94420 + Interview Questions in Chemistry in Class 12

1.	The boiling point of glycerol is more than propanol because of
Answer» HYDROGEN bonding Hybridisation Resonance All the above Answer :B

Discussion

2.	The boilingpoint of glycerol is more than propanolbecause of
Answer» HYDROGEN BONDING HYBRIDIZATION resonance metamerism ANSWER :A

Discussion

3.	The boiling point of glycerol is 563K but it decomposes below 563K. It is purified by :
Answer» Sublimation Vacuum distillation Steam distillation Fractional distillation Answer :B

Discussion

4.	The boiling point of ethyl alcohol should be less than that of __________
Answer» PROPANE formic acid DIMETHYL ether none of the above ANSWER :B

Discussion

5.	The boiling point of ethyl alcohol is much higher than that of dimethyl ether , though both have the same molecular weight. The reason for this is :
Answer» Ether is insoluble in water Methyl groups are attached to oxygen in ether Dipole MOMENT of ethyl ALCOHOL is less Ethyl alcohol shows hydrogen BONDING ANSWER :D

Discussion

6.	The boiling point of ethanol is 78^(@)C and its nolal boiling point elevation constant per 1000 g is 1.15 K. A solution of 1.12 g of camphor in 32 g of ethanol has a boiling point of 78.28^(@)C. Cakculate the molecular mass of camphor.
Answer» Solution :`T_(b)^(@)=78^(@)C,T_(b)=78.28^(@)C,DeltaT_(b)=T_(b)-T_(b)^(@)=78.28-780.28^(@)C=0.28 K` `K_(b)=1.15" K KG mol"^(-1), W_(B)=1.12 g, W_(A)=0.032 kg, M_(B)= ?` `M_(B)=(K_(b)xxW_(B))/(DeltaT_(b)xxW_(A))=((1.5" K kg mol"^(-1))xx(1.12 g))/((0.28 K)xx(0.032 Kg ))=143.75" g mol"^(-1)`.

Discussion

7.	The Boiling point of ether is very low compared to that of same molecular weight of alcohol, because -
Answer» Ether having intermolecular VANDERWALL's bond. Ether is a WEAK POLAR solvent. Ether can not FORM intermolecular hydrogen bond. All of these Solution :Ether can not form intermolecular hydrogen bond.

Discussion

8.	The boiling point of ethanol is …..
Answer» 334 K 351 K 360 K 313 K Solution :351 K

Discussion

9.	The boiling point of carbon tetrachloride is 77^(@)C and its heat of vaporisation is "31 kJ mol"^(-1). Calculate the vapour pressure of carbon tetrachloride in atmospheres at 25^(@)C.
Answer» Solution :Applying Clausius - Clapeyron equation `log(P_(2))/(P_(1))=(Delta_("vap.")H)/(2.303R)[(T_(2)-T_(1))/(T_(1)T_(2))]` `T_(1)=25^(@)C=25+273=298K,P_(1)=?` `T_(2)=77^(@)C=77+273=350K, p_(2)="1 atm (PRESSURE at the b. pt.)"` `Delta_("vap")H="31 kJ MOL"^(-1)="31000 J mol"^(-1),R=8.314"J K"^(-1)"mol"^(-1)` `therefore""log.(1)/(P_(1))=(31000)/(2.303xx8.314)[(350-298)/(298xx350)]` `"or"log.(1)/(P_(1))=0.8072"or"-logP_(1)=0.8072"or"logP_(1)=-0.8072=bar1.1928` `therefore""P_(1)="Antilog "bar1.1928=0.1559 atm.`

