InterviewSolution
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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.
| 1. |
`K_(f)` for water is `1.86 K kg mol^(-1)`. IF your automobile radiator holds `1.0 kg` of water, how many grams of ethylene glycol `(C_(2)H_(6)O_(2))` must you add to get the freezing point of the solution lowered to `-2.8^(@)C` ?A. 27 gB. 72 gC. 93 gD. 39 g |
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Answer» Correct Answer - C `Delta T_(f) = (1000K_(f) xx W_(2))/(M_(2) xx W_(1))` `:. 2.8 = (1000 xx 1.86 xx W_(2))/(1000 xx 62) or W_(2) = 93g.` |
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| 2. |
For a dilute solution containing `2.5 g` of a non-volatile non-electrolyte solution in `100g` of water, the elevation in boiling point at `1` atm pressure is `2^(@)C`. Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure (mm of `Hg)` of the solution is: (take `k_(b) = 0.76 K kg mol^(-1))`A. 724B. 740C. 736D. 718 |
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Answer» Correct Answer - A `M_(2) = (K_(b) xx W_(2))/(W_(1) xx Delta T_(b)) xx 1000 = (1000 xx 0.76 xx 2.5)/(100 xx 2) =9.5` `(p_(1)^(o)-p)/p_(1)^(o) = (W_(2)//M_(2))/(W_(1)//M_(1))` `:. (760-p)/760 = (2.5 // 9.5)/(100//18)=0.047` or 760 - p = `760 xx 0.047` = 35.7 or p=724.3 mm |
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| 3. |
29.2% (W/W) HCI stock solution has a density of 1.25 g `mL^(-1)`. The molecular mass of HCI is 36.5 g `mol^(-1)` Calculate the volume (mL) of stock solution required to prepare 200 mL of 0.4 M HCIA. 2 mLB. 4 mLC. 8 mLD. 6 mL |
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Answer» Correct Answer - C Stock solution of HCI = 29.2 % (w/w) Thus, 29.2 g HCI are present in 100 g of the solution. As density of solution = 1.25 g `mL^(-1)`, Volume of 100 g of the solution = 100/1.25 mL Molar mass of HCI = 36.5 g `mol^(-1)`, `"Molarity of the solution" = (W_(2) xx 1000)/(M_(2) xx V (cm^(3)))` `29.2/36.5 xx 1.25/100 xx 1000 = 10 M` `{:(M_(1) V_(1),=,M_(2) V_(2)),(("stock solution"), ,("solution required")),(10 xx V_(1),=,0.4 xx 200):}` ` or V_(1) = 8 mL` |
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| 4. |
The freezing point among the following equimolal aqueous solutions will be highest forA. `C_(6)H_(5)NH_(3)Cl` (aniline hydrochloride)B. `Ca(NO_(3))_(2)`C. `La(NO_(3))_(3)`D. `C_(6)H_(12)O_(6)` (glucose) |
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Answer» Correct Answer - D Glucose, being non-electrolyte, gives minimum no. of particles and hence minimum `Delta T_(f)` or maximum F. pt. |
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| 5. |
The amount of urea to be dissolved in 500 ml of water (K = 18.6 K `"mole"^(-1)` in 100 g solvent) to produce a depression of `0.186^(@)C` in freezing point isA. 9 gB. 6 gC. 3 gD. 0.3 g |
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Answer» Correct Answer - C `Delta T_(f)=(100xx K xx w)/(m xx W) therefore 0.186 = (100xx18.6xx w)/(60xx500)` w = 3 g |
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| 6. |
The normality of `0.3 M` phosphorous acid `(H_(3) PO_(3))` isA. `0.1`B. `0.9`C. `0.3`D. `0.6` |
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Answer» Correct Answer - D Basicity of `H_(3)PO_(3)` is 2. Hence 0.3 `M H_(3)PO_(3) = 0.6 N`. |
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| 7. |
Benzoic acid dissolved in benzene showsA. its normal molecular massB. Double of its normal molecular massC. Half of its normal molecular massD. Not definite. |
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Answer» Correct Answer - B Benzoic acid dimerises in benzene. |
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| 8. |
When benzoic acid dissolve in benzene, the observed molecular mass isA. 244B. 61C. 366D. 122 |
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Answer» Correct Answer - A Benzoic acid in benzene undergoes association through intermolecular hydrogen bonding. |
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| 9. |
Four solutions of `K_(2)So_(4)` with the concentrations 0.1m ,0.001 m, and 0.0001 mare available . The maximum value of colligative property corresponds to :A. 0.0001 m solutionB. 0.001 m solutionC. 0.01 m solutionD. 0.1 m solution |
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Answer» Correct Answer - A Electrolytes have maximum degree of dissociation in diluted solution. |
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| 10. |
