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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.
| 801. |
What will happen when red blood corpuscles `(RBC_(s))` ae placed in (a) 1% NaCI solution (b) 0.6% NaCI solution ? |
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Answer» We all know that `RBC_(s)` are isotonic with 0.9% Naci solution. If `RBC_(s)` are placed in contact with 1% NaCI solution, then the osmotic pressure of 1% NaCI will be higher than thet of `RBC_(s)`. As a result, water present inside the call moves into the NaCI solution through cell walls acting as semi-permeable membrane. The `RBC_(s)` will therefore, shrink. However, reverse will take place in case these are kept in contact with 0.6% NaCI solution which has less osmotic pressure. Water will now move into the `RBC_(s)` and they will swell. Keeping this in view, the injectable medicines are generally dissolved in saline water (0.9% NaCI solution) before being injected. This prevants the cell walls form either undergoing thrinkage or bursting. |
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| 802. |
Out of 1 M glucose and 2 M glucose, which one has a higher boiling point why ? Wha happens when the external pressure applied becomes more than the osmotic pressure of solution ? |
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Answer» 2 M glucose solution is more concetrated and has less vapour p-ressure than 1 M glucose solution. It has therefore. Higher boiling point than 1 M solution. Ravers Osmosis takes place. For details consult section 12. |
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| 803. |
When fruits and vegetables are dried and placed in water, they slowly swell and return to original form, Why? Does an increase in temperature accelerate the process? Explain. |
| Answer» When fruits and vegetables that have dried in water, osmosis takes place, i.e., water molecular pass through semipermeable membrane present in cell-walls, therefore, they swell. If temperature is increased osmosis will be faster. | |
| 804. |
Blood cells are isotonic with 0.5% sodium chloride solution. What happens if we place blood cells in a solution containing (i)1.2% NaCl (ii) 0.4% NaCl solution. |
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Answer» (i) 1.2 % Sodium chloride is hypertonic than blood cells, hence cells will shrink. Plasmolysis will take place. (ii) 0.4% Sodium chloride solution is hypotonic than blood cell, so cells will swell. Endo osmosis will take place. |
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| 805. |
Blood cells are isotonic with 0.9% sodium chloride solution. What happens if we place blood cells in a solution containing 1.2% sodium chloride 0.4% sodium chloride 1.2% sodium chloride 0.4% sodium chloride. |
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Answer» 1.2% sodium chloride solution is more concentrated and is hypertonic in nature. Osmosis takes place from 0.9% sodium bloride solution (hypotonic). As a blood colls shrind. 0.4% sodium chloride solution is less concentrateed (hypotonic) and osmosis takes place into the blood cells (hyphotonic). As a result, there is swelling in the blood cells. |
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| 806. |
Blood cells are isotonic with 0.9% sodium chloride solution. What happens if we palace blood cells in a solution containing 1.2% sodium chloride solution (ii) 0.4% soldium chloride solution |
| Answer» (i) Blood cells shring (ii) Blood cells swell. | |
| 807. |
Out of the following solutions which has the maximum freezing point and why ? (i) 1 M urea (ii) 1 M potassium chloride (iii) 1 M aluminium chloride |
| Answer» Correct Answer - 1 M urea solution. | |
| 808. |
Why does NaCI solution freeze at a temperature lower than the freezing point of water ? |
| Answer» The addition of non-volatile solvent decreses the freezing point of the solution. Fro details, conslt section 11. | |
| 809. |
Define azeotrope. What type of azeotrope is formed by negative deviation from Raoult’s law? Give an example. |