Discussion

10.	The boiling point ofbenzene rises from 80.1^(@)C to 13.76g of biphenyl (C_(6)H_(5)-C_(6)H_(5)) is dissolved into 100 g of benzene. Calculate latent heat of vaporisation of benzene.
Answer» Solution :Calculation of molal ELEVATION constant `(K_(B))` for benzene `DeltaT_(b)=(1000k_(b)w_(2))/(w_(1)M_(2))or K_(b)=(DeltaT_(b)xxw_(1)xxM_(2))/(1000xxw_(2))=((82.4-80.1)Kxx100gxx"154 g mol"^(-1))/("1000 g kg"^(-1)xx13.76g)="2.574 K kg mol"^(-1)` Calculation of `Delta_("vap")H` `K_(b)=(M_(1)RT_(0)^(2))/(1000Delta_("vap")H)or Delta_("vap")H=(M_(1)RT_(0)^(2))/(1000K_(b))=("78 g mol"^(-1)xx8.314" jK"^(-1)"mol"^(-1)xx(8.01+273)^(2)K^(2))/("1000 g kg"^(-1)xx2.574"K kg mol"^(-1))` `="31411 J mol"^(-1)=31.411" kJ mol"^(-1)`

Discussion

11.	The boiling point of C_6H_6, CH_3OH, C_6H_5NH_2 and C_6H_5NO_2 are 80^@C,65^@C,184^@C, and 212^@C respectively. Which will show highest vapour pressure at room temperature?
Answer» `C_6H_6` `CH_3OH` `C_6H_5NH_2` `C_6H_5NO_2` ANSWER :B

Discussion

12.	The boiling point of benzene is 353.23 K. When 1.80 gm of a nonvolatile silute was dissolved in 90 gm of benzene, the boiling point is raised to 354.11 K. The molar mass of the solute is [K_(b) for benzene = 2.53 K mol^(-1)]
Answer» `5.8 g mol^(-1)` `0.58 g mol^(-1)` `58 g mol^(-1)` `0.88 g mol^(-1)` Solution :The ELEVATION `(Delta T_(b))` in the BOILING point `= 354.11 K - 353.23 K = 0.88 K` Substituting these values in expression `M_("SOLUTE")=(K_(b)xx1000xxw)/(Delta T_(b)xx W)` Where, w = WEIGHT of solute, W = weight of solvent `M_("solute")=(2.53xx1.8xx1000)/(0.88xx90)=58 gm mol^(-1)` Hence, MOLAR mass of the solute `= 58 gm mol^(-1)`

Discussion

13.	The boiling point of benzene is 353.23 K. When 1.80 g of a non - volatile solute was dissolved in 90 g of benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of the solute. K_(b) for benzene is 2.53 K kg mol^(-1).
Answer» Solution :The elevation `(Delta T_(b))` in the boiling point `= 354.11 K - 353.23 K = 0.88 K` Sdubstituting these VALUES in expression `M_(2)=(1000 xx w_(2))xx (K_(b))/(Delta T_(b)xx w_(1))` we get, `M_(2)=(2.53"K kg MOL"^(-1)xx1.8 g xx1000 g kg^(-1))/(0.88K xx 90 g)` `= 58 gmol^(-1)` Therefore, molar mass of the SOLUTE, `M_(2)=58 g mol^(-1)`

Discussion

14.	The boiling point of benzene is 353.23 K. When 1.80 g of a non-volatile solute was dissolved in 90 g benzene, the boiling point is raised to354.11 K. Calculate the molar mass of the solute. (K_(b) for benzene is 2.53 K kg mol^(-1))
Answer» Solution :Given values are: `T_(("benzene"))^(@) = 353.00 K , K_(b) = 2.53.00 K kg mol_(-1)` `T_(b("solution")) = 354.00 K` `W_("SOLUTE") = 1.80 g` `W_("solvent") = 90 g` The elevation in boiling point ,`DeltaT_(b) = T_(b("solution")) - T_(b("solvent"))^(@)` `= 354.11 - 353.23` `=0.88 K` Molar mass of solute is given as `Mw_("solute") = (K_(b) xx 1000 xx W_("solute"))/(DeltaT_(b) xx W_("solvent")` `Mw_("solute") = (2.53 xx 1000 xx 1.80)/(0.88 xx 90) = 58.0 g mol^(-1)` Hence, the molar mass of solute is `58.0 g mol^(-1)`.