Dry air was passed successively through a solution of 5 gm of a solute in 80 gm of water and then through pure ware. The loss in weight of solution was 2.50 gm and that of pure solvent 0.04 gm. What is the molecular weight of the soluteA. `70.31`B. `7.143`C. `714.3`D. 80 |
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Answer» Correct Answer - A `m=(5xx18xx2.5)/(0.04xx80)=70.31` |
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| 11. |
One litre of a solution of N/2 HCl was heated in beaker and it was observed that when the volume of solution got reduce to 600 mL, 3.25 g of HCl was lost. Calculate the normality of the resulting solution.A. 0.50 NB. 0.60 NC. 0.68 ND. 0.70 N |
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Answer» Correct Answer - c 1 L of N/2 HCl contains `= (1)/(2)xx36.5 = 18.25 g` HCl now present `= 18.25 - 3.25 = 15 g` Now, volume = 600 mL = 0.600 L Normality `= (15)/(36.5)xx(1)/(0.600 L) = 0.685 N` |
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| 12. |
The wt. of unhydrous sodium carbonate needed to prepare 500 ml of a decinormal solution would beA. 5.3 gB. 10.6gC. 1.06 gD. 2.65 g |
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Answer» Correct Answer - D Wt. per `dm^(3) = N xx eq. Wt.` `= 0.1 xx 53 = 5.3 g` Wt. in 500 ml `= (5.3)/2= 2.65 g. |
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| 13. |
How many grams of a dibasic acid (Mol. Wt. =200) should be present in 100 ml of its aqueous solution to give decinormal strengthA. 1gB. 2gC. 10gD. 20g |
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Answer» Correct Answer - A `N=(w)/(E xxV(l))rArr 0.1=(w)/(100xx0.1)rArr w=1 gm` |
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| 14. |
The elevation in boiling point would be highest forA. `0.08 m BaCl_(2)`B. `0.10 m` glucoseC. `0.10 m KCl`D. `0.06 m` calcium nitrate |
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Answer» Correct Answer - a More the number of particles and concentration higher is the elevation in boiling point. Thus, 0.08 m KCl `= -0.082xx2 = 0.24m` (highest) |
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| 15. |
The latent heat of vaporisation of water is `9700 "Cal/mole"` and if the b.p.is `100^(@)C`, ebullioscopic constant of water isA. `0.516^(@)C`B. `1.026^(@)C`C. `10.26^(@)C`D. `1.832^(@)C` |
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Answer» Correct Answer - a `K_(b) = (R.T_(b)^(2))/(1000.L_(V)) = (2xx373xx373xx18)/(1000xx9700) = 0.516^(@)C` |
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| 16. |
Which one has the highest boiling pointA. `0.1 N Na_(2)SO_(4)`B. `0.1 N MgSO_(4)`C. `0.1 M Al_(2)(SO_(4))_(3)`D. `0.1 M BaSO_(4)` |
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Answer» Correct Answer - C `Al_(2)(SO_(4))_(3)` gives maximum ion hence it will show highest boiling point. |
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| 17. |
The latent heat of vapourisation of water is 9700 Cal/mole and if the b.p. is `100^(@)C`, equllioscopic constant of water isA. `0.513^(@)C` kg/molB. `1.026^(@)C` kg/molC. `10.26^(@)C` kg/molD. `1.832^(@)C` kg/mol |
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Answer» Correct Answer - A `K_(b)=(M_(1)RT_(b)^(2))/(1000 Delta H_(V))=(18xx1.987xx(373)^(2))/(1000xx9700)=0.513^(@)C` |
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| 18. |
At a boiling point of pure solvent ,solution will not boil becauseA. V.P. of solvent is less than that of solutionB. V.P. of solvent is equal to that of solutionC. V.P. of solution is less than that of solutionD. all |
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Answer» Correct Answer - C It is a fact. |
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| 19. |
If the density of the solvent is 2.5 kg `dm^(-3)` . The 2 molal solution of a solute in this solvent will be molal solution of a solution of a solute in this solvent will beA. 5 MB. 2.5 MC. 4 MD. 1.25 M |
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Answer» Correct Answer - A `M = m xx d = 2 xx 2.5 = 5` |
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| 20. |
The molal elevation constant is the ratio of the elevation in boiling point to :A. molarityB. molalityC. mole fraction of soluteD. mole fraction of solvent. |
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Answer» Correct Answer - B `Delta T_(b)=K_(b) xx m ` or ` K_(b)=Delta T_(b)//m.` |
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| 21. |
