InterviewSolution
Saved Bookmarks
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.
| 1051. |
Lowering of vapour pressure of solution : (a) is a property of solute (b) is a property of solute as well as solvent (c) is a property of solvent (d) is a colligative property |
|
Answer» Option : (d) is a colligative property. |
|
| 1052. |
Colligative property depends only on ………………. in a solution. (a) Number of solute particles (b) Number of solvent particles (c) Nature of solute particles (d) Nature of solvent particles |
|
Answer» Option : (a) Number of solute particles |
|
| 1053. |
The addition of the nonvolatile solute into the pure solvent ……………..(a) increases the vapour pressure of solvent (b) decreases the boiling point of solvent (c) decreases the freezing point of solvent (d) increases the freezing point of solvent |
|
Answer» Option : (c) decreases the freezing point of solvent. |
|
| 1054. |
A solvent and its solution containing a nonvolatile solute are separated by a semipermable membrane. Does the flow of solvent occur in both directions? Comment giving reason. |
|
Answer» 1. When a solvent and a solution containing a non-volatile solute are separated by a semipermeable membrane, there arises a flow of solvent molecules from solvent to solution as well as from solution to solvent. 2. Due to higher vapour pressure of solvent than solution, the rate of flow of solvent molecules from solvent to solution is higher. 3. As more and more solvent passes into solution due to osmosis, the solvent content increases, and the rate of backward flow increases. 4. At a certain stage an equilibrium is reached where both the opposing rates become equal attaining an equilibrium. |
|
| 1055. |
Partial pressure of solvent in solution of non-volatile solute is given by equation, (a) P = x2P0(b) P0 = xP (c) P = x1P0(d) P = x1P0 |
|
Answer» Option : (c) P = x1P0 |
|
| 1056. |
Components of a binary mixture of two liquids A and B were being separated by distillation. After some time separation of components stopped and composition of vapour phase became same as that of liquid phase. Both the components started coming in the distillate. Explain why this happened. |
|
Answer» Since both the components are appearing in the distillate and composition of liquid and vapour is same, this shows that liquids have formed azeotropic mixture and hence cannot be separated at this stage by distillation. |
|
| 1057. |
Components of a binarey mixture of two liquids A and B were being separted by distillation. After some time separation of components stopped and composition of vapour phase vecame same as that of liquid phase. Both the components stated coming in the distillate. Explain why this happened ? |
| Answer» When the copsition of a sbinary mizture of two liquids A and B in the liquid state and in the vapour phase in the same, the mixture is known as azenotropic mixtue, In any axetropic mixture, the compoinents connot be separeted any more. For details, consults section 7. | |
| 1058. |
Benzene and toluene forms nearly an ideal solution. At 300 K, `P_("toluene")^(@)=32.06" mm"` and `P_("benzene")^(@)=103.01" mm"` (of Hg) (i) A liquid mixture is conposed of 3 mole of tolune and 2 mole of benzene. If the pressure over the mixture at 300 K is reduced, at what pressure does the first vapour form ? (ii) What is the composition of the first trace of vapour formed ? (iii) If the pressure is reduced futher,at what pressure does the last trace of liquid dissappear ? (iv) What is the composition of last trace of liquid ? |
|
Answer» (i) 60.44 mm Hg (ii) `x_("toluene")=0.3181` (iii) P=44.25 mmHg (iv)`x_("toluene")=0.8281` |
|
| 1059. |