Discussion

15.	The boiling point of benzene is 353.23 K. When 1.80 g of a non-volatile solute was dissolved in 90 g of benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of the solute. K_(b) for benzene is 2.53 K kg mol^(-1).
Answer» Solution :Here, we are GIVEN `"" w_(2)=1.80 g, w_(1)=90 g, "" Delta T_(B)=354.11-353.23 K=0.88 K` `"" K_(b)=2.53 " K kg mol"^(-1)` Substituting these VALUES in the formula, `M_(2)=(1000 K_(b)w_(2))/(w_(1)Delta T_(b))`, we get `M_(2)=(1000 " g kg"^(-1)xx2.53" K kg mol"^(-1)xx1.80 g)/(90g XX 0.88 K)=58 "g mol"^(-1)`

Discussion

16.	The boiling point of benzene is 353.23 K. When 1.80 g of a non - volatile non - ionisable solute was dissolved in 90 g of benzene, the boiling point raised to 354.11 K.
Answer» Solution :Given `T_(b)^(0)=353.2,""w_(2)=1.80g,""w_(1)=90G,""T_(b)=354.11K,""K_(b)=2.54"K Kg mol"^(-1)` `M_(2)=(K_(b)xxw_(2)xx1000)/(DeltaT_(b)xxw_(1))=(2.53xx1.8xx1000)/(0.88xx9)="58 g/mol"`

Discussion

17.	The boiling pointof benzene is353 .23 K. When1.80gramof non- volatile solutewasdissolvedin 90gramof benzene , theboilingpoint is raisedto 354.11 K. Calculatethe molarmass ofsolute . [K_(b)for benzene = 2.53 K kg mol^(-1)]
Answer» Solution :Given : `T_(b)^(o) =353 K , T_(b)= 354 . 11 K` `W_(1) = 90 g , W_(2)= 1.8 , K_(b)= 2.53 K kg mol^(-1)` Molar mass`=M_(2) = ?` `DELTA T_(b) = T_(b)- T_(b)^(o) = 354.11 - 353 .23 = 0.88K` `Delta T_(b) = K_(b) XX (W_(2) xx 1000)/(W_(1) xx M_(2))` `:. M_(2)= (K_(b) xx W_(2) xx 1000)/(Delta T_(b)xx W_(1))` ` = (2.53 xx1.8 xx 1000)/(0.88 xx90)` ` =57.5 g "mol"^(-1)`

Discussion

18.	The boiling point of benzene at 1 atm is 80.2^@C. Calculate the enthalpy of vaporisation of benzene at its b. pt.
Answer» SOLUTION : `30.022 kJ.mol^(-1)`

Discussion

19.	The boiling point of an azeotropic mixture of water-ethanol is less than that of both water and ethanol. Then :-
Answer» The MIXTURE will SHOW negative deviation from RAOULT's law. The mixture will show positive deviation from Raoult's law The mixture will show no deviation from Raoult's law This mixture considered as colloidal solution. Answer :B

Discussion

20.	The boiling point of an azeotropic mixture of water and ethyl alcohol is less than that of theoretical value of water and alcohol mixture. Hence the mixture shows:
Answer» That SOLUTION is highly saturated Positive deviation from RAOUL.s law Negative deviation from Raoul. law Nothing can be said Answer :B

Discussion

21.	The boiling point of an aqueous solution of a non-volatile solute is 100.15^(@) C. What is the freezing point of an aqueous solution obtained by diluting the above solution with an equal volume of water? The value of K_(b) and k_(l) for water are 0.512^(@)C and 1.86^(@) C " molality"^(-1) respectively.
Answer» `-0.544^(@)`C `-0.52^(@)` C `-0.272^(@)` C `-1.86^(@)` C ANSWER :C