Boiling point of a solution is independent ofA. amount of solutionB. pressureC. nature of solventD. concentration of solution |
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Answer» Correct Answer - A B.pt of a solution depends upon concentration, pressure and nature of solvent. |
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| 22. |
Elevation in boiling point was `0.52^(@)C` when 6 g of a compound X was dissolved in 100 g of water. Molecular weight of X is (`K_(b)` of water is `5.2^(@)C` per 100 g of water)A. 120B. 60C. 180D. 342 |
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Answer» Correct Answer - B `M_(2) = (K_(b) xx W_(2))/(Delta T_(b) xx W_(1)) xx 1000 = (5.2 xx 6 xx 10 ^(-3))/(0.52 xx 100 xx 10^(-3))=60` |
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| 23. |
The elevation in boiling point for one molal solution of a solute in a solvent is calledA. Cryoscopic constantB. Boiling point constantC. Molal Ebullioscopic constantD. None |
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Answer» Correct Answer - C `Delta T=K_(b) xx "molality" ` `"Molatity = 1. xx Delta T = K_(b)` |
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| 24. |
A solution containing `4 g` of a non-volatile organic solute per `100 mL` was found to have an osmotic pressure equal to `500 cm` of mercury at `27^(@)C`. The molecular weight of solute isA. 14.97B. 149.7C. 1697D. 1.497 |
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Answer» Correct Answer - B `pi V = W_(2)/M_(2)RT,` `500/76 xx 100/1000 = 4/M_(2) xx 0.0821 xx 300,` `M_(2) = 149.6` |
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| 25. |
The molal elevation constant for water is 0.52 K. `"molality"^(-)`. The elevation caused in the boiling point of water by dissolving 0.25 mole of a non volatile solute in 250 g of water will be :A. `52^(@)`B. `5.2 ^(@)C`C. `0.52^(@)C`D. `0.052 ^(@)C` |
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Answer» Correct Answer - C `(Delta T_(f)=1000 xx K_(f))/W_(1) xx W_(2)/M_(2)=(1000 xx K_(f))/W_(1)xx n` `Delta T_(f) = (1000 xx 0.52)/250 xx 0.25 =0.52^(@)C` |
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| 26. |
The freezing point of aqueous solution that contains `5%` by mass urea. `1.0%` by mass `KCl` and `10%` by mass of glucose is: `(K_(f) H_(2)O = 1.86 K "molality"^(-1))`A. 290.2 KB. 285.5 KC. 269 93 KD. 250 K |
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Answer» Correct Answer - C `Delta T_(f) = Delta T_(f) ("glucose") + Delta T_(f) (KCl) +Delta T_(f) ("urea")` But `Delta T_(f)=(K_(f) xx W_(2))/(M_(2) xx W_(2)) xx 1000` `:. (1000 xx1.86 xx10)/(100 xx 180) +(1000 xx 1.86 xx 1 xx 2)/(74.5 xx 100)+(1000+1.86 xx5)/(100 xx 60)=3.069` f.pt. =273 - 3.069 = 269.93 K |
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| 27. |
The osmotic pressure of a solution containing `40 g` of solute `("molecular mass 246")` per litre at `27^(@)C` is `(R=0.0822 atm L mol^(-1))`A. 0.1 atmB. 0.4 atmC. 0.2 atmD. 0.8 atm |
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Answer» Correct Answer - B `piV = "Mass"/("Molar Mass") xx R xxT ` `pi xx 1 = 4/246 xx 0.082 xx 300` `pi = 0.4 atm` |
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| 28. |
The molal elevation constant for water is `0.56 K kg mol^(-1)`. Calculate the boiling point of a solution made by dissolving 6.0g of urea `(NH_(2)CONH_(2))` in 200g of water.A. `10.028^(@)C`B. `100.28 ^(@)C`C. `50.14 ^(@)C`D. none of these. |
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Answer» Correct Answer - B `Delta T_(b) = (1000 xx K_(b) xx W_(2))/(W_(1)M_(2))=(1000 xx 0.56 xx 6.0)/(200 xx 60)=0.28^(@),` `T^(@) + Delta T_(b) = 100^(@) + 0.28=100.28^(@) C` |
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| 29. |
0.01 molar solutions of glucose, phenol and potassium chloride were prepared in water. The boiling points ofA. Glucose solution = Phenol solution = Potassium chloride solutionB. Potassium chloride solution gt Glucose solution gt Phenol solutionC. Phenol solution gt Potassium chloride solution gt Glucose solutionD. Potassium chloride solution gt Phenol solution gt Glucose solution |
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Answer» Correct Answer - D `underset("Boiling point decreasing order"rarr.)(KCl gt C_(6)H_(5)OH gt C_(6)H_(12)O_(6))` Potassium chloride is ionic compound and phenol formed phenoxide ion hence it shows greater boiling point then glucose. |
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| 30. |