What concentration of nitrogen should be present in a glass of water at room frmperatrure ? Assume a temperature of `25^(@)C`, a total pressure ofone atmosphere and molefraction of nitrogen in air as 0.78 `(K_(H)"for ni trogen"=8.42x10^(-7)M/mm Hg)`. |
|
Answer» Acoording to available information. `p_(N_2)=x_(N_2)"is air"xx P_("total")=0.78xx760 mm Hg=592.8 mm Hg,` `K_(H)=8.42xx10^(-7)M//mm Hg` Mole fraction of `N_(2)(x_(N_(2)))=K_(H)xxP_(N_2))` `=(8.42xx10^(-7)M/mm Hg)xx(592.8mmHg)` `=4991.376xx10^(-7)M=4.99xx10^(-4)M=4.99xx10^(-4) ` `x_(N_2)=n_(N_2)/(n_(N_2)+n_(H_(2)O))=n_(N_2)/n_(H_(2)O)or n_(H_2)=x_(N_2)xxn_(H_(2)O)` `(n_(N_2) "has been neglected in this expression since the gas isvery little soluble in water")` `n_(N_2)=(4.99xx10^(-4))xx(1000g//18 g mol^(-1))` ` =4.99xx10^(-4)xx55.55 `mol `=0.0277 mol=2.77xx10^(-4)` mol |
|
| 1060. |
The vapour pressure of pure benzene is `639.7 mm Hg` and the vapour pressure of solution of a solute in benzene at the temperature is `631.9 mm Hg`. Calculate the molality of the solution. |
|
Answer» Correct Answer - 0.162 m Solute-B, Benzene-A `"Let molality of solution=X"` `(P^(0)-P_(solution))/(P^(0))=X_(solute)` `(639.7-631.7)/(639.7)=X_(solute)` `m=(X_(solute))/((1-X_(solute)))xx1000/78` `"therefore molality of solution=0.162 m"` |
|
| 1061. |
Why are some solution processes exothermic whereas others are endothermic? |
| Answer» Correct Answer - The overall energy change associated with dissolution depends on the rolative magnitude of the solutesolute, solvent-solvent and solute-solvent interactions. The process is exothermic if the new interaction release more energy than disrupting the old interactions requires, it is endothermic if opoosite is true. | |
| 1062. |
Define mass percentage solution. |
|
Answer» The mass percentage of a component of a solution is defined as the `"mass "%=("Mass of the component in the solution")/("Total mass of the solution")xx100` |
|
| 1063. |
What is ppm of a solution ? |
|
Answer» ppm - parts per million : It is a convenient method of expressing concentration when a solute is present in trace quantities. Parts per million is defined as the `"ppm"=("Number of parts of the component")/("Total no. of parts of all components of the solution")xx10^(6)` |
|
| 1064. |
What is Ebullioscopic constant ? |
| Answer» Ebullioscopic constant : The elevation of boiling point observed in one molal solution containing non-volatile solute is called Ebullioscopic constant (or) molal elevation constant. | |
| 1065. |
Which of the following gas minimum equivalent volume?A. one equivalent of hydrogenB. one equivalent of oxygenC. one equivalent of nitrogenD. all have volumes |
|
Answer» Correct Answer - 3 Equivalent volume of any gas is the volume occupied by one equivalent of gas at NTP (or STP). We all know that 1 mole of molecules of any gas occupies 22.4 L at NTP and 1 mole of atoms any diatomic gas `(H_(2), O_(2)` and `N_(2))` occupies 11.2 litres at NTP. Since 1 equivalent of `H_(2) = 1 mol` of `H` atoms or `1//2` mol of `H_(2)` molecules 1 equivalent of `O_(2)=1/2` mol of `O` atoms or `1//4` mol of `O_(2)` molecules 1 equivant of `N_(2)=1/3` mol of `N` atoms or `1//6` mol of `N_(2)` molecules We have `(at NTP)` Volume of 1 equivalent of `H_(2)(g)=11.2L` Volume of 1 equivalent of `O_(2)(g)=5.6 L` Volume of 1 equivalent of `N_(2)(g)=3.73 L` through the knowledge for a gas as follows: No of equivalents of gas `=("Volume of gas at NTP")/("Equivalent volume of gas")` Note that different number of molecules are present in 1 equivalent of each of these gases: i equivalent of `H_(2)=(6.022xx10^(23))/2 H_(2)` molecules 1 equivalent of `O_(2)=(6.022xx10^(23))/4 O_(2)` molecules 1 equivalent of `N_(2)=(6.022xx10^(23))/6 N_(2)` molecules |