Discussion

22.	The boiling point of an aqueous solution of a nonvolatile solute is 100.15^@C. What is the freezing point of an aqueous solution obtained by diluting the above solution with an equal volume of water? The value of K_b and K_f for water are 0.512^@C and 1.86^@C K molality^-1:
Answer» `-0.544^@C` `-0.512^@C` `-0.272^@C` `-1.86^@C` ANSWER :C

Discussion

23.	The boiling point of an aqueous solution of a non-volatile solute is 100, 15^(@) C. What is the freezing point of an aqueous solution obtained by diluting the above solution with an equal volume of water? The value of K_(b) and K_(f) for water are 0.512^(@) C and 1.86^(@) "molality"^(-1) respectively.
Answer» `-0.544^(@)` C `0.52^(@)` C `-0.272^(@)` C `-1.86^(@)` C ANSWER :C

Discussion

24.	The boiling point of an aqueous solution of a non-volatile is 100.156^(@) C. What is the freezing point of an aqueous solution obtained by diluting the above solution with an equal-voilume of water? The values of K_(b) and K_(f) for water are 0.512 and 1.86 K/molality.
Answer» `-0.544^(@)` C `-0.512^(@)` C `-0.272^(@)` C `-1.86^(@)` C ANSWER :C

Discussion

25.	Theboiling pointof anaqueoussolutionis 100.18.^(@)C. Find thefreezingpointof the solution . (Given: K_(b)= 0.52K kg mol^(-1),K_(f) = 1.86K kgmol^(-1))
Answer» Solution :Given : `T_(b)=100.18.^(@)C +273= 373 .18 K` `K_(o) =0.52 K kg MOL^(-1), K_(f) =1.86 K kg mol^(-1)` Boiling point ofwater`= T_(o)= 373 K` `T_(f)= ?` `Delta T_(b)= T_(b) - T_(o) =373 . 18 - 373= 0.18 K` If m is themolalityof thesolutionthen `Delta T_(b)= K_(b ) XX m " and"Delta T_( f) = K_( f) xx m` `:.(DeltaT_(f))/(Delta T_(b))= (K_(f) xx M)/(K_(b) xx M) = (K_(f))/(K_(b))` `:.Delta T_(f) = Delta T _(b) xx .(K_(f))/(K_(b))= 0.18 xx .(1.86)/(0.52) =0.6438 K` Freezingpoint ofwater = `T_(o)= 273 K` `Delta T_(f) = T_(o)- T_(f)` HENCETHE freezingpoint of thesolutionis `T_(f)=T_(o)- Delta T_(f) =273-0.6438= 272. 3562 K` ORFreezingpoint ofsolutionis - 0.6438 `.^(@)C`

Discussion

26.	The boiling point of an aqueous solution is found to be 100.28^(@)C . Then the freezing point of the same solution is -X^(0)C . What is the value of 'X' ?
Answer» ANSWER :1

Discussion

27.	The boiling point of alcohol is higher than that of isomeric alkane and alkyl halide, because....
Answer» They are in LIQUID state Intermolecular ATTRACTIONS are more They have H-bond Not given Solution :They have H-bond

Discussion

28.	The boiling point of alcohol, gradually decrease as moving from 1^(@) (Primary) to 3^(@) (Tertiary) alcohol because -
Answer» INTERMOLECULAR DISTANCE decreases Formation of H-bond Intermolecular distance INCREASES All of these Solution : Intermolecular distance increases

Discussion

29.	The boiling point of a water containing non-valatile solute is 101.04 ^(@)C of 2 molal solution, the ebullioscopic constant of water is
Answer» 0.52 K . Kg `"mol"^(-1)` 1.04 K . Kg `"mol"^_1)` 10.4K . Kg `"mol"^(-1)` 5.2 k . Kg `"mol"^(-1)` SOLUTION :`T = 101.04^(@) C = 374.04 K` `T^(@) = 100^(@)C = 373 K, m=2 "MOLAL"` `:. Delta T_(b) = K_(b) XX m` `K_(b) = (Delta T_(b))/m = (374.04-373)/2=1.01/2=0.52` K. Kg `"mol"^(-1)`