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Answer» Some liquids on mixing form azeotropes which are binary mixtureshaving the same composition in liquid and vapour phase and boil ata constant temperature. In such cases, it is not possible to separate the components by fractional distillation. From negative deviation maximum boiling azeotropes are formed. Example phenol and aniline. |
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| 810. |
What type of azeotrope is formed by mixing nitric acid and water ? |
| Answer» Maximum boiling azetrope is formed. | |
| 811. |
Why does sodium chloride freeze at a lower temperature than water? |
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Answer» It is a colligative property. When a solute (NaCl) is added to a solvent (Water) there is an elevation in boiling point and a lowering in freezing point. |
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| 812. |
Type of deviation in mixture of ethanol and acetone? Type of azeotrope is formed by mixing ethanol and acetone? |
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Answer» They show positive deviation and hence minimum boiling azeotrope. |
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| 813. |
What type of deviation is shown by a mixture of ethanol and acetone ? What type of axetrope is formed by mixing the two ? |
| Answer» `+ve` deviation, maximum boiling azeotrope. | |
| 814. |
A storage battery contains a solution of `H_(2)SO_(4) 30%` by weight. Find out (Given density of solution`=1.2gm//cm^(3)`)A. MolalityB. MolarityC. NormalityD. Mole fraction of `H_(2)SO_(4)` |
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Answer» 30% by weight `implies W_("solute")=30gm,W_("solution")=100gm,W_("solvent")=70gm` `V_("solution")=100/1.2mL` `n_("solution")=30/98=0.306 " "n_("solvent")=70/18=3.889` (i) Molality`=(n_("solute")xx1000)/w_("solvent")=(0.306xx1000)/70=4.37m` (ii) Molarity`=(n_("solute")xx1000)/V_("solution")=(0.306xx1000)/(100/1.2)=3.67M` (iii)`"Normality" =(w_("solute")xx1000)/(E_("solute")xxV_("solution"))=(30xx1000)/(49xx100/1.2)=7.37N` Alternately `Normality=("n-factor"xx"Molarity")` `=2xx3.67=7.34` (iv)` X_(H_(2)SO_(4))=X_(H_(2)SO_(4))/(n_(H_(2)SO_(4))+n_(H_(2)O))=0.306/(0.306+3.889)=0.07` |
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| 815. |
An aqueous solution of glucose boils at `100.01^(@)C`.The molal elevation constant for water is `0.5 kmol^(-1)kg`. The number of molecules of glucose in the solution containing `100g` of water isA. `6.023xx10^(23)`B. `6.023xx10^(22)`C. `12.046xx10^(20)`D. `12.046xx10^(23)` |
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Answer» Correct Answer - C `Delta"T"_(b)="K"_(b).m` Or `"m"=(DeltaT_(b))/(K_(b))=0.01/0.5=.02 " mole Kg" ^(-1) "of water"` lt brgt So, the number of moles of glucose in 100g of water `=(0.02xx100)/1000=0.002` mole of glucose `=0.002xx6.023xx10^(23)=2xx6.023xx10^(20)` |
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| 816. |
At`25^(@)"C"`, the vapour pressure of methyl alcohol is 96.0 torr. What is the mole fraction of `"CH"_(3)"OH"` in a solution in which the (partial) vapor pressure of `"CH"_(3)"OH"` is 23.0 torr art `25^(@)"C"`? |
| Answer» `"X"_(CH_(3)OH)=(P_(CH_3OH))/(P_(CH_(3)OH)^(0))=23/96=0.24` | |
| 817. |
One mole of non-volatile solute is dissolved in two moles of water. The vapour pressure of the solution relative to that water ofA. `2/3`B. `1/3`C. `1/2`D. 3/2` |
| Answer» `("P"_("solvent")^(@)-"P"_("solution"))/"P"_("solvent")^(@)=1/(1+2)=1/3 " (mole fraction of solute in solution)" " Or " "P"_("solution")/"P"_("solvent")^(@)=1-1/3=2/3` | |
| 818. |
The freezing point of equimolal soution will be highest for :A. `C_(6)H_(5)NH_(3)Cl`B. `Ca(No_(3))_(2)`C. `La(NO_(3))_(3)`D. `C_(6)H_(12)O_(6)` |
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Answer» Correct Answer - D `La(NO_(3))_(3)`will furnish four ions and thus will develop more lowering in freezing point where as glucose gives only one particle and thus lowering in freezing point is minimum. |
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| 819. |