The values of observed and calculated molecular weights of silver nitrate are `92.64` and `170` respectively.The degree of dissociation of silver nitrate is:A. 0.6B. 0.835C. 0.467D. 0.6023 |
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Answer» Correct Answer - B `alpha = (i-1)/(n^(l) - 1)` For `AgNO_(3), n^(l) = 1 :. alpha = (i-1)/(2-1) :. alpha = i-1``"But" i="normal molar mass."/"observed molar mass."` `:. alpha = 170/92.64 - 1 = 0.835` `:. % alpha = 0.835 xx 100 = 83.5 %` |
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| 31. |
For an aqueous solution freezing point is `-0.186^(@)C`. The boiling point of the same solution is `(K_(f) = 1.86^(@)mol^(-1)kg)` and `(K_(b) = 0.512 mol^(-1) kg)`A. 0.186B. 0.512C. 0.512//1.86D. 0.0512 |
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Answer» Correct Answer - D `Delta T_(b) = K_(b) xx "molality"` `Delta T_(b) = K_(b) xx"molality" ` `(Delta T_(b))/(Delta T_(f))=K_(b)/K_(f) ` or `Delta T_(b) = (Delta T_(f) xx K_(b))/K_(f)` `=(0.186 xx 0.512)/1.86=0.0512` |
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| 32. |
What will be the molecular weight of `NaCl` determined experimentally following elevation in the boiling point or depression in freezing point method?A. lt 58.5B. gt58.5C. 58.5D. None |
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Answer» Correct Answer - A Normal molar mass of electrolyte gt Exp. Molar mass . |
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| 33. |
The boiling point of 0.1 molal aqueous solution of urea is `100.18^(@)C` at 1 atm. The molal elevation constant of water isA. `1.8`B. `0.18`C. 18D. `18.6` |
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Answer» Correct Answer - A `K_(b)=(0.18)/(0.1)=1.8` |
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| 34. |
The molecular mass of a solute cannot be calculated by one of the following relationsA. `M_(2)=(K_(b) xx 1000 xx w_(2))/(Delta T_(b) xx w_(1))`B. `M_(2)=(w_(2) xx RT )/(pi V)`C. `M_(2)=(P_(0)^(1) xx W_(2) xxM_(1))/((P_(0)^(1)-P) xx W_(1))`D. `M_(2)=(Delta T_(b))/K_(b) xx1000 W_(2)/W_(1)` |
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Answer» Correct Answer - D All relation are correct except (d). |
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| 35. |
Assertion :- Molecular mass of polymers cannot be calculated using boiling point or freezing point method. Reason : Polymers solutions do not possess a constant boiling point or freezing point.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 the assertion and reason both are false. |
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Answer» Correct Answer - C The polymer solutions possess very little elevation in boiling point or depression in freezing point. |
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| 36. |
Assertion :- Camphor is used as solvent in the determination of moleclar masses of naphthalene, anthracene etc. Reason :- Camphor has high molal elevation constant.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 the assertion and reason both are false. |
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Answer» Correct Answer - C Camphor has high molal depression constant. |
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| 37. |
What is the effect of the addition of sugar on the boiling and freezing points of waterA. Both boiling point and freezing point increasesB. Both boiling point and freezing point decreasesC. Boiling point decreases and freezing point decreasesD. Boiling point decreases and freezing point increases |
| Answer» Correct Answer - C | |
| 38. |
The molal depression constant for a solvent is 4.9. The depression in freezing point for a millimolal solution of a non-electrolyte in the solvent isA. 0.49B. 4.9C. 4.9 + 0.001D. 0.0049. |
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Answer» Correct Answer - D `Delta T_(f) = K_(f) xx m=4.9 xx 10^(-3)=0.0049` |
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| 39. |
What happens to freezing point of benzene when naphthalene is addedA. IncreasesB. DecreasesC. Reamains unchangedD. First decreases and then increases |
| Answer» Correct Answer - B | |
| 40. |
The molecular mass of acetic acid dissolved in water is 60 and when dissolved in benzene it is 120. This difference in behaviour of `CH_(3)COOH` is becauseA. Water prevents association of acetic acidB. Acetic acid does not fully dissolve in waterC. Acetic acid fully dissolves in benzeneD. Acetic acid does not ionize in benzene |