Discussion

30.	The boiling point of alcohol are…… than corresponding thiols
Answer» More Same Either of these Less Answer :A

Discussion

31.	The boiling point of a solvent containing non volatile solute :
Answer» is depressed is elevated does not change None of the above Answer :B

Discussion

32.	The boiling point of a solvent containing a non-volatile solute
Answer» is DEPRESSED is elevated does not change none of the above. Answer :B

Discussion

33.	The boiling point of a solution of 0.11 of a substance in 15g of ether was found to be 0.1^(@)C higher than that of pure ether. The molecular weight of the substance will be (K_(b)=2.16) :
Answer» 148 158 168 178 Solution :`m=(K_(B)XX W xx1000)/(Delta T_(b)xx W)` `K_(b)=2.16, w=0.11, W=15 g, Delta T_(b) =0.1` `m=(2.16xx0.11xx1000)/(0.1xx15)=158.40 ~= 158`.

Discussion

34.	The boiling point of a solution containing 2.62 g of a substance A in 100 g of water is higher by 0.0512^(@)C than the boiling point of pure water. The molar mass of the substance ( K_(b)=5.12Km^(-1) ) is :
Answer» 131 262 26.2 2620 Solution :`M_(B)=(K_(b)xx1000 xx w_(B))/(w_(A)xx DELTA T_(b))` `=(5.12xx1000xx2.62)/(100 xx 0.0512)=2620`

Discussion

35.	The boiling point of a compound is raised by
Answer» volatility of compound non-polarity in the MOLECULES intermolecular HYDROGEN BONDING intramolecular hydrogen bonding Solution : We know that, the boiling point of a compound, evidently, is a consequence of strong intermolecular forces due to hydrogen BOND. THEREFORE the boiling point of a compound is raised by intermolecular hydrogen bonding

Discussion

36.	The boiling point of 0.2molkg^(-1)solution of X in water is greater than equimolal solution of Y in water.Which one of the following statement is true in the case?
Answer» Molecular MASS of X is less than the molecular mass of Y Y is undergoing dissociationin water while X undergoes no change. X is undergoes dissociation in water. Molecular mass of X is greater then the molecular mass of Y. SOLUTION :`DeltaT_(B)=iKb_(m)` For equilbrium solutions, elevation in boiling point will be higher of solution undergoes dissociation i.e.,`igt1`

Discussion

37.	The boiling point of 0.2 mol "kg"^(-1) solution of X in water is greater than equimolal solution of Y in water, Which one of the following statements is true in the case ?
Answer» MOLECULAR mass of X is less than the molecular mass of Y. Y is undergoing dissociation in waer while X UNDERGOES no change X is undergoing dissociation in water Molecular mass of X is greater than the molecular mass of Y Answer :C

Discussion

38.	The boiling point of 0.2 mol kg^(-1) solution of X in water is greater than equimolal solution of Y in water. Which one of the following statements is true in this case ?
Answer» X is undergoing dissociation in water. Molecular mass of X is greater than the molecular mass of Y. Molecular mass of X is less than the molecular mass of Y. Y is undergoing dissociation in water while X undergoes no CHANGE. SOLUTION :`(Delta T_(B))_(x)gt (Delta T_(b))_(y)` same SOLVENT so, `K_(b)` is same m is same (given) `i_(x).K_(b).m gt i_(y).K_(b).m rArr i_(x)gt i_(y)` so, X is undergoing dissociation in water.

Discussion

39.	The boiling point of "0.2 mol kg"^(-1) solution of X in water is greater than equimolal solution of Y in water. Which one of the following statements is true in this case?
Answer» MOLECULAR mass of X is less than molecular mass of Y Y is undergoing dissociation in water while X UNDERGOES no change X is undergoing dissociation in water Molecular mass of X is greater than the molecular mass of Y SOLUTION :For EQUIMOLAL solution of two substances, the one that undergoes dissociation produces greater number ofparticles (ions) in the solution and HENCE has greater boilling point.

