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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. |
Which of the following is wrong:A. Enthalpy (numerical value) of physisorption is greater than that of chemisorptionB. Physisorption is not very specific but chomisorption is highly specificC. Chemisorption takes place at relatively high temperaturesD. In physisorption generally multi-molecular layers are formed on the adsorbent |
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Answer» Correct Answer - A `DeltaH=20-40 kJ mol^(-1)` for physisorption and `DeltaH=200-400s kJ mol^(-1)` for chemisorption. |
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| 252. |
0.15g of a substance dissolved in 15g of a solvent boiled at a temp. higher by `0.216^(@)C` than that of the pure solvent. Find out the molecular weight of the substance `(K_(b)` for solvent is `2.16^(@)C`)A. 1.01B. 10.1C. 100D. 10 |
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Answer» Correct Answer - C `M = (K_(b).W_(b).1000)/(DeltaT_(b).W_(A))` `=(2.16 xx 0.15 xx 1000)/(216 xx 15) = 100` |
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| 253. |
The arsenous sulphide sol has -ve charge. The maximum power of precipitating is of:A. `H_(2)SO_(4)`B. `Na_(3)PO_(4)`C. `CaCl_(2)`D. `AlCl_(3)` |
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Answer» Correct Answer - D Power of precipitating `prop` charge on cation. |
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| 254. |
`0.15`g of a subatance dissolved in `15`g of solvent boiled at a temperature higher at `0.216^(@)`than that of the pure solvent. Calculate the molecular weight of the substance. Molal elecation constant for the solvent is `2.16^(@)C`A. 216B. 100C. 178D. None of these |
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Answer» Correct Answer - B Here it is given that `"w"=0.15"g, " Delta"T"_("b")=0.216^(@)"C"` `"W"=15"g""K"_("b")=2.16^(@)"C ""m"?` Substituting values in the expression, `"m"=(1000xx"K"_("b")xx"w")/(Delta"T"_("b")xx"")" m"=(1000xx2.16xx0.15)/(0.216xx15)=100` |
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| 255. |
Liquid-liquid sol is known asA. aersolB. foamC. emulsionD. gel |
| Answer» Correct Answer - C | |
| 256. |
Which of the following is (are) lyophobic colloids ?A. Gold solB. `As_(2) S_(3)` solC. Starch solD. `Fe(OH)_(3)` sol |
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Answer» Correct Answer - A::B::D Gold sol, `As_(2)S_(3)` and `Fe(OH)_(3)` are lyophobic colloid. Therefore, (A, B, D) are correct options. |
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| 257. |
For coagulating `As_(2)S_(3)` colloidal sol, which of the following will have the lowest coagulation valueA. NaClB. KClC. `BaCl_(2)`D. `AlCl_(3)` |
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Answer» Correct Answer - D `As_(2)S_(3)` is negatively charged sol so more positively charged ion will have minimum coangulating value. |
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| 258. |
Lyophobic colloids are :A. Reversible colloidsB. Irreversible colloidsC. Protective colloidsD. Gum proteins |
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Answer» Correct Answer - B Lyophobic colloids are irreversible colloids. |
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| 259. |
Which of the following ions will be most effective in coagulating the `As_(2)S_(3)` sol?A. `Fe^(3+)`B. `Ba^(2+)`C. `Cl^(-)`D. `PO_(4)^(3-)` |
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Answer» Correct Answer - A According Hardy-Schulze rule |
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| 260. |
Positive sol is :A. GoldB. GelatinC. `As_(2)S_(3)`D. None |
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Answer» Correct Answer - B Gelatin is positive sol. |
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| 261. |
Smoke is an example of :A. Gas dispersed in liquidB. Gas dispersed in solidC. Solid dispersed in gasD. Solid dispersed in solid |
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Answer» Correct Answer - C Smoke is an example of solid dispersed in gas |
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| 262. |