|
| 1066. |
If `m_(1)` and `m_(2)` are masses of two reactants in any reaction, having their gram equivalent masses `E_(1)` and `E_(2)` respectively, which of the following equatios represents the law of equivence correctly?A. `m_E_=m_(2)E_(2)`B. `m_(1)m_(1)=E_(1)E_`C. `m_(1)E_(2)=m_(2)E_(1)`D. `sqrt(m_(1)//E_(1))=sqrt(m_(2)//E_(2))` |
|
Answer» Correct Answer - 3 According to the law of equivalence, the number of equivalent of different substances (reactants and products) involved in a given chemical reaction are always equal. Thus `m_(1)/E_(1)=m_(2)/E_(2)` or `m_(1)E_(2)=m_(2)E_(1)` |
|
| 1067. |
Two liquids X and Y from an ideal solution at`300K`, Vapour pressure of the Solution containing `1`mol of X and `3`mol of Y is `550`mmHg . At the same temperature, if `1` mol of Y is further added to this solution ,vapour pressur of the solutions increases by `100`mmHg Vapour pressure (in mmHg) of X and Y in their pure states will be,respectivelyA. `300` and 400`B. `400` and `600`C. `500` and `600`D. `200` and `300` |
|
Answer» Correct Answer - B `P_("total")=P_(A)^(@)x_(A)+P_(B)^(@)x_(A)` `760=520x_(A)+1000(1-x_(A))` `760=520x_(A)+1000-1000x_(A)` `x_(A)=0.5` mol `%=50%` |
|
| 1068. |
Match the items given in Column I and Column II.Column IColumn II(i) Saturated solution(a) Solution having same osmotic pressure at a given temperature as that of given solution.(ii) Binary solution(b) A solution whose osmotic pressure is less than that of another.(iii) Isotonic solution(c) Solution with two components.(iv) Hypotonic solution(d) A solution which contains maximum amount of solute that can be dissolved in a given amount of solvent at a given temperature.(v) Solid solution(e) A solution whose osmotic pressure is more than that of another.(vi) Hypertonic solution(f) A solution in solid phase. |
|
Answer» (i) → (d) (ii) → (c) (iii) → (a) (iv) → (b) (v) → (f) (vi) → (e) |
|
| 1069. |
Match the items given in Column I with the type of solutions given in Column II.Column IColumn II(i) Soda water(a) A solution of gas in solid(ii) Sugar solution(b) A solution of gas in gas(iii) German silver(c) A solution of solid in liquid(iv) Air(d) A solution of solid in solid(v) Hydrogen gas in palladium(e) A solution of gas in liquid(f) A solution of liquid in solid |
| Answer» (i) → (e) (ii) → (c) (iii) → (d) (iv) → (b) (v) → (a) | |
| 1070. |
Match the laws given in Column I with expresions given in Column II.Column IColumn II(i) Raoult’s law(a) ΔTf = Kfm(ii) Henry’s law(b) Π = CRT(iii) Elevation of boiling point(c) p = x1p10 + x2p20(iv) Depression in freezing point(d) ΔTb = Kbm(v) Osmotic pressure(e) p = KH.x |
| Answer» (i) → (c) (ii) → (e) (iii) → (d) (iv) → (a) (v) → (b) | |
| 1071. |
Compartment `A` and `B` have the following combinations of solution: `{:(,A,B),(1,0.1 M KCl,0.2 M KCl),(2,0.1%(m//V) NaCl,10% (m//V) NaCl),(3,18 gL^(-1) "glucose",34.2 gL^(-1) "sucrose"),(3,20% (m//V) "glucose",10% (m//V) "glucose"):}` Indicate the solution(s) in which compartment `B` will show an increases in volume.A. `1,2,4`B. `1,2`C. `2,3`D. `3,4` |
|
Answer» Correct Answer - B In hypertonic solution osmotic pressure is higher, therefore, volume rise is higher in `B`. |
|
| 1072. |
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 to ethanol `(K_(f)^(ethanol))=2.0 K kg mol^(-1))` Boiling point elevation constant 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 = `351.5 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 anwering the following questions consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative. The freezing point of the solution `M` isA. `268.7 K`B. `268.5 K`C. `150.9 K`D. `268.7 K` |
|
