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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.
| 251. |
The two isomers X and Y with the formul `Cr(H_(2)O_(5))ClBr_(2)` were taken for experiment on depression in freezing point. It was found that one mole of X gave depression corresponding to 2 moles of particles and one mole of Y gave depression due to 3 moles of particles. The structural formulae of X and Y respectively areA. `[Cr(H_(2)O)_(5)Cl]Br_(2) , [Cr(H_(2)O)_(4)Br_(2)]Cl.H_(2)O`B. `[Cr(H_(2)O)_(5)Cl]Br_(2) , [Cr(H_(2)O)_(3)ClBr_(2)].2H_(2)O`C. `[Cr(H_(2)O)_(5)Br]BrCl , [Cr(H_(2)O)_(4)ClBr]Br.H_(2)O`D. `[Cr(H_(2)O)_(4)Br_(2)]Cl.H_(2)O, [Cr(H_(2)O)_(5)Cl]Br_(2)` |
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Answer» Correct Answer - D Depression in freezing point is a colligative property that depends upon the number of moles of ions which a comples may give in solution. X gave depression corresponding to 2 moles of particles. It means 2 mol of ions of a complex is present in solution, which is only possible when comples is in the form `[Cr(H_(2)O)_(4)Br_(2)]Cl. H_(2)O` or `[Cr(H_(2)O)_(4)Br_(2)]^(+)Cl^(-)`. Similarly, Y complex will be `[Cr(H_(2)O)_(5)Cl]Br_(2)` or `[Cr(H_(2)O)_(5)Cl]^(+)+2Br^(+)`. |
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| 252. |
To prepare a solution of concentration of 0.03 g/ml. of `AgNO_(3)`. What amount of `AgNO_(3)` should be added in 60mL of solution?A. `1.8`B. `0.8`C. `0.18`D. None of these |
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Answer» Correct Answer - A Amount of `AgNO_(3)` added in 60 ml of solution `=60xx0.03=1.8g` |
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| 253. |
If `5.85`g of `NaCl`(molecular weight`58.5)`is dissolved in water and the solution is made up to `0.5`litre , the molarity of the solution will beA. `0.2`B. `0.4`C. `1.0`D. `0.1` |
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Answer» Correct Answer - A `M=(w)/(m.wt. xx "Volume in litre")=(5.85)/(58.5xx0.5)=0.2M` |
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| 254. |
A solution containing 1.8 g of a compound (empirical formula `CH_(2)O`) in 40 g os water is observed to freeze at `-0.465^(@)C`. The molecular formula of the compound is `(K_(f) "of water" =1.86 kg K mol^(-1))`A. `C_(2)H_(4)O_(2)`B. `C_(3)H_(6)O_(3)`C. `C_(4)H_(8)O_(4)`D. `C_(6)H_(12)O_(6)` |
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Answer» Correct Answer - D `Delta T_(f)=K_(f)xxm` `0.465=1.86xx(1.8)/(M)xx(1000)/(40)rArr M =180` Molecular formula = `("empirical formula")_(n)` `n=("Molecular mass")/("Empirical formula mass")=(180)/(30)=6` Molecular formaula `= (CH_(2)O)_(6)=C_(6)H_(12)O_(6)`. |
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| 255. |
The osmotic pressure of 0.4% urea solution is 1.66 atm and that of a solution of sugar of 3.42% is 2.46 atm. When both the solution are mixed then the osmotic pressure of the resultant solution will beA. 1.64 atmB. 2.46 atmC. 2.06 atmD. 0.82 atm |
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Answer» Correct Answer - C `pi=(1.66+2.46)/(2)=2.06` atm |
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| 256. |
Mole fraction of glucose in 18% (wt./wt.) solution of glucose isA. 0.18B. 0.1C. 0.017D. 0.021 |
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Answer» Correct Answer - D Glucose =18 g, Solution =100g :. Water = 100-18=82g Mole freaction of glucose `=(18//180)/(18//180+82//18` `=0.1/(0.1+4.56)=0.1/4.66=0.021 |
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| 257. |
What is the normlity of a 1 M solution of `H_(3)PO_(4)`?A. 0.5 NB. 1.0 NC. 2.0 ND. 3.0 N |
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Answer» Correct Answer - D `H_(3)PO_(4)` is tribasic so `N = 3M = 3xx1=3`. |
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| 258. |
Normality of `2`M sulphuric acid isA. 2 NB. 4 NC. N/2D. N/4 |
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Answer» Correct Answer - B N = M `xx` basicity , `N=2xx2=4`. |
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| 259. |