Calculate the molal elevation constant of water evaporates at `100^(@)"C"` with the absorption of 536 calories per gm `("R"=2 "cals")`A. `0.519^(@)"C"`B. `0.0519^(@)"C"`C. `1.519^(@)"C"`D. `2.519^(@)"C"` |
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Answer» Correct Answer - A Molal elevation constant of the solvent. `"K"_(b)=("RT"_(b)^(2))/(l_("v")xx100)=(2xx373xx373)/(536xx1000)=0.519^(@)"C"` |
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| 263. |
An example of extrinsic colloid (lyophobic colloids) is :A. `As_(2)S_(3)` solB. `Fe(OH)_(3)` solC. Egg albuminD. Au sol |
| Answer» Correct Answer - A::B::D | |
| 264. |
which of the following will have the highest coagulating power for `As_(2)S_(3)` colloid?A. `PO_(4)^(3-)`B. `SO_(4)^(2-)`C. `Na^(+)`D. `Al^(3+)` |
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Answer» Correct Answer - D `As_(2)S_(3)` is -ve charged colloidal solution. Coagulation power `prop` charged on cation |
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| 265. |
Which one is a lyophobic colloid?A. GelatinB. StarchC. SulphurD. Gum arabic |
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Answer» Correct Answer - C Sulphur is a lyophobic colloid. |
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| 266. |
Which of the following is property of colloid?A. Scattering of lightB. Shows attractionC. DialysisD. Emulsion |
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Answer» Correct Answer - A Lyophilic colloidal particles are solvated by dispersion medium. |
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| 267. |
The clouds consist of charged particles of water dispersed in air. Some of them are +vely charged, other are -vely charged. When +vely charged clouds come closer they cause lightening and thundering whereas when +ve and -ve charged colloid come closer they cause heavy rain by aggregation of minute particles. It is possible to cause artificial rain by throwing electrified sand or silver iodic from an aeroplane and thus coagulating the mist hanging in air. Smoke screen is a cloud of smoke used to hide military, naval police etc. It consists of fine particles of `TiO_(2)`. When excess of `AgNO_(3)` is treated with KI solution, AgI forms :A. `+ve` charges solB. `-vely` charged solC. neutral solD. true solution |
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Answer» Correct Answer - A `AgNO_(3)("excess")+KI rarr Agl overset(+AgNO_(3) ("remaining"))(rarr)AgI//Ag^(+)` |
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| 268. |
Which of the following statement is not correct?A. A colloidal solution is a heterogeneous two-phase systemB. Silver sol in water is an example of lyophilic solutionC. Metal hydroxides in water are examples of lyophobic solutionD. Liquid-liquid colloidal solution is not a stable system |
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Answer» Correct Answer - B Silver sol in water is an example of lyophobic solution. |
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| 269. |
Lyophilic sols areA. Irrebersible solsB. They are prepared from inorganic compoundC. Coagulated by adding electrolytesD. Self-stabilizing |
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Answer» Correct Answer - D In lyophilic sols the dispersed phase have great affnity (attraction) towards dispersion medium. So they are self stabilizing. |
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| 270. |
The stability of lyophilic is due to which of the following:A. Charge on their particlesB. Large size of their particlesC. Small size of their particlesD. Solvation by dispersion medium |
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Answer» Correct Answer - D Lyophilic colloid is solvated by dispersion medium and becomes more stable. |
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| 271. |