Discussion

40.	The boiling point of 0.2 mol kg^(-1) solution of X in water is greater than equimolal soltuion of Y is water. Which one of the following stamtements is true in this case
Answer» MOLECULAR mass of X is greater than the molecular of Y Molecular mass of X is LESS than the moelcular mass of Y Y is undergoing DISSOCIATION in water while X UNDERGOES no change X is undergoing dissociation in water Solution :When SOLUTE undergoes dissociation than vant Hoff factor `i gt Delta T_(b)= i K_(b) m`

Discussion

41.	The boiling point elevation and the freezing point depression of solutions have a number of practical applications. Ethylene glycol (CH_(2)OH. CH_(2)OH) is used in automobile radiators 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 antifreeze. For the boiling point elevation to occur the solute must be non-volatile, but no such restriction applies to freezing point depression. For example mthanol (CH_(3)OH), a fairly volatile liquid that boils only at 65^(@)C is sometimes used as antifreeze in automobile radiators. If cost of glycerol, glycol and methanol are same, then the sequence of economy to use these compounds as antifreeze will be:
Answer» GLYCEROL `GT` GLYCOL `gt` methanol methanol `gt` glycol `gt` glycerol methanol = glycol = glycerol methanol `gt` glycol `LT` glycerol Answer :B

Discussion

42.	The boiling point of 0.1 molal aqueous solution of urea is 100.18^(@)C at 1 atm. The molal elevation constant of water is
Answer» `1.8` `0.18` 18 `18.6` SOLUTION :`K_(B)=(0.18)/(0.1)=1.8`

Discussion

43.	The boiling point of 0.1 m K_(4)[Fe(CN)_(6)] is expected to be (K_(b) for water =0.52K kg "mol" ^(-1))
Answer» `100.52^(@)C` `100.10^(@)C` `100.26^(@)C` `102.6^(@)C` ANSWER :C

Discussion

44.	The boiling point of 0.1 molal aqueous solution of urea is 100.18 ^(@)C atm. The molal elevation constant of water is
Answer» 1.8 0.18 18 18.6 Solution :`K_(B) = (DELTA T_(b))/m=(T_(b)-T^(@))/m=(100.18 -100)/0.1=0.18/0.1=1.8^(@)`

Discussion

45.	The boiling point elevation for toluene is 3.32 K kg "mol"^(-1) . The normal boiling point of toluence is 110.7^(@)C . The enthalpy of vapourization of the toluene would be nearly :
Answer» `17.oKJ "MOL"^(-1)` `21.33KJ "mol"^(-1)` `51.KJ "mol"^(-1)` `68.0KJ "mol"^(-1)` ANSWER :B

Discussion

46.	The boiling point elevation and the freezing point depression of solutions have a number of practical applications. Ethylene glycol (CH_(2)OH. CH_(2)OH) is used in automobile radiators 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 antifreeze. For the boiling point elevation to occur the solute must be non-volatile, but no such restriction applies to freezing point depression. For example mthanol (CH_(3)OH), a fairly volatile liquid that boils only at 65^(@)C is sometimes used as antifreeze in automobile radiators. 620g glycol is added to 4kg water in the radiator of a car. What amount of ice will separate out at -6^(@)C ? K_(f) = 1.86K kg mol^(-1).
Answer» 800g 900g 600g 1000g Answer :B

Discussion

47.	The boiling point elevation and the freezing point depression of solutions have a number of practical applications. Ethylene glycol (CH_(2)OH. CH_(2)OH) is used in automobile radiators 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 antifreeze. For the boiling point elevation to occur the solute must be non-volatile, but no such restriction applies to freezing point depression. For example mthanol (CH_(3)OH), a fairly volatile liquid that boils only at 65^(@)C is sometimes used as antifreeze in automobile radiators. 124g each of the two reagents glycol and glycerol are added in 5kg water of the radiators in the two cars. Which of the following statements is wrong?
Answer» Both will act as antifreeze GLYCOL will be BETTER Glycerol is better because its molar mass is GREATER than glycol Glycol is more volatile than glycerol. Answer :C