A 500 g toothpaste sample has `0.2` g fluoride concentration. What is the concentration of `F^(Theta)` in ppm ?A. 250B. 200C. 400D. 1000 |
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Answer» Correct Answer - C `F^(-)` ions in PPM `=(0.2)/(500)xx10^(6)=400` |
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| 260. |
An aqueous solution of glucose is `10%` in strength ,The volume in which `1`g mole of it dissolved will beA. 18 litreB. 9 litreC. 0.9 litreD. 1.8 litre |
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Answer» Correct Answer - D 10% glucose solution means `10g=(10)/(180)`mole in 100 cc. i.e., 0.1 litre Hence, 1 mole will be present in `(0.1xx180)/(10)=1.8` litre. |
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| 261. |
An aqueous solution of glucose is `10%` in strength ,The volume in which `1`g mole of it dissolved will beA. 18 litresB. 9 litresC. 0.9 litresD. 1.8 litres. |
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Answer» Correct Answer - D 10% glucose sol means `10g = 10/180` mole in 100 cc equiv 0.1 L. Hence,1 mole will be present in `(0.1 xx 1 )/(10//180) =1.8L` |
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| 262. |
Which of the following stsrement is corrent about steel (S) and tungsten carbide (WC)?A. S is a substitutional solid while WC is an interstitial solid solitionB. S is an interstitial solid while WC is an substitutional solid solutionC. Both are interstitial solid solutionsD. Both are substitutional solid solutions |
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Answer» Correct Answer - B It is a fact. |
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| 263. |
Assertion : `Ca^(++)` and `K^(+)` ions are responsible for maintaining proper osmotic pressure balance in the cells of organism. Reason : Solutions having the same osmotic pressure at same temp are called isotonic solutions.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If asserrion is false but reason is true. |
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Answer» Correct Answer - D Sodium ion, `Na^(+)` and potassium ion, `K^(+)` are responsible for maintaining proper osmotic pressure balance inside and outside of the cells of organisms. |
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| 264. |
One gm equivalent of substance present in :-A. It shows molar concentrationB. It shows molal concentrationC. It shows normalityD. It shows strength gm/gm |
| Answer» Correct Answer - B | |
| 265. |
A mixture of ethane and ethene occupies 41 L at atm and 500 K. The mixture reacts compeletly with 10/3 mole of oxygen to produce `CO_(2)` and water. The mole fraction of ethane and ethene in the mixture are (R=0.0821L atm `K^(-1)mol^(-1)` respectivelyA. `0.50, 0.50`B. `0.75, 0.25`C. `0.67, 0.33`D. `0.25, 0.75` |
| Answer» Correct Answer - C | |
| 266. |
Ammonia undergoes self dissociation according to the reaction `2 NH_(3(l)) hArr NH_(2(am))^(+)+NH_(2(am))^(-)` where am, stands for ammoniated. When 1 mol of `NH_(4)` CI is dissolved in 1 kg of liquid ammonia, the b.p. at 760 torr is observed as `-32.7^(@)C` (normal boiling b.p. of `NH_(3(l))` is `33.4^(@)C`). What conclusion is reached about the nature of the solutionA. `NH_(4)` CI is completely dissociated in `NH_(3)`B. `NH_(4)` CI is partially dissociated in `NH_(3)`C. `NH_(4)`CI is not dissociated in `NH_(3)`D. Boiling point is not raised |
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Answer» Correct Answer - A `NH_(4)Cl` is a strong acid in liq. `NH_(3)`. Since, it ionizes to give `NH_(4)^(+)` (which is cation of the the solvent). `NH_(4)Cl hArr NH_(4)^(+)+Cl^(-)` |
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| 267. |
An aqueous solution of a weak monobasic acid containing 0.1 g in 21.7g water freezes at 272.813 K. If the value of `K_(f)` for water is 1.86 K-kg/mol, what is the molecular mass of the monobasic acidA. 50g/moleB. 46g/moleC. 55g/moleD. 60g/mole |
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Answer» Correct Answer - D `m=(K_(f)xx wxx1000)/(Delta T_(f)xx W)=60` g/mole. |
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| 268. |