Many lyophilic sols and few lyophobic sols when coagulated under some special conditions changes into semi rigid mass, enclosing whole amount of liquid within of liquid within itself, it is called gel and the process is called gelation. Gelatin, agar-agar, gum-Arabic can be converted into gels by cooling them under moderate concentration conditions. Hydrophobic sols like silicic acid. `Al(OH)_(3)` are prepared by double decomposition and exchanged of solvent method. Types of gel : 1. Syneresis/weeping of gel : The spontaneous liberation of liquid from a gel is called syneresis or weeping of gels. It is reverse of swelling. e.g., geletin, agar-agar show syneresis at low concentration while silicic shows it at high concentration. 2. Imbibition or swelling of gel : When gel is kept in a suitable liquid (water) it absorb large volume of liquid. The phenomenon is called imbibition or sweeling of gel. 3. Thixotropic : Some gels when shaken to form a sol, on keeping changes into gel are termed as thixotropic gel and phenomenon is called thixotropy. e.g., gelatin and silica liquify on shaking changing into corresponding sol and the sol on keeping changes back into gel. Some types of gels like gelatin and silica liquify on shaking thereby changing into sols. The sols on standing change back into gels. This process is know as :A. synersisB. thixotropyC. double decompostionD. peptization |
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Answer» Correct Answer - B Interconversion of sol and gel is known as thiixotropy. |
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| 272. |
Which of the following statement is correct for a lyophilic solution?A. It is not easily solvatedB. It is unstableC. The coagulation of this sol is irreversible in natureD. It is quite stable in a solvent |
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Answer» Correct Answer - D Lyophilic solution is easily solvated and quite stable in solvent. |
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| 273. |
Graph between `log x//m` and `log p` is a straight line inclined at an angle of `45^@` . When pressure is `0.5 atm` and `1n k = 0.693`, the amount of solute adsorbed per gram of adsorbent will be:A. `1`B. `1.5`C. `0.25`D. `2.5` |
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Answer» Correct Answer - A logx/M=log K+1/n log P` `1/n=tan 45^(@)" "ln k=0.69` `n=1" "k=2` `x/m=2xx(0.5)^(1)` `x=1` |
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| 274. |
Match the boiling point with `K_(b)` for x,y and z, if molecular weight of x,y and z are same. `{:(,b.pt.,"K"_("b")),(x,100,0.68),(y,27,0.53),(z,253,0.98):}` |
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Answer» Molal elevation constant may be calculated as, `K_(b_(1000))=(RT_(0)^(2))/(1000L_(V))" "("where", T_(0)="boiling point of pure solvent "L_(V)= "latent heat of vaporization per gram "L_(V) =(DeltaH_(V))/(m_(B)))` `= (RT_(0)^(2))/(1000(DeltaH_(V))/(m_(B)))` `=(RT_(0)^(2)m_(B))/(1000DeltaH_(V))" "("here",DeltaH_(V)= "molar latent heat of vaporisation "m_(B)="molar mass of solute")` `K_(b_(1000))=(RT_(0)m_(B))/(1000DeltaS_(V))` Since, `DeltaS_(V)=(DeltaH_(V))/(T_(0))` here, `DeltaS_(V)=` entropy of vaporization By considering `Delta S_(V)` as almost constant, `K_(b) prop T_(0)` `:. K_(b)(x)=0.68,K_(b)(y)=0.53andK_(b)(z)=0.98`. |
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| 275. |
Two solutions of `KNO_(3)` and `CH_(3)COOH` are prepared separately. The molarity of both is `0.1 M` and osmotic pressure is `P_(1)` and `P_(2)`, respectively. The correct relationship between the osmotic pressure isA. `P_(2)gtP_(1)`B. `P_(1)=P_(2)`C. `P_(1)gtP_(2)`D. `P_(1)/(P_(1)+P_(2))=P_(2)/(P_(1)+P_(2))` |
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Answer» Correct Answer - C `KNO_(3)` is `100%` ionized while `CH_(3)COOH` is a weak elecrolyte. |
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| 276. |
`18g` glucose `(C_(6)H_(12)O_(6))` is added to `178.2g` water. The vapour pressure of water (in torr) for this aqueous solution is:A. 7.6B. `76.0`C. 752.4D. `759.0` |
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Answer» Correct Answer - C `(Deltap)/(p_(0))=x_(B)` `Deltap=x_(B)xxp_(0)` `=(n_("glucose"))/(n_("water")+n_("glucose"))xxp_(0)` `Deltap=(0.1)/(9.9+0.1)xx760` `p_(0)-p=7.6` `p=p_(0)-7.6` `=760-7.6` `=752.4` torr] |
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| 277. |
`18g` of glucose `(C_(6)H_(12)O_(6))` is added to `178.2g` of water. The vapour pressure of water for this aqueous solution at `100^(@)C` is-A. `7.60` TorrB. `76.00` TorrC. `752.40` TorrD. `759.00` Torr |
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Answer» Correct Answer - C `PA = P_(A)^(@) x_(A) = 760 xx (178.2//18)/((178.2)/(18)+(18)/(180))` `= 760 xx (9.9)/(10) = 752.4` torr |
