)A. `0.24 dm^(3)`B. `0.12 dm^(3)`C. `1.2 dm^(3)`D. `0.2 dm^(3)` |
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Answer» Correct Answer - D Wt. of solvent = Wt. of solution- Wt. of solute `=100 - 20 =80g` 20 g of solute is present in 80 g of solvent. 60 g of solute will be present in `(80 xx 60)/20=240 g `:."Volume of solvent" = (Wt)/"density"=240/1.2=200 cm ^(3)` = `0.2 dm^(3)` |
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| 389. |
`N_(2)` exerts a pressure of 0.987 bar. The mole fraction of `N_(2)` ia `(K_(H)=74.48 k `bar)A. `7.648 xx 10^(-3)`B. `9.87 xx 10^(-5)`C. `1.3 xx 10^(-5)`D. `2.6 xx 10^(-5)` |
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Answer» Correct Answer - C `P_(N2)=K_(H) xx X_(N2)` `:. X_(N2)=P_(N2)/K_(H)=(0.987"barr")/(76480"bar")=1.29 xx 10^(-5)` |
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| 390. |
A sugar syrup of weight `214.2 g` contains `34.2 g` of sugar `(C_(12) H_(22) O_(11))`. Calculate a. the molal concentration. b. the mole fraction of the sugar in the syrup.A. 0.55B. 5.5C. 55D. 0.1. |
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Answer» Correct Answer - A Water present = 214.2-342=180g Sugar =34.2 g `m=(W_(2) xx 1000)/(M_(2) xx W_(1)(g))=(34.2 xx 1000)/(342 xx 180)=0.55` |
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| 391. |
The normality of 2.3 M `H_(2)SO_(4)` solution isA. 2.3 NB. 4.6 NC. 0.46 ND. 0.23 N |
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Answer» Correct Answer - B Normality of `2.3 M H_(2)SO_(4)=M xx` Valency `=2.3xx2=4.6 N` |
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| 392. |
When `7.1gm Na_(2)SO_(4)` (molecular mass 142) dissolve in `100 ml H_(2)O`, the molarity of the solution isA. 2.0 MB. 1.0 MC. 0.5 MD. 0.05 M |
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Answer» Correct Answer - C Molarity `=(w xx 1000)/(ml wt. xx "Volume ml.")=(7.1xx1000)/(142xx100)=0.5 M`. |
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| 393. |
When we add a non-volatile solute to a solvent lowers its vapour pressure. Therefore, the vapour pressure of a solution is lower than that of pure solvent, at the same temperature. A higher temperature is needed to raise the vapour pressure upto one atmosphere pressure, when boiling point is attained. However, b.pt. increases small. For example, 0.1 molal aqueous sucrose solution boils at `100.05^(@)C`. Sea water, which is rich in `Na^(+)` and `Cl^(-)` ions, freezes about `1^(@)C` lower than freezes pure water. At the freezing point of a pure-solvent, the rates at which two molecule stick together to form the solute is present. Few solvent molecules are in constant with surface of solid. However, the rate at which the solvent molecules leave the surface of solid remains unchanged. That is why temperature is lowered to restore the equilibrium. The depression is lowered to restore the equilibrium. The depression in freezing point in a dilute solution is proportional to molality of the solute. An aqueous solution of 0.1 molal concentration of sucrose should have freezing point `(K_(f)=1.86 K mol^(-1)kg)`A. `0.186^(@)C`B. `1.86^(@)C`C. `-1.86^(@)C`D. `-0.186^(@)C` |
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Answer» Correct Answer - D `Delta T_(f)=K_(f)m=1.86xx0.1=0.186`, Freezing point `=0-Delta T_(f)=-0.186^(@)C` |
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| 394. |
In a solution of 7.8 g benzene `(C_(6)H_(6))` and 46.0 g toluene `(C_(6)H_(5)CH_(3))` the mole fraction of benzene is :A. `1//6`B. `1//5`C. `1//2`D. `1//3` |
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Answer» Correct Answer - A Mole fraction of `C_(6)H_(6)=(7.8//78)/(7.8/78+46/92)=1/6` |
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| 395. |