Which one is not the characteristic of chemisorption:A. Multilayer adsorptionB. Exothermic natureC. Strong adsorption by adsorption sitesD. Irreversible |
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Answer» Correct Answer - A Monolayer is formed during chemisorption. |
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| 820. |
Soaking of water by a sponge is an example of :A. Simple adsorptionB. Physical adsorptionC. ChemisorptionD. Adsorption |
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Answer» Correct Answer - D Sponge will be completely soaked by water, so it is adsorption. |
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| 821. |
Which one of the following statements is false for hydrophilic sols?A. they do not require electrolytes for stabilityB. their viscosity is of the order of that of waterC. their surface tension is usually lower than that of dispersion medium.D. none of these |
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Answer» Correct Answer - B Viscosity of lyophilic colloid is less than water. |
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| 822. |
Why the measurement of osmotic pressure is more widely used for determining molar masses of macromolecules than the elevation in boiling point or depression in freezing point of their solutions? |
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Answer» "The Osmotic pressure method has the advantage over other methods as pressure measurement is around room temperature and the molarity of the solution is used instead of molarity.As compared to other colligative properties, its magnitude is large even for very dilute solutions. The technique of osmotic pressure for determination of molar mass of solutes is particularly useful for biomolecules as they are generally not stable at high temperatures and polymers have poor solubility." |
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| 823. |
Of which of the following colloidal systems, fog is an example?A. Liquid dispersed in gasB. Gas dispersed in gasC. Solid dispersed in gasD. Solid dispersed in liquid |
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Answer» Correct Answer - A Fog is an example of quild dispersed in gas |
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| 824. |
Assertion : Molarity of the solution changes with temperature. Reason : Molarity is a colligative property.A. If both assertion are reason are correct and reason is correct explanation for assertion.B. If both assertion and reason are corrct but reason is not correct explanation for assertion.C. If assertion is corrct but reason is incorrect.D. If assertion and reason both are incrrect. |
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Answer» Correct Answer - c Correct Reason. Molarity depends upon volume which changes with change in temperature. |
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| 825. |
The colloidal solutions of gold prepared by different methods have different colors due to :A. different diameters of colloidal gold particlesB. variable valency of goldC. different concentration of gold particlesD. impurities produced by different methods |
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Answer» Correct Answer - A Different colloidal particle will provide different colour to the sol. |
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| 826. |
Which of the following sols is positively charged?A. Arsenious sulphideB. Aluminium hydroxideC. Ferric hydroxideD. Silver iodide in silver nitrate solution |
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Answer» Correct Answer - B::C::D `As_(2)S_(3)` is negatively charged. |
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| 827. |
`NaCl` is added to`1` litre water to such an extent that `DeltaT_(f)//K_(f)` becomes to `(1)/(500)`,wt.of `NaCl` added toA. `5.85g`B. `0.585g`C. `0.0585g`D. `0.0855g` |
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Answer» Correct Answer - C For `NaCl i=2` or`DeltaT_(f)=i.K_(f).m` `DeltaT_(f)//K_(f)=2.m.` Or =`2m` Or molarity `=(1)/(1000)or(1)/(1000)` moles are dissolved `Kg^(-1)`of water or approximately `(1)/(1000)`moles are present in `1.0` litre of water(considering molarity and molality to be the same) `:.` Weight of `NaCl=(1)/(1000)xx58.5 g` of `NaCl` `= 0.0585 g` of NaCl. |