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| 278. |
Elevation in boiling point of a molar `(1M)`glucose solution`(d=1.2 g ml_(-1))` isA. `1.34K_(b)`B. `0.98K_(b)`C. `2.40K_(b)`D. `K_(b)` |
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Answer» Correct Answer - B Molarity =`1M=1moL^(-1)` `1000`mL of solution has glucose =`180g` `(1000xx1.2)g` of solution has glusose =`180g` `:.` Solute`(w_(1))=180g` solution =`1200g` solvent ,`H_(2)O(w_(2))=1200-180=1020g` `:. DeltaT_(b)=(1000K_(b)w_(1))/(m_(1)w_(2))` `(1000xxk_(b)xx180)/(180xx1020)=0.98K_(b)` |
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| 279. |
Elevation of boiling point of 1 molar aqueous glucose solution `("density"=1.2g//ml)` isA. `"K"_("b")`B. 1.20 `"K"_("b")`C. 1.02`"K"_("b")`D. 0.98`"K"_("b")` |
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Answer» Correct Answer - D Molality=1 `therefore "n"_("glucose")=1` `"W"_("glucose")=180"gm"` `"V"_("solution")=1000"mL"` `"W"_("solution")=1000xx1.2` =1200 gm `"W"_("solution")=1200-180=1020"gm"` `Delta"T"_("b")="K"_("b")xx("n"_("solute"xx1000))/("W"_("solvent"))=(1xx1000)/1020` `Delta"T"_("b")=0.98"K"_("b")` |
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| 280. |
The specific conducitivity of a 0.5 M aq. Solution of monobasic acid HA at `27^(@)"C"` is 0.006 `"Scm"^(1)`. It’s molar conductivity at infinite dilution is 200 S `"cm"^(2)"mol"^(-1)`. Calculate osmotic pressure (in atm) of 0.5 M HA(aq) solution at `27^(@)"C"`. Given R`=0.08"L""-atm"//"K""-mol"` |
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Answer» Correct Answer - 12.72 `"A"_("m")^("C")=("K"xx1000)/"C"=(0.006xx1000)/0.500=12" S cm"^(3)"mol"^(-1)` `"A"_("m")^("0")=200"scm"^(2)"mol"^(-1)` `alpha=("A"_("m")^("C"))/("A"_("m")^("0"))=0.06` For HA, `"n"=2` `"i"=1+("n"-1)alpha` `=1+(2-1)xx0.6` `"i"=1.06` `pi="i"xx"CRT"` `=1.06xx0.5xx0.082xx300` `pi=12.72"atm"` |
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| 281. |
Assertion : Greater the molal depression constant of the solvent used, less is the freezing point of the solution are different. Reson : Depression in freezing point depends upon the natur of the solvent.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 - a Reason is the correct explanation for assertion. |
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| 282. |
The values of Van’t Hoff factors for KCl, NaCl and K2SO4, respectively, are _____________.(i) 2, 2 and 2(ii) 2, 2 and 3(iii) 1, 1 and 2(iv) 1, 1 and 1 |
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Answer» (ii) In reverse osmosis, solvent molecules move through a semipermeable membrane from a region of lower concentration of solute to a region of higher concentration. |
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| 283. |
Which of the following statements is false? (i) Two different solutions of sucrose of same molality prepared in different solvents will have the same depression in freezing point. (ii) The osmotic pressure of a solution is given by the equation Π = CRT ( where C is the molarity of the solution). (iii) Decreasing order of osmotic pressure for 0.01M aqueous solutions of barium chloride, potassium chloride, acetic acid and sucrose is BaCl2 > KCl > CH3COOH > sucrose. (iv) According to Raoult’s law, the vapour pressure exerted by a volatile component of a solution is directly proportional to its mole fraction in the solution. |
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Answer» (i) Two different solutions of sucrose of same molality prepared in different solvents will have the same depression in freezing point. |
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| 284. |
At a given temperature, osmotic pressure of a concentrated solution of a substance __A. is higher than that at a dilute solutionB. is lower than that of a dilute solutionC. is same as that of a dilute solutionD. cannot be compared with osmotic pressure of dilute solution |
| Answer» Correct Answer - A | |
| 285. |
At a given temperature, osmotic pressure of a concentrated solution of a substance _____________.(i) is higher than that at a dilute solution.(ii) is lower than that of a dilute solution.(iii) is same as that of a dilute solution.(iv) cannot be compared with osmotic pressure of dilute solution. |