As a result osmosis the volume of solutionA. IncreasesB. DecreasesC. Remains contantD. Increases or decreases |
| Answer» Correct Answer - A | |
| 269. |
What would happen if a thin slice of sugar beet is placed in a concentrated solution of NaClA. Sugar beet will lose water from its cellsB. Sugar beet will absorb water from solutionC. Sugar beet will neither absorb nor lose waterD. Sugar beet will dissolve in solution |
| Answer» Correct Answer - A | |
| 270. |
If a thin slice of sugar beet is placed in concentrated solution of `NaCl`, thenA. sugar beet will lose water from its cellsB. sugar beet will absorb water from solutionC. sugar beet will neither absorb nor lose waterD. sugar beet will dissolve in solution |
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Answer» Correct Answer - A Osmosis occurs from dilute solution to concentrated solution i.e. exo-osmosis. |
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| 271. |
Elevation of boiling point is directly proportional toA. molality of the solutionB. depression of freezing point in the same solutionC. both of theseD. none of these |
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Answer» Correct Answer - C It is a fact. |
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| 272. |
The empirical formula of a non-electrolyte is `CH_(2)O`. A solution containing 3 g `L^(-1)` of the compound exerts the same osmotic pressure as that of 0.05 M glucose solution. The molecular formula of the compound is :A. `C_(2)H_(4)O_(2)`B. `C_(3)H_(6)O_(3)`C. `C_(5)H_(10)O_(5)`D. `C_(4)H_(8)O_(4)` |
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Answer» Correct Answer - D For isotonic solution : `(w_(1))/(m_(1)V_(1))=(w_(2))/(m_(2)V_(2))` `w_(1` = mass of glucose `= 0.05xx180g = 9g` `m_(1)` = molecular mass of glucose = 180 g Assuming `V_(1)=V_(2)=1L` `w_(2)` = mass of compound = 6g `m_(2)` = molecular mass of compound = ? `therefore (9)/(180)=(6)/(x)rArr x=120g` `therefore ("Molecularmass")/("Empiricalmass")=n rArr=(120)/(30)=4` `therefore` Molecular formula `= C_(4)H_(8)O_(4)`. |
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| 273. |
The average osomotic pressure of human blood is 7.8 bar at `37^(@)C`. What is the concentration of an aqueous `NaCI` solution that could be used in the blood stream?A. 0.16 mol/LB. 0.32 mol/LC. 0.60 mol/LD. 0.45 mol/L |
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Answer» Correct Answer - B `pi=CRT , C=(pi)/(RT)=(7.8)/(082xx310)=0.31` mol/litre |
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| 274. |
200 mL of an aqueous solution of a protein contains its 1.26g. The osmotic pressure of this solution at 300K is found to be `2.57 xx 10^(-3)` bar. The molar mass of protein will be `(R = 0.083 L bar mol^(-1)K^(-1))`A. `31011g mol^(-1)`B. `61038 g mol^(-1)`C. `51022 g mol^(-1)`D. `122044 g mol^(-1)` |
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Answer» Correct Answer - B `pi=CRT=(wt xx 1000)/(GMM xx V)RT` `2.57xx10^(-3)=(1.26xx1000)/(GMM xx200)xx0.083xx300` GMM = 61038 g |
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| 275. |
The molarity of a solution obtained by mixing 800 ml of 0.5 M HCl with 200 ml of 1 M HCl will beA. 0.8 MB. 0.6 MC. 0.4 MD. 0.2 M |
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Answer» Correct Answer - B `underset(("Initial"))(M_(1)V_(1))+M_(2)V_(2)=underset(("Final"))(M_(3)V_(3))` `V_(3)=V_(1)+V_(2)=800+200=1000` `M_(3)xx1000=0.5xx800+1xx200` `M_(3)=(600)/(1000)=0.6M`. |
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| 276. |
Solute when dissolved in water:A. decreases the vapour pressure of waterB. increases the boiling point of waterC. decreases the freezing point of waterD. all of the above |
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Answer» Correct Answer - D Addition of solute to a solvent lowers the vapour pressure and freezing point but increases the boiling point of solution. |
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| 277. |
The osmotic pressure of a dilute solution is given byA. `P = P_(0)^(1) xx N_(1)`B. `pi V = nRT`C. `Delta P = P_(0)^(1) N_(2)`D. `(Delta P)/P_(0)=(P_(0)^(1)-P)/P_(0)^(1)` |