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| 828. |
To observe an elevation of boiling point of `0.0^(@)C`, the amount of th solte (Molar mass 100 g `mol^(-1)`) to be added to 100 g of water `(K_(b)=0.5 K mol^(-1))` is:A. 2.0 gB. 0.5 gC. 1.0 gD. 0.75 g |
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Answer» Correct Answer - a `W_(B)=(M_(B)xxW_(A)xxDeltaT_(b))/K_(b)` `=((100g mol^(-1))xx(0.1 kg)xx0.5 K)/((0.5"K kg" mol^(-1)))=1.0g` |
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| 829. |
The vapour pressure of acetone at `20^(@)C` is 185 torr. When 1.2 g of a non-volatile substance was dissolved in 100 g of acetone at `20^(@)C`, its vapour pressure was 183 torr. The molar mass of the substance is :A. 128B. 488C. 32D. 64 |
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Answer» Correct Answer - d `(P_(A)^(@)P_(S))/P_(S)=(W_(B)xxW_(A))/(W_(A)xxM_(B))` `((185-183))/((183))=((1.2 g)xx(58 g mol^(-1)))/((100g)xxM_(B))` `M_(B)=((1.2g)xx(58g mol^(-1))xx(183))/(2xx(100g))=63.68g mol^(-1)` =64 g `mol^(-1)` |
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| 830. |
A motor vehicle raditor was filled with `8 L`of water to which `2 L`of methyl alcohol `(density 0.8 g mL^(-1))` was added. What is the lowest temperature at which the vehicle can be parked outdoors without the danger that the water in the raditor will freeze? Given that `K_(f)` for water is `1.86 K kg mol^(-1)` |
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Answer» Correct Answer - `T_(f)=-11.625^(@)C` Mass of `8 L` water = `8 kg` or `8000 g` because density is `1`. Mass of density alcohol added =Density`xx`Volume =`0.8 xx 2000` =`1600 g` `DeltaT_(f)=K_(f)m` =`1.86 xx 1600/32 xx 1/8000 xx1000` =`11.625^(@)C` Therefore, the lowest temperature at which the vehicle can be parked outdoors without the danger that the raditor will freeze is `-11.625^(@)C`. |
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| 831. |
The vapour pressure of water is 12.3 k `P_(a)` at 300 K. Calculate the vapour pressure of 1 molal solution of a non-volatile solute in it. |
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Answer» Given molality of solution = 1m Vapour pressure of water `P_(0)=12.3" kP"_(a)` Number of moles of solute `=1.(n_(s))` Number of moles of water `=(1000)/(18)` `=55.55(n_(0))` Mole fraction of solute `(X_(s))=(n_(s))/(n_(0)+n_(s))` `=(1)/(55.55+1)` `=(1)/(56.55)=0.0177` Mole fraction of water `(X_(0))=1-X_(s)` `=1-0.0177=0.9823` Vapour pressure of solution `(P_(s))=P_(0)X_(0)` `=12.3xx0.9823=12.08" kP"_(a)` |
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| 832. |
Consider separate solutions of `0.500 M C_(2)H_(5)OH(aq)`,`0.100 M Mg_(3)(PO_(4))(aq)`,`0.250 M KBr(aq)`, and `0.125 M Na_(3)PO_(4)(aq)` at `25^(@)C`. Which statement is true about these solutions, assuming all salts to be strong electrolytes?A. `0.125 M Na_(3)PO_(4) (aq)` has the highest osmotic pressureB. `0.500 M C_(H_(5))OH (aq)` has the highest osmotic pressureC. they all have the same osmotic pressureD. `0.100 M Mg_(3) (PO)_(2) (aq)` has the highest osmotic pressure. |
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Answer» Correct Answer - 3 we have `pi=iCRT` For `C_(2)H_(5)OH` which neither associates nor dissociates `pi=(1)(0.500 M)RT` For magbesium phosphate which dissociates `Mg_(3)(PO_(4))_(2)(aq) rarr 3 Mg^(2+)(aq)+2PO_(4)^(3-)(aq)` `pi=(5)(0.100 M) RT` For potassium bromide, which dissociates `KBr(aq) rarr K^(+)(aq) pi=(2)(0.250 M)RT` For sodium phosphate which dissociates `Na_(3)PO_(4)(aq) rarr 3Na^(+)(aq)+PO_(4)^(3-)(aq)` `pi=(4)(0.125 M) RT` Thus, all have same effective molarity and are isotonic solution. They all have some osmotic pressure at the same temperature. |
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| 833. |