Answer» `chi_("Ethanol")=0.9` ` chi_("water")=0.1` `("Solvent" =-B)` `("Solvent" =-A)` `DeltaT_(f)=K_(f) .m=K_(f) . (chi_(B))/(1-chi_(B)) xx (1000)/(Mw_(A))` `=2 xx 0.1/0.9 xx 1000/46=4.86 K` `T_(f)=T_(f)^(@)-DeltaT_(f)=155.7 + 4.83 =150.87 K` |
|
| 1073. |
The boiling point elevation and freezing point depression of solutions have a number of partical applications. Ethylene glycol `(CH_(2)OH-CH_(2)OH)` is used in automobile radiatiors as an antifreeze because it lowers the freezing point of the coolant. The same substance also helps to prevent the radiator coolant from boiling away by elevating the boiling point. Ethylene glycol has low vapour pressure. We can also use glycerol as an antifreeze. In order for the boiling point elevation to occur, the solute must be non-volatile, but no such restriction applies to freezing point depression. For example, methanol `(CH_(3)OH)`, a fairly volatile liquid that boils only at `65^(@)C`, is sometimes used as an antifreeze in automobile radiators. If the cost of glycerol, glycol, and methanol is same, then the sequence of economy to use these compounds as antifreeze will beA. Glycerol gt Glycol gt MethanolB. Methanol gt Glycol gt GlycerolC. Methanol = Glycol = GlycerolD. Methanol gt Glycol `lt` Glycerol |
|
Answer» Correct Answer - B Methanol is lighter and has better depression in freezing point. |
|
| 1074. |
A `1.24 M` aqueous solution of `KI` has density of `1.15 g cm^(-3)`. Answer the following questions about this solution: What is the freezing point of the solution if `KI` is completely dissociated in the solution?A. `-4.87^(@)C`B. `-3.22^(@)C`C. `-1.22^(@)C`D. None of these |
| Answer» Correct Answer - A | |
| 1075. |
A `1.24 M` aqueous solution of `KI` has density of `1.15 g cm^(-3)`. Answer the following questions about this solution: The percentage composition of solute in the solution isA. `17.89`B. `27.89`C. `37.89`D. `47.89` |
| Answer» Correct Answer - A | |
| 1076. |
A `1.24 M` aqueous solution of `KI` has density of `1.15 g cm^(-3)`. Answer the following questions about this solution: The molality of this solution will beA. `2.61`B. `1.31`C. `4.12`D. `3.12` |
| Answer» Correct Answer - B | |
| 1077. |
A `1.24 M` aqueous solution of `KI` has density of `1.15 g cm^(-3)`. Answer the following questions about this solution: The experimental freezing point of the solution is `-4.46^(@)C`. What percentage of `KI` is dissociated?A. `82%`B. `90%`C. `83%`D. None |
| Answer» Correct Answer - C | |
| 1078. |
The boiling point elevation and freezing point depression of solutions have a number of partical applications. Ethylene glycol `(CH_(2)OH-CH_(2)OH)` is used in automobile radiatiors as an antifreeze because it lowers the freezing point of the coolant. The same substance also helps to prevent the radiator coolant from boiling away by elevating the boiling point. Ethylene glycol has low vapour pressure. We can also use glycerol as an antifreeze. In order for the boiling point elevation to occur, the solute must be non-volatile, but no such restriction applies to freezing point depression. For example, methanol `(CH_(3)OH)`, a fairly volatile liquid that boils only at `65^(@)C`, is sometimes used as an antifreeze in automobile radiators. `620 g` glycol is added to `4 kg` water in the radiator of car. What amount of ice will separate out at `-6^(@)C`?A. `800 g`B. `900 g`C. `600 g`D. `1000 g` |
| Answer» Correct Answer - B | |
| 1079. |