Discussion

48.	The boiling of an aqueous of non volatile & non electrolytic solute is 100.15^(@)C. How many times the solution is diluted by water to get freezing point of resultant solution as -0272^(@)C given K_(b) and K_(f) of water are 0.512 and 1.86 K kg "mol"^(-1)
Answer» ANSWER :B

Discussion

49.	The boiling point and melting point of inert gases are
Answer» high low very high very low Answer :D

Discussion

50.	The boiling point of 0.2 mol kg^(-1) solution of X in water is greater than equimolal solution of Y in water. Which of the following statements is true in this case?
Answer» Molecular mass of X is less than that of Y Y is UNDERGOING DISSOCIATION in water wheres X dose not undergoing dissociation in water. X is undergoing dissociation in water. Molecular mass X is greater than that of Y. Solution :Scince bot solution are equimolal, the solution undergoing dissociation has higher boiling point, THEREFORE, solution X is undergoing dissociation in WASTER.

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

The boiling point of glycerol is more than propanol because of

The boilingpoint of glycerol is more than propanolbecause of

The boiling point of glycerol is 563K but it decomposes below 563K. It is purified by :

The boiling point of ethyl alcohol should be less than that of __________

The boiling point of ethyl alcohol is much higher than that of dimethyl ether , though both have the same molecular weight. The reason for this is :

The boiling point of ethanol is 78^(@)C and its nolal boiling point elevation constant per 1000 g is 1.15 K. A solution of 1.12 g of camphor in 32 g of ethanol has a boiling point of 78.28^(@)C. Cakculate the molecular mass of camphor.

The Boiling point of ether is very low compared to that of same molecular weight of alcohol, because -

The boiling point of ethanol is …..

The boiling point of carbon tetrachloride is 77^(@)C and its heat of vaporisation is "31 kJ mol"^(-1). Calculate the vapour pressure of carbon tetrachloride in atmospheres at 25^(@)C.

The boiling point ofbenzene rises from 80.1^(@)C to 13.76g of biphenyl (C_(6)H_(5)-C_(6)H_(5)) is dissolved into 100 g of benzene. Calculate latent heat of vaporisation of benzene.

The boiling point of C_6H_6, CH_3OH, C_6H_5NH_2 and C_6H_5NO_2 are 80^@C,65^@C,184^@C, and 212^@C respectively. Which will show highest vapour pressure at room temperature?

The boiling point of benzene is 353.23 K. When 1.80 gm of a nonvolatile silute was dissolved in 90 gm of benzene, the boiling point is raised to 354.11 K. The molar mass of the solute is [K_(b) for benzene = 2.53 K mol^(-1)]

The boiling point of benzene is 353.23 K. When 1.80 g of a non - volatile solute was dissolved in 90 g of benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of the solute. K_(b) for benzene is 2.53 K kg mol^(-1).

The boiling point of benzene is 353.23 K. When 1.80 g of a non-volatile solute was dissolved in 90 g benzene, the boiling point is raised to354.11 K. Calculate the molar mass of the solute. (K_(b) for benzene is 2.53 K kg mol^(-1))

The boiling point of benzene is 353.23 K. When 1.80 g of a non-volatile solute was dissolved in 90 g of benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of the solute. K_(b) for benzene is 2.53 K kg mol^(-1).

The boiling point of benzene is 353.23 K. When 1.80 g of a non - volatile non - ionisable solute was dissolved in 90 g of benzene, the boiling point raised to 354.11 K.

The boiling pointof benzene is353 .23 K. When1.80gramof non- volatile solutewasdissolvedin 90gramof benzene , theboilingpoint is raisedto 354.11 K. Calculatethe molarmass ofsolute . [K_(b)for benzene = 2.53 K kg mol^(-1)]

The boiling point of benzene at 1 atm is 80.2^@C. Calculate the enthalpy of vaporisation of benzene at its b. pt.