| Answer» (i) is higher than that at a dilute solution. | |
| 286. |
The unit of ebulioscopic constanat isA. `"K kg "mol^(-1)"or "K^(-1)("molality")^(-1)`B. `"mol kg "K^(-1)"or "K("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 is the correct answer. |
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| 287. |
The unit of ebulioscopic constant is _______________.(i) K kg mol–1 or K (molality)–1(ii) mol kg K–1 or K–1(molality)(iii) kg mol–1 K–1 or K–1(molality)–1(iv) K mol kg–1 or K(molality) |
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Answer» (i) K kg mol–1 or K (molality)–1 |
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| 288. |
An unripe mango placed in a concentrated salt solution to prepare pickle, shrivels because _____________.(i) it gains water due to osmosis.(ii) it loses water due to reverse osmosis.(iii) it gains water due to reverse osmosis.(iv) it loses water due to osmosis. |
| Answer» (iv) it loses water due to osmosis. | |
| 289. |
Why is freezing point depression of 0.1 M sodium chloride solution nealy twice than that of 0.1 M glucose solution ? |
| Answer» Depression in freezing point temperature `(DeltaT_(f))` is a colligative progative propety and it depends upon the number of the particles of the solute in the solution.Sodium chloride (NaCL) being a strong electrolyte almost completely dissociates in aqueous solution into ions. On the other hand, glucose `(C_(6)H_(12)O_(6))` being a molecular solid remains as such. Therefore. depresssion in freezing point in sodium chlution is expected to be twice as compered to glucose solution. | |
| 290. |
In comparison to a 0.01M solution of glucose, the depression in freezing point of a 0.01M MgCl2 solution is _____________.(i) the same(ii) about twice(iii) about three times(iv) about six times |
| Answer» (iii) about three times | |
| 291. |
In comparison to a 0.01 M solution of glucose, the depression in freezing point of a 0.01 M `MgCI_(2)` solution isA. the sameB. about twiceC. about three timesD. about six times |
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Answer» Correct Answer - c is the correct answer. Glucose is a molecular solid while `MgCI_(2)` being lonic, provides (0.03 M) moles of inos per litre. It is thrice the number os inos procided by glucose (0.01 M). |
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| 292. |
In coparison to a 0.01 M solution of glucose, the depression in freezing point of a 0.01 M `MgCl_(2)` solution is……A. the sameB. about twiceC. about three timesD. about six times |
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Answer» Correct Answer - C `DeltaT_f=iK_fm` For glucose , i=1 For `MgCl_2` , i=3 Thus, the depression in freezing point `(DeltaT_f)` of a 0.01 M glucose solution. |
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| 293. |
The exprimental values of colligative properties of many solute in solution resemble calculatede value of colligative properties. However in same cases, the expermental value of colligative property differ widely form those obtained by calculation. Such experimental values of colligative properties are known as abnormal values of colligative propertiy. Cause for abnormal values of colligative properties are : Dessiation of solute : It increases the colligative properties. Association of solute : It decreases the colligative properties. Choose the correct answer. If degree of dissociation of an electrolyte, `A_(2)B_(3) "is" 25%` in a solvent, thenA. Normal boiling point = Expermental boiling point.B. Normal freezing point `gt` Experimental freezing pointC. Normal osmotic pressure = 1/2 Experimantal osmotic pressureD. Normal molecular mass = 1/4 Exprimental molecular mass |
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Answer» Correct Answer - c it is corrct answer. |
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| 294. |