The boiling point elevation and freezing point depression of solutions have a number of partical applications. Ethylene glycol `(CH_(2)OH-CH_(2)OH)` is used in automobile radiatiors as an antifreeze because it lowers the freezing point of the coolant. The same substance also helps to prevent the radiator coolant from boiling away by elevating the boiling point. Ethylene glycol has low vapour pressure. We can also use glycerol as an antifreeze. In order for the boiling point elevation to occur, the solute must be non-volatile, but no such restriction applies to freezing point depression. For example, methanol `(CH_(3)OH)`, a fairly volatile liquid that boils only at `65^(@)C`, is sometimes used as an antifreeze in automobile radiators. Which among the following is the most volatile and the best antifreeze?A. `CH_(3)OH`B. `C_(2)H_(5)OH`C. GlycolD. Glycerol |
| Answer» Correct Answer - A | |
| 1080. |
A `1.24 M` aqueous solution of `KI` has density of `1.15 g cm^(-3)`. Answer the following questions about this solution: The normality of the solution isA. `0.62`B. `1.24`C. `2.48`D. `3.72` |
| Answer» Correct Answer - B | |
| 1081. |
Two liquids, `A` and`B` form an ideal solution. At the specified temperature, the vapour pressure of pure `A` is `20 mm Hg` while that of pure `B` is `75 mm Hg`. If the vapour over the mixture consists of `50` mol percent `A`, what is the mole percent `A` in the liquid? |
| Answer» Correct Answer - `27.3` mole `%` | |
| 1082. |
During the preparation of solution for intravenous injection which essential factor should be kept in mind? |
|
Answer» The solution should be isotonic with RBC. |
|
| 1083. |
Which of the following concentration terms is independent on temperature ?A. NormalityB. Mass-volume percentC. MolalityD. Molarity |
| Answer» Correct Answer - C | |
| 1084. |
Name the substance used as cell membrane in reverse osmosis. |
|
Answer» Cellulose acetate |
|
| 1085. |
Solubility curve of `Na_(2)SO_(4).10H_(2)O` in water with temperature is given as A. Solution process is exothermicB. Solution process is exothermic till `32.4^(@)C` and endothermic after `32.4^(@)C`C. Solution process is endothermic till `32.4^(@)` and exothermic thereafterD. Solution process is endothermic |
|
Answer» Correct Answer - C Below `34^(@)C, Na_(2)SO_(4).10H_(2)O` exists as such, since in this part solubility increases. This implies that process must be endothermic. Thereafter, solubility decrease with rise in temperature. It means that the solution process is exothermic. In fact after `34^(@)C` the substance exists in anhydrous state `Na_(2)SO_(4)`. |
|
| 1086. |
When `5` gram of `BaCI_(2)` is dissolved in water to have `10^(6)` gram of solution. The concentration of solution is:A. 2.5 ppmB. 5 ppmC. 5MD. `5gm L^(-1)` |
|
Answer» Correct Answer - B ppm `= (5)/(10^(6)) xx 10^(6) = 5` |
|
| 1087. |
Cryoscopic constant of a liquidA. is the decrease in freezing point when `1`g of solute is dissolved per kg of the solventB. is the decrease in the freezing point when `1`mole of solute of dissolved per kg of the solventC. is the elevation for `1` molar solutionD. is a factor used for calculation of depression in freezing point |
| Answer» Correct Answer - B | |
| 1088. |
Cryoscopic constant of a liquid is:A. decrease in freezing point when 1 g of solute is dissolved per kg of the solventB. decrease in freezing point when 1 mole of solute is dissolved per kg of the solventC. the elevation for 1 molar solutionD. a factor used for calculation of elevation in boiling point. |
| Answer» Correct Answer - B | |
| 1089. |
Why is Anoxia disease very common at higher altitudes? |
|
Answer» As concentration of oxygen is less in air at high altitude. |
|
| 1090. |
Assertion (A): `0.1 M` solution of glucose has same increment in freezing point than has `0.1 M` solution of urea. Reason (R ): `K_(f)` for both has different value.A. If both `(A)` and `(R )` are correct, and `(R )` is the correct explanation of `(A)`.B. If both `(A)` and `(R )` are correct, but `(R )` is not the correct explanation of `(A)`.C. If `(A)` is correct, but `(R )` is incorrect.D. If `(A)` is correct, but `(R )` is correct. |