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Answer» Correct Answer - B `pi V = W/m RT` or `pi V = nRT` |
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| 278. |
The osmotic pressure of a solution isA. Directly proportional to the concentrationB. Inversely proportional to the concentrationC. Directly proportional to the square of the concentrationD. Directly proportional to the square of the root of the concentration |
| Answer» Correct Answer - A | |
| 279. |
Following solutions at the same temperature will be isotonicA. 3.42 g of cane sugar in one litre water and 0.18 g of glucose in one litre waterB. 3.42 g of cane sugar in one litre water and 0.18 g of glucose in 0.1 litre waterC. 3.42 g of cane sugar in one litre water and 1.17 g of NaCl in one litre waterD. 3.42 g of cane sugar in one litre water and 1.17 g of NaCI in one litre water |
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Answer» Correct Answer - B For isotonic solution, `pi_(1) = pi_(2)` (and for nonelectrolytes also `C_(1) = C_(2))` `pi_(1) = W/(M xx V) xx RT = 3.42 /(342 xx 1)RT = 0.01 RT` `pi_(2) = 0.18/(180 xx 0.1) RT = 0.01 RT` |
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| 280. |
Isotonic solutions are those which haveA. same osmotic pressureB. same molarityC. same densityD. same normality |
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Answer» Correct Answer - A A solution of M/2 NaCI is isotonic with M glucoge. The required condition is `pi_(1) = pi_(2)` |
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| 281. |
Which statement if wrong regarding osmotic pressure (P), volume (V) and temperature (T)A. `P prop (1)/(V)` id T is constantB. `F prop T` is V is constantC. `P prop V` is T is constantD. PV is constant if T is constant |
| Answer» Correct Answer - C | |
| 282. |
Blood has been found to be isotonic withA. normal saline solutionB. saturated NaCI solutionC. saturated KCI solutionD. saturated solution of a 1:1 mixture of NaCI and KCI |
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Answer» Correct Answer - A Normal saline is 0.16 M NaCI solution. |
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| 283. |
At the same temperature, following solution will be isotonicA. 3.24 g of sucrose per litre of water and 0.18 gm glucose per litre of waterB. 3.42 gm of sucrose per litre and 0.18 gm glucose in 0.1 litre waterC. 3.24 gm of surcose per litre of water and 0.585 gm of sodium chloride per litre of waterD. 3.42 gm of sucrose per litre of water and 1.17 gm of sodium chloride per litre of water |
| Answer» Correct Answer - B | |
| 284. |
Which of the following associated with isotonic solutions is not correctA. They will have the same osmotic pressureB. They have the same weight concentrationsC. Osmosis does not take place when the two solutions are separated by a semipermeable membraneD. They will have the same vapour pressure |
| Answer» Correct Answer - B | |
| 285. |
Isotonic solutions have sameA. Equal temperatureB. Equal osmotic pressureC. Equal volumeD. Equal amount of solute |
| Answer» Correct Answer - B | |
| 286. |
Blood is isotonic with:A. 0.16 M NaClB. Conc. NaClC. 50% NaClD. 30% NaCl |
| Answer» Correct Answer - A | |
| 287. |
Isotonic solution have the sameA. DensityB. Molar concentationC. NormalityD. None of these |
| Answer» Correct Answer - B | |
| 288. |
0.1 M NaCl and 0.05 M `BaCl_(2)` solutons are separated by a semi-permeable membrane in a container. For this system choose the correct answerA. There is no movement of any solution across the membraneB. Water flows from `BaCl_(2)` solution towards NaCl solutionC. Water flows from NaCl solution towards `BaCl_(2)` solutionD. Osmotic pressure of 0.1 M NaCl is lower than the osmotic pressure of `BaCl_(2)` (Assume complete dissociation) |
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Answer» Correct Answer - B Applying `pi V = iCRT` i=2 for NaCl i=3 for `BaCl_(2)` `pi V_(NaCl)=2xx0.1 RT=0.2 RT` `pi V_(BaCl_(2))=3xx0.05 RT =0.15 RT` As `pi V_(NaCl)gt pi V_(BaCl_(2))`. So, water from `BaCl_(2)` towards `NaCl_(2)` solution. |
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| 289. |