The boiling point of an azeotropic mixture of water-ethanol is less than that of both water and ethanol. Then :-

The boiling point of an azeotropic mixture of water and ethyl alcohol is less than that of theoretical value of water and alcohol mixture. Hence the mixture shows:

The boiling point of an aqueous solution of a nonvolatile solute is 100.15^@C. What is the freezing point of an aqueous solution obtained by diluting the above solution with an equal volume of water? The value of K_b and K_f for water are 0.512^@C and 1.86^@C K molality^-1:

The boiling point of an aqueous solution of a non-volatile is 100.156^(@) C. What is the freezing point of an aqueous solution obtained by diluting the above solution with an equal-voilume of water? The values of K_(b) and K_(f) for water are 0.512 and 1.86 K/molality.

Theboiling pointof anaqueoussolutionis 100.18.^(@)C. Find thefreezingpointof the solution . (Given: K_(b)= 0.52K kg mol^(-1),K_(f) = 1.86K kgmol^(-1))

The boiling point of an aqueous solution is found to be 100.28^(@)C . Then the freezing point of the same solution is -X^(0)C . What is the value of 'X' ?

The boiling point of alcohol is higher than that of isomeric alkane and alkyl halide, because....

The boiling point of alcohol, gradually decrease as moving from 1^(@) (Primary) to 3^(@) (Tertiary) alcohol because -

The boiling point of a water containing non-valatile solute is 101.04 ^(@)C of 2 molal solution, the ebullioscopic constant of water is

The boiling point of alcohol are…… than corresponding thiols

The boiling point of a solvent containing non volatile solute :

The boiling point of a solvent containing a non-volatile solute

The boiling point of a solution of 0.11 of a substance in 15g of ether was found to be 0.1^(@)C higher than that of pure ether. The molecular weight of the substance will be (K_(b)=2.16) :

The boiling point of a solution containing 2.62 g of a substance A in 100 g of water is higher by 0.0512^(@)C than the boiling point of pure water. The molar mass of the substance ( K_(b)=5.12Km^(-1) ) is :

The boiling point of a compound is raised by

The boiling point of 0.2molkg^(-1)solution of X in water is greater than equimolal solution of Y in water.Which one of the following statement is true in the case?

The boiling point of 0.2 mol "kg"^(-1) solution of X in water is greater than equimolal solution of Y in water, Which one of the following statements is true in the case ?

The boiling point of 0.2 mol kg^(-1) solution of X in water is greater than equimolal solution of Y in water. Which one of the following statements is true in this case ?

The boiling point of "0.2 mol kg"^(-1) solution of X in water is greater than equimolal solution of Y in water. Which one of the following statements is true in this case?

The boiling point of 0.2 mol kg^(-1) solution of X in water is greater than equimolal soltuion of Y is water. Which one of the following stamtements is true in this case

The boiling point of 0.1 molal aqueous solution of urea is 100.18^(@)C at 1 atm. The molal elevation constant of water is

The boiling point of 0.1 m K_(4)[Fe(CN)_(6)] is expected to be (K_(b) for water =0.52K kg "mol" ^(-1))

The boiling point of 0.1 molal aqueous solution of urea is 100.18 ^(@)C atm. The molal elevation constant of water is

The boiling point elevation for toluene is 3.32 K kg "mol"^(-1) . The normal boiling point of toluence is 110.7^(@)C . The enthalpy of vapourization of the toluene would be nearly :

The boiling of an aqueous of non volatile & non electrolytic solute is 100.15^(@)C. How many times the solution is diluted by water to get freezing point of resultant solution as -0272^(@)C given K_(b) and K_(f) of water are 0.512 and 1.86 K kg "mol"^(-1)

The boiling point and melting point of inert gases are

The boiling point of 0.2 mol kg^(-1) solution of X in water is greater than equimolal solution of Y in water. Which of the following statements is true in this case?