The mole of fraction of nitrogen, in a mixture of 7 g of `N_(2)` and 16 g of `O_(2)` isA. 0.5B. 0.75C. 0.66D. 0.33 |
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Answer» Correct Answer - D Moles of `N_(2) = 7/28=0.25` and Moles of `O_(2) = 16/32 = 0.5` Mole fraction of `N_(2) = 0.25/(0.25 + 0.5)=0.33` |
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| 396. |
If we take 44g of `CO_(2)` and 14g of `N_(2)` what will be mole fraction of `CO_(2)` in the mixtureA. `1//5`B. `1//3`C. `2//3`D. `1//4` |
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Answer» Correct Answer - C Mole fraction of `CO_(2)=(n_(CO_(2)))/(n_(CO_(2))+n_(N_(2)))=((44)/(44))/((44)/(44)+(14)/(28))=(2)/(3)`. |
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| 397. |
`2N HCI` solution will have the same molar concentration as aA. `4.0N H_(2)SO_(4)`B. `0.5 N H_(2)SO_(4)`C. `1N H_(2)SO_(4)`D. `2N H_(2)SO_(4)` |
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Answer» Correct Answer - A From the relation `("Normality")/("Molarity")=("Molecular mass")/("Equivalent mass")=n` For 2 N HCl Molarity `=("Normality" xx "Equivalent weight")/("Molecular weight")` Molarity `=(2xx36.5)/(36.5)=2` For 4 N `H_(2)SO_(4)` Molarity `=("Normality" xx "Equivalent weight")/("Molecular weight")` Molarity `=(4xx49)/(98)=2` Hence 4 N `H_(2)SO_(4)` and 2 N HCl solution will have same molar concentration. |
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| 398. |
The molarity of 90% `H_(2)SO_(4)` solution is [density = 1.8 gm/ml]A. `1.8`B. `48.4`C. `91.83`D. `94.6` |
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Answer» Correct Answer - C The density of solution = 1.8 gm/ml Weight of one litre of solution = 1800 gm `therefore` Weight of `H_(2)SO_(4)` in the solution `=(1800xx90)/(100)=1620 gm` `therefore` Weight of solvent = 1800 - 1620 = 180 gm `therefore` Molality `=(1620)/(98)xx(1000)/(180)=91.83` |
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| 399. |
When we add a non-volatile solute to a solvent lowers its vapour pressure. Therefore, the vapour pressure of a solution is lower than that of pure solvent, at the same temperature. A higher temperature is needed to raise the vapour pressure upto one atmosphere pressure, when boiling point is attained. However, b.pt. increases small. For example, 0.1 molal aqueous sucrose solution boils at `100.05^(@)C`. Sea water, which is rich in `Na^(+)` and `Cl^(-)` ions, freezes about `1^(@)C` lower than freezes pure water. At the freezing point of a pure-solvent, the rates at which two molecule stick together to form the solute is present. Few solvent molecules are in constant with surface of solid. However, the rate at which the solvent molecules leave the surface of solid remains unchanged. That is why temperature is lowered to restore the equilibrium. The depression is lowered to restore the equilibrium. The depression in freezing point in a dilute solution is proportional to molality of the solute. When 250 mg of eugenol is added to 100 g of camphor `(k_(f)=39.7 "K molality"^(-1))`, it lowered the freezing point by `0.62^(@)C`. The molar mass of eugenol isA. `1.6xx10^(2)g//mol`B. `1.6xx10^(4)g//mol`C. `1.6xx10^(3)g//mol`D. 200 g/mol |
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Answer» Correct Answer - A `0.62=(250xx10^(-3))/(M xx100)xx1000xx39.7xx1` M = 160 or `1.6xx10^(2)` g/mol |
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| 400. |
In a solution of 7.8 gm benzene `C_(6)H_(6)` and 46.0 gm toluene `(C_(6)H_(5)CH_(3))`, the mole fraction of benzene in this solution isA. `1//6`B. `1//5`C. `1//2`D. `1//3` |
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Answer» Correct Answer - A Mole fraction at `C_(6)H_(6)=((7.8)/(78))/((7.8)/(78)+(46)/(92))=(1)/(6)`. |
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