Diffusion of solvent through a semi permeable membrane is calledA. DiffusionB. OsmosisC. Active absorptionD. Plasmolysis |
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Answer» Correct Answer - B Diffusion of solvent through a semi permeable membrane is Osmosis. Hence option (B) is the right answer |
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| 290. |
Molarity of a solution prepared by dissolving 75.5 g of pure KOH in 540 ml solution isA. 3.05 MB. 1.35 MC. 2.50 MD. 4.50 M |
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Answer» Correct Answer - C `M=(w xx1000)/(m.wt. xx V" in ml")=(75.5xx1000)/(56xx540)=2.50M`. |
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| 291. |
10ml of conc. `H_(2)SO_(4)` (18 molar) is diluted to 1 litre. The approximate strength of dilute acid could beA. 0.18 NB. 0.09 NC. 0.36 ND. 1800 N |
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Answer» Correct Answer - C `(H_(2)SO_(4))N_(1)V_(1)=N_(2)V_(2)` (dilute acid) `N_(2) = (10xx36)//1000=0.36 N`. |
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| 292. |
The normality of 10% (weight/volume) acetic acid isA. 1 NB. 10 NC. 1.7 ND. 0.83 N |
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Answer» Correct Answer - C `N=(w xx1000)/(Eq.wt. xx "Volume")=(10xx1000)/(60xx100)=1.66 N`. |
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| 293. |
The molarity of pure water isA. `55.6`B. `5.56`C. 100D. 18 |
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Answer» Correct Answer - A Molarity of pure water `=(1000)/(18)=55.6 M`. |
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| 294. |
The molarity of pure water isA. 55.6B. 50C. 100D. 18 |
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Answer» Correct Answer - A 1 L of `H_(2)O`=1000 cc = 1000 g `"Molar mass of" H_(2)O = 18 g "mol"^(-1)` `" Molarity of pure water" = 1000/18 = 55.6` |
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| 295. |
The molarity of 720 g of pure water isA. 40 MB. 4 MC. 55.5 MD. unpredictable |
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Answer» Correct Answer - c Molarity `= (w)/(m.V) = (720)/(180x720//1000) = 55.5 M` (720 g water = 720 mL water) |
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| 296. |
An excess of `AgNO_(3)` is added to `100mL` of a `0.01M` solution of dichlorotetraaquachromium(III) chloride The number of moles of `AgCI` precipitated would be .A. `0.01`B. `0.001`C. `0.002`D. `0.003` |
| Answer» Correct Answer - B | |
| 297. |
`0.2` mole of `HCI and 0.2` mole of barium chloride were dissolved in water to produce a `500mL` solution. The molarity of the `CI^-` ions is :A. .04 MB. 0.8 MC. 0.4 MD. 0.08 M |
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Answer» Correct Answer - B `0.2 "mole" HCI = 0.2 "mole" CI^(-)` `0.1 "mole" CaCI_(2) = 0.2 "mole" CI^(-)` `"Totel"CI^(-) = 0.4 "mole".` `"Hence,Molarity of " CI^(-) =(n xx 1000)/(V(mL))` `(0.4 xx 1000)/500=0.8` |
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| 298. |
The molarity of a solution of `Na_(2)O_(3)`having`10.6g//500ml`of solution isA. 0.2 MB. 2 MC. 20 MD. 0.02 M |
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Answer» Correct Answer - A `M=(w xx1000)/(m.wt. xx "Volume in ml.") =(10.6xx1000)/(106xx500)=0.2M`. |
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| 299. |
If `25 ml` of `NaCl ` solution is diltuted with water to a volume of `500ml` the new concentration of the solution isA. 0.167 MB. 0.0125 MC. 0.833 MD. 0.0167 M |
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Answer» Correct Answer - B `M_(1)V_(1)=M_(2)V_(2), M_(2)=(0.25xx25)/(500)=0.0125`. |
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| 300. |
Twenty grams of a substance were dissolved in 500ml. Of water and the osmotic pressure of the solution was found to be 600mm of mercury at `15^(@)"C".` Determine the molecular weight of the substance.A. 1000B. 1200C. 1400D. 1800 |
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Answer» Correct Answer - B `pi=(n)/(V)RT=(m//MRT)/(V)` `(600)/(760)=(20xx0.0821xx288xx1000)/(500xx M) , M = 1200` |
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