Consider separate solutions of `0.500 M C_(2)H_(5)OH(aq)`,`0.100 M Mg_(3)(PO_(4))(aq)`,`0.250 M KBr(aq)`, and `0.125 M Na_(3)PO_(4)(aq)` at `25^(@)C`. Which statement is true about these solutions, assuming all salts to be strong electrolytes?A. `0.125 M Na_(3)PO_(4)(aq)` has the highest osmotic pressure.B. `0.500 M C_(2)H_(5)OH` has the highest osmotic pressure.C. They all have the same osmotic pressure.D. `0.100 M Mg_(3)(PO_(4))_(2)(aq)` has the highest osmotic pressure. |
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Answer» Correct Answer - C For `C_(2)H_(5)OH`, `pi=1 xx 0.5 xx RT =0.5 RT` For `KBr`, `pi=2 xx 0.25 xx RT =0.5 RT` For `Mg_(3)(PO_(4))_(2)`, `pi=5 xx 0.1 xx RT =0.5 RT` For `Na_(3)PO_(4)`, `pi=4 xx 0.125 xx RT =0.5 RT` So, all are isotonic solutions. |
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| 834. |
If the osmotic pressure of glucose solution is 1.52 bar at 300 K. What would be its concentration if R = 0.083L bar `"mol"^(-1)"K"^(-1)` ? |
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Answer» `pi = "CRT"` `R=0.0836."bar. mole"^(-1)"K"^(-1)` T = 300 K `pi = 1.52" bar"` `C = (pi)/("RT")=(1.52)/(0.083xx300)=(1.52)/(24.9)=0.061" M"` |
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| 835. |
A mixture of benzene and toluence formsA. An ideal solutionB. Non-ideal solutionC. SuspensionD. Emulsion |
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Answer» Correct Answer - A Mixing of benzene and toluene does not involve any kind of decrease and increase in interaction forces in between molecules. |
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| 836. |
The unit of ebillioscopic constant isA. K kg `mol^(-1) or ("molality")^(-1)`B. mol kg `K^(-1)or K^(-1)("molality")`C. kg `mol^(-1)K^(-1)or K^(-1)("molality")^(-1)`D. K `mol kg^(-1)or K ("molality")` |
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Answer» Correct Answer - A As we know from elevation in boiling point that `DeltaT_(b)K_(b)m` `K_(b)=(DeltaT_(b))/(m)` Unit of `K_(b)=("unit of "DeltaT_(b))/("unit of"m)=(K)/("molality")` `=(K)/(molkg^(1))=Kmol^(-1)kg` |
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| 837. |
Which of the following solutions will have highest boiling point:A. 1% glucose in waterB. 1% sucrose in waterC. `1%NaCI` in waterD. 1% urea in water |
| Answer» Correct Answer - C | |
| 838. |
Colligative properties depend on ____________.(i) the nature of the solute particles dissolved in solution.(ii) the number of solute particles in solution.(iii) the physical properties of the solute particles dissolved in solution.(iv) the nature of solvent particles. |
| Answer» (ii) the number of solute particles in solution. | |
| 839. |
The decreases in the freezing point of an aqueoues solution of a substance is `1.395K` and that in the freezing point of benzene solution of the same substance is `1.280K`. Explain the difference in `DeltaT`. The substance:A. dissociates in the aqueous solution as well as in the benzene solutionB. forms complexes in solutionC. associates in the benzene solutionD. dissociates in the aqueous solution and not in the benzene solution |
| Answer» Correct Answer - D | |
| 840. |
The colligative properties of a solution depend onA. The number of solute particles present in itB. The chemical nature of the solute particles present in itC. The nature of the solvent usedD. None of these |
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Answer» Correct Answer - A Colligative properties are properties of solution which depend on the number of particles present in solution. |
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| 841. |
The unit of ebillioscopic constant isA. K kg `mol^(-1)` or K `"(molality)"^(-1)`B. mol kg `K^(-1)` or `K^(-1)` (molality )C. kg `mol^(-1) K^(-1)` or `K^(-1) "(molality)"^(-1)`D. K mol `kg^(-1)` or K (molality ) |
| Answer» Correct Answer - A | |
| 842. |
Which of the following property does nto depend upon the number of solute particles only?A. Boiling point elevationB. Lowering in a vapour pressureC. Osmotic pressureD. Depression in freezing point |
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Answer» Correct Answer - B It is Relative lowering in vapour pressure which is a colligative property and not lowering in vapour pressure. |
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| 843. |