|
Answer» Correct Answer - C `K_(f)` is same for both solutions. |
|
| 1091. |
The degree of dissociation `alpha` of a week electrolyte is where `n` is the number of ions given by `1 mol` of electrolyte.A. `(i-1)/(n+1)`B. `(i-1)/(n-1)`C. `(n-1)/(i-1)`D. `(n+1)/(i-1)` |
|
Answer» Correct Answer - B `nA hArr A_n " " A_n hArr nA` `1 , " " 1` `(1- alpha, alpha//n) " " (1-alpha , nalpha)` `i=1-alpha+alpha//n " " i=1-alpha+nalpha` `rArr alpha=(i-1)/(n-1)` |
|
| 1092. |
The freezing point depression of a 0.10 m solution of HF (aq) solution is -0.201 °C. Calculate the percent dissociation of HF (aq). |
|
Answer» ΔTf = i x Kf x m 0.201 = i x 1.86 x 0.1 i = 0.201/0.186 = 1.08 α = (i-1)/( n-1) =1.08 – 1 = 0.08 % dissociation of HF is 8% |
|
| 1093. |
Calculate the freezing point depression expected for 0.0711 m aqueous solution of Na2SO4. If this solution actually freezes at -0.320°C, what would be the value of Van’t Hoff factor? |
|
Answer» ΔTf = Kf x m = 1.86 x .0711 = .132°C i = .320/.132 = 2.42 |
|
| 1094. |
A `5.25%` solution of a substance is isotonic with a `1.5%` solution of urea (molar mass=`60gmol^(-1)`)in the same solvent.If the densities of both the solutions are assumed to be equal to `1.0gcm^(-3)`,molar mass of the substance will be:A. `90.0 g "mol"^(-1)`B. `115.0 g "mol"^(-1)`C. `105.0g "mol"^(-1)`D. `210.0g "mol"^(-1)` |
|
Answer» Correct Answer - D Solution with the same osmotic pressure are isotonic Let the molar mass of the substance be M `pi_(1)=C_(1)RT=C_(2)RT=pi_(2)` `so,C_(1)=C_(2)` As density of the solution are same So,`(5.25)/(M)=(15)/(60)` `:.M=(5.25xx60)/(1.5)=210` |
|
| 1095. |
300 mL solution at 27 °C contains 0.2 mol of a nonvolatile solute. Calculate osmotic pressure of the solution. |
|
Answer» Given : V = 300 ml = 0.3 dm3 ; T = 273 + 27 = 300 K π = 0.2 mol, π = ? π = \(\frac{nRT}{V}\) = \(\frac{0.2\times0.08206\times 300}{0.3}\) = 16.41 atm ∴ Osmotic pressure = π = 16.41 atm. |
|
| 1096. |
Which statement is false?A. Two sucrose solutions of same molality prepare different solvents have same`DeltaT_(f)`B. Osmotic pressure ,`Pi=MRT`C. Osmotic pressure for `0.01`M aqueous solutions:`BaCl_(2)gtKClgtCh_(3)COOHgt` SucroseD. The vapour pressure of a component over a solution proportional to its mole fraction. |
|
Answer» Correct Answer - A `K_(f)`of solvents is altogether different. |
|
| 1097. |
The osmotic pressure of 0.1 M HCl solution at 27 °C will be : (a) 2.46 atm (b) 0.164 atm (c) 4.92 atm (d) 0.0082 atm |
|
Answer» Option : (c) 4.92 atm |
|
| 1098. |
A solution of glucose `(C_(6)H_(12)O_(6))`is isotonic with `4`g of urea `(NH_(2)-CO-NH_(2))` per litre of solution .The concentration of glucose is :A. `4g//L`B. `8g//L`C. `12g//L`D. `14g//L` |
|
Answer» Correct Answer - C Isotonic solution has same conc. `pi_(1)=pi_(2)C_(1)=C_(2)n_(1)=n_(2)(W_(1))/(M_(1))=(W_(2))/(M_(2))` `So ,(x)/(180)=(4)/(60)rArr x=12g` |
|
| 1099. |
If a, b, c and d are the van’t Hoff factors for Na2SO4, glucose and K4[Fe(CN)6] then,(a) a > b > c(b) a < b < c (c) b < a < c (d) c < a < b |
|
Answer» Option : (c) b < a < c c)
|
|
| 1100. |
Two solution (A)containing`Fecl_(3)(aq)`and(B) semipermeable membrance as shown below.If `FeCl_(3)`on reaction with `K_(4)[Fe(CN)_(6)]`,produces blue colour of `Fe_(4)[Fe(CN)_(6)]`, the clue colour will be noticed in : `|((A),,(B)),(FeCl_(3),,K_(4)[Fe(CN)_(6)])|`A. (A)B. (B)C. in Both (A)and(B)D. neither in (A)nor in(B) |
|
Answer» Correct Answer - D In osmosis only solvent particles move. |
|