The exprimental values of colligative properties of many solute in solution resemble calculatede value of colligative properties. However in same cases, the expermental value of colligative property differ widely form those obtained by calculation. Such experimental values of colligative properties are known as abnormal values of colligative propertiy. Cause for abnormal values of colligative properties are : Dessiation of solute : It increases the colligative properties. Association of solute : It decreases the colligative properties. Choose the correct answer. If degree of dissociation of an electrolyte, `A_(2)B_(3) "is" 25%` in a solvent, thenA. `(NH_(2))_(2)CO"solution"gt"NaCI solution "gtNa_(2)SO_(4)" solution "gt K_(4)[Fe(CN)_(6)]" solution"`B. `"NaCI solution "gt Na_(2)SO_(4)" solution "gt NH_(2)_(2)CO" solution"gt K_(4)[Fe(CN)_(6)]" solution"`C. `K_(4)[Fe(CN)_(6)" solution" gt Na_(2)SO_(4)" solution "gt MaCI" solution "gt (NH_(2))_(2)CO" solution"`D. `NaSO_(4)" solution " gt (NH_(2))_(2)CO" solution " gt "NaCI solution "gt K_(4)[Fe(CN)_(6)]" solution"` |
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Answer» Correct Answer - c it is corrct order of decresing osmotic pressures. |
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| 295. |
The exprimental values of colligative properties of many solute in solution resemble calculatede value of colligative properties. However in same cases, the expermental value of colligative property differ widely form those obtained by calculation. Such experimental values of colligative properties are known as abnormal values of colligative propertiy. Cause for abnormal values of colligative properties are : Dessiation of solute : It increases the colligative properties. Association of solute : It decreases the colligative properties. Choose the correct answer. (19) One mole `I_(2)` (solid) is added in !M, 1 litre KL solution, ThenA. Osmotic pressure of solution increasesB. Freezing point of solution increasesC. Rrlative lowering in vapour pressure decreasesD. No change in boiling point of solution. |
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Answer» Correct Answer - d There is no chane in the boiling point of the solution. |
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| 296. |
Colligative properties depend upon the amount fraction of solute in solution.A. relative number of solute molecules irrespective of the nature of the solute and solventB. relative number of solute molecules in solution and nature of solventC. relative number of solute molecules and the nature of solute and solventD. relative number of solvent molecules. |
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Answer» Correct Answer - B Colligative properties depend upon no of particles in solution i.e., no of solute particles. It also depends on the nature of solvent. |
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| 297. |
Colligative properties depend on ___A. the nature of the solute particles dissolved in solutionB. The number of solute particles in solutionC. the physical properties of the solute particles dissolved in solutionD. the nature of solvent particles |
| Answer» Correct Answer - B | |
| 298. |
Colligative properties of the solution depend on:A. Nature of soluteB. Nature of solventC. Number of particles present in the solutionD. Number of moles of solvent only |
| Answer» Correct Answer - C | |
| 299. |
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. 0.100 M `Mg_(3)(PO_(4))_(2)`(aq) has the highest osmotic pressure.B. 0.125 M `Na_(3)PO_(4)` (aq) has the highest osmotic pressure.C. 0.500 M`C_(2)H_(5)OH(aq)` has the highest osmotic pressure.D. They all have the same osmotic pressure. |
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Answer» Correct Answer - D |
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| 300. |
A 5% solution (by mass) of cane suger in water has freezing point of 271K. Calculate the freezing point of 5% glucose in water if freezing point of water is 273.15 K. |
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Answer» Given 5% solution of cane sugar w = 5 gms, W = 95 gms m = 342, `DeltaT_("f")=273.15-271=2.15` `DeltaT_("f")=(K_("f")xxw)/(m xx W)` `2.15=("K"_("f")xx5)/(342xx95)" …(1)` 5% solution of glucose w = 5 gms, W = 95 gms ltbrrgt m = 180, `DeltaT_("f")=` ? `DeltaT_("f")=(K_("f")xx5)/(180xx95)" "...(2)` Dividing equation (2) by Equation (1) `(DeltaT_("f"))/(2.15)=(K_("f")xx5)/(180xx95)xx(342xx95)/(K_("f")xx5)` `DeltaT_("f")=(342xx2.15)/(180)=4.085"K"` `:.` The freezing point temperature for `5%` glucose solution `=273.15-4.085=269.07" K"` |
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