| Answer» Correct Answer - B | |
| 41. |
Assertion :- The freezing point is the temperature at which solid crystallizzes from solution. Reason :- The freezing point depression is the difference between that temperature and freezing point of pure solvent.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 the assertion and reason both are false. |
| Answer» Correct Answer - B | |
| 42. |
Statement - The water pouch of instant cold pack for treating athletic injuries breakes when squeezed and `NH_(4)NO_(3)` dissolves lowering the temperature. Explanation - Addition of non-volatile solute into solvent results into depression of freezing point of solvent.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 the assertion and reason both are false. |
| Answer» Correct Answer - A | |
| 43. |
Assertion (A): The increasing pressure on water decreases its freezing point. Reason (R ):The density of water is maximum at `273 K`.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 the assertion and reason both are false. |
| Answer» Correct Answer - C | |
| 44. |
Ethylene glycol is used as an antifreeze in a cold climate. Mass of ethlene glycol which should be added to 4 kg of water to prevent it form freezing at `-6^(@)C` will be : (`K_(f)` for water `=1.86K kg mol^(-1)`, and molar mass of ethylene glycol `= 62 g mol^(-1)`)A. 80.32 gB. 204.30 gC. 400.00 gD. 304.60 g |
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Answer» Correct Answer - A `Delta T_(f)=iK_(f)m, Delta T_(f)=6^(@)C, i=1` `6=1xx1.86xx(w)/(62xx4)rArr w=804.32 g` |
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| 45. |
Which one of the following aqueous solution will have the lowest freezing point?A. `0.2 M Na_(2)SO_(4)`B. `0.1 M "Urea"`C. `0.2 M NaNO_(3)`D. `0.1 M BaCI_(2)` |
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Answer» Correct Answer - A Both (a) and (b) gives three moles of ions but molarity of (a) is more hence it has highest depression in freezing point, shows lowest freezingg point . |
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| 46. |
Ethylene glycol is used as an antifreeze in a cold climate. Mass of ethylene glycol which should be added to 4 kg of water to prevent it from freezing at `-6^(@)C` will be `(K_(f)` for water `= 1.86 Kkg mol^(-1)`, and molar mass of ethylene glycol `=62 g mol^(-1))`A. 204.30gB. 400.00 gC. 304.60 gD. 800.00 g |
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Answer» Correct Answer - D `Delta T_(f) xx (1000K_(f) xx W_(2))/(M_(2) xx W_(1))` `:. 6 = (1000 xx 1.86 xx W_(2))/(4000 xx 62)` or `W_(2) = 800g` |
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| 47. |
At a suitable pressure near the freezing point of ice, there existsA. only iceB. ice and waterC. ice and steamD. ice, water and steam ,all existing side by side |
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Answer» Correct Answer - D At triple point all the three phases exist together. `(P=2.56mm, T = 0.0098^(@)C)` |
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| 48. |
The melting point of most of the solid substances increases with an increase of pressure acting on them . However , ice melts at a temperature lower than its usual melting point when the pressure increases . This is because :A. ice is lass denser than waterB. pressure generates heatC. the bonds break under pressureD. ice is not a true solid |
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Answer» Correct Answer - A `V_(ice) rarr V_(water)` and thus increase in pressure favours forward reaction. |
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| 49. |
0.1 molal aqueous solution of sodium bromide freezes at `-0.335^(@)C` at atmospheric pressure. `K_(f)` for water is `1.86^(@)C.` The percentage of dissociation of the salt in solution isA. 90B. 80C. 58D. 98 |
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Answer» Correct Answer - B `Delta T_(f) = i K_(f) m or 0.335 = i xx 1.86 xx 0.1 or i= 1.80 ` `alpha = (i-1)/(n-1) = (i-1)/(2-1)` For NaBr `i = 1 + alpha or alpha = 0.80 = 80%` |
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| 50. |
The temperature at which the vapour pressure of a liquid becomes equals to the external (atmospheric) pressure is itsA. b.pt.B. f.pt.C. sublimation pointD. none |
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Answer» Correct Answer - B It is the definition of freezing point. |
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