Considering the formation, breaking and strength of hydrogen bond, predict which of the following mixtures will show a positive deviation from Raoult’s law?(i) Methanol and acetone.(ii) Chloroform and acetone.(iii) Nitric acid and water.(iv) Phenol and aniline. |
| Answer» (i) Methanol and acetone. | |
| 844. |
In non-ideal solutions, at one of the intermediate compositions, the total vapour pressure is highest and the boiling point is lowest. At this point, the composition of the liquid and vapour phase is same. So, if liquid mixture vapourises at this point and vapours are condensed, the condensate contains same composition as present in orginal liquid mixture. It means that at this point liquid behaves like a pure and is called as Azeotropic mixture. Choose the correct answer : Choose the correct statemnentA. Ideal solutions connot be separated into their components by frectional distillationB. For ideal solution, entghalpy of mixing is always greater than zeroC. Only non-ideal solution showing positive devialtion cannot be separeated out by fration distillation.D. Non-ideal solution showing both positive and ngative deviation cannot be separated out by fractional distillation. |
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Answer» Correct Answer - d it is corrct answer. |
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| 845. |
Low concentration of oxygen in the blood and tissues of people living at high altitude is due to ____________.(i) low temperature(ii) low atmospheric pressure(iii) high atmospheric pressure(iv) both low temperature and high atmospheric pressure |
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Answer» (ii), [Hint : Body temperature of human beings remains constant.] |
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| 846. |
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. (7) water is added to the solution M such that the mole fraction so wter in solution becomes 0.9. The boiling point of solution is :A. `380.4 K`B. `376.2 K`C. 357.5 K`D. 245.7 K` |
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Answer» Correct Answer - b In this case, water is solvent and ethanol is solute `"Molality of solution (m)"=("No. of moles ethanol")/("Mass of water in kg")` `n_("ethanol")=0.1 `, `W_("water")=((0.9mol)xx(18" g mol"^(-1)))/1000=0.0162 kg.` `m=((0.1mol))/((0.0162 kg))=6.173 "mol kg"^(-1)` =6.173m `DeltaT_(b)=K_(b)xxm=(0.52"k kg mol"^(-1))xx(6.173" mol kg"^(-1))=3.2 K` b.p. of solution = 373+3.2= 376.2 K |
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| 847. |
Low concentration of oxygen in the blood ndtissues of people living at high altitude is due to…….A. low temperatureB. low atmospheric pressureC. high atmospheric pressureD. both low temperature and high atmospheric pressure |
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Answer» Correct Answer - B At high altitudes, the partial pressure of oxygen is less than that at the ground level. This leads to low concentrations of oxygen in the blood and tissues of people living at high altitudes or climbers. Low blood oxygen causes climbers to become weak and unable to think clearly , symptoms of a condition known as anoxia . |
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| 848. |
Normal freezing of a solvent is `5^(@)C`. A 0.5 molal solution of urea in this solvent causes a freezing point depression of two degree. Calculate molal depression constant. `(K_(f))` |
| Answer» `K_(f)=(DeltaT_(f))/m=(2K)/(0.5m)=4Km^(-1)` | |
| 849. |
Depression in freezing point for 1 M urea, 1 M NaCl and 1 M `CaCl_(2)` are in the ratio ofA. `1:2:3`B. `1:1:1`C. `3:2:1`D. Data insufficient |
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Answer» Correct Answer - 1 |
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| 850. |
Why is the vapous pressure of an aqueous solution of gulucose lower than that of water ? |
| Answer» Glucose occupies some surface area of water. However, id does not contribute to vapour pressure due to its non-volatile nature. Therefore, the vapour pressure of the solution is less than that of pure solvent i.e. water. | |