94420 + Interview Questions in Chemistry in Class 12 Page 616 InterviewSolution

30751.	The full name of the catalyst ZSM-5 used in petroleum industry for getting a mixture of hydrocarbons by dehydration of alcohols is……….
Answer» SOLUTION :ZEOLITE SIEVE of MOLECULAR Porosity-5

Discussion

30752.	The fuel in atomic pile is :
Answer» carbon sodium petroleum uranium Answer :d

Discussion

30753.	The fuel gas having volume composition equal to 34% CH_4+48% H_2 +12%O_2+3% COis :
Answer» OIL gas Water gas Coal gas Petrol gas Answer :C

Discussion

30754.	The froth-floation process is based upon
Answer» magnetic properties of gange specific GRAVITY of ORE particles prefential wetting of ore particles PREFERENTIAL wetting of gangue Solution :The froth-floation process is BASED upon preferential wetting of ore particles.

Discussion

30755.	The frequency (v )of a certain lineof the lyman series of atomic spectrum of hydrogn satisfies the following conditions (i) Is is the sum of the frequencies of another lyman line and a balmer line (ii) It is the sum of the frequecnies of a certain line a lym,an line and a paschen line (iii) It has no brackett ine Towhat transition does (v )correspond ?
Answer» `n_(2)=3 to n_(1)=1` `n_(2)=3 to n_(1)=2` `n_(2)=2 to n_(1)=1` `n_(2)=4 to n_(1) =1`] ANSWER :D

Discussion

30756.	The frequency of the radiation emitted when the electron falls from n = 4 to n = 1 in a hydrogen atom will be (Given ionization energy of H = 2.18 xx 10^(-18) J atom^(-1) and h = 6.625 xx 10^(-34) Js)
Answer» `1.54xx10^(15)s^(-1)` `1.03xx10^(15)JS^(-1)` `3.08xx10^(15)s^(-1)` `2.0xx10^(15)s^(-1)` Solution :`I.E.=E_(oo)-E_(1)=0-E_(1)` `=2.18xx10^(18)J` ATO`m^(-1)` Thus, `E_(n)=-(2.18xx10^(-18))/(n^(2))Jmol^(-1)` `DELTAE=E_(4)-E_(1)=-2.18xx10^(-18)(1/(4^(2))-1/(1^(2)))` `=2.044xx10^(-18)J"ato"m^(-1)`. `DeltaE=hv` or `v=(DeltaE)/h=(2.044xx10^(-18)J)` `(6.625xx10^(-34)Js)` `-3.085xx10^(15)s^(-1)`

Discussion

30757.	The frequency of radiation emitted when the electron falls from n=4 to n=1 in a hydrogen atom will be (given ionisation energy of H=2.18xx10^(18)J"atom"^(-1) and h=6.625xx10^(-25)Js)
Answer» `1.03xx10^(3)s^(-1)` `3008xx10^(15)s^(-1)` `2.00xx10^(15)s^(-1)` `1.54xx10^(-15)s^(-1)` Solution :`E_(1)=2.18xx10^(-18)J"ATOM"^(-1)` `E_(4)=(2.18xx10^(-18))/16=0.136xx10^(-18)J` atom Energy released `=(2.18-0.136)xx10^(-18)J"atom"^(-1)` `=2.044xx10^(-18)J"atom"^(-1)` Now `hv=E` `v=E/h=(2.044xx10^(-18))/(6.625xx10^(-34))=3.08xx10^(15)s^(-1)`

Discussion

30758.	The frequency of radiation emitted when the electron falls from n = 4 to n = 1 in a hydrogen atom will be (given ionization energy of H=2.18xx10^(-18)J"atom"^(-1)andh=6.625xx10^(-34)Js
Answer» `1.54xx10^(15)s^(-1)` `1.03xx10^(15)s^(-1)` `3.08xx10^(15)s^(-1)` `2.00xx10^(15)s^(-1)` SOLUTION :`DeltaE=hv=E_(1)(1/n_(1)^(2)-1/n_(2)^(2))`

Discussion

30759.	The frequency of light emitted for the transition n=4 to n=2 of He^(+) is equal to the transition in H atom corresponding to which of the following?
Answer» n=2 to n=1 n=3 to n=2 n=4 to n=3 n=3 to n=1 Solution :`v(H)=v(HE^(+))` `[RZ^(2)(1/(n_(1)^(2))-1/(n_(2)^(2)))]_(H)=RZ^(2)(1/(n_(1)^(2))-1/(n_(2)^(2)))_(He^(+))` `1/(n_(1)^(2))-1/(n_(2)^(2))=4(1/4-1/16)` `1/(n_(1)^(2))-1/(n_(2)^(2))=1/1-1/4` `:.n_(1)=1` and `n_(2)=2`

Discussion

30760.	The frequency of the first line of the Lyman series in the hydrogen atom is nu. What will be the frequency of the corresponding line for the singly ionised helium atom ?
Answer» `v_0//4` `4v_0` `v_0//2` `2v_0` ANSWER :B

Discussion

30761.	The frequency of first line of Balmer series in hydrogen atom is V_0. The frequency of corresponding line emitted by singly ionised helium atom is:
Answer» `2_(v_0)` `4_(v_0)` `V_0/2` `V_0/4` ANSWER :B

Discussion

30762.	The French physical Louis de Broglie in 1924 postulated that matter, like radiation, should exhibit a dual behaviour. He proposed the following relationship between the wavelength lambda of a material particle, its linear momentum p and planck constant h. lambda=h/p=h/(mv) The de Broglie relation implies that the wavelength of a particle should decrease as its velocity increases. It also implies that for a given velocity heavier particles should have shorter wavelength than fighter particles.The wave associated with particles in motion are called matter waves or de Broglie waves.These waves differ from the electromagnetic waves as they (I)have lower velocities (II)have no electrical and magnetic fields and (III)are not emitted by the particle under consideration The experiment confirmation of the de Broglies relation was obtained when Davisson and Germer, in 1927, observed that a beam of electronsis diffrated by a nickel crystal.As diffraction is a characteristic property of waves hence the beam of electron behaves as a wave, as proposed by de broglie. De-Broglie wavelength of an electron travelling with speed equal to 1% of the speed of light
Answer» 400 pm 120 pm 242 pm 375 pm Solution :`lambda=h/(MV)=(6.6xx10^(-34))/(9.1xx10^(-31)xx3xx10^6)=242` pm

Discussion

30763.	The frequency of a green light is 6 xx 10^(14)Hz. Its wavelength is
Answer» 500nm 5nm 5000nm none of these Solution :1NM `=10^(-9)m`

Discussion

30764.	The French physical Louis de Broglie in 1924 postulated that matter, like radiation, should exhibit a dual behaviour. He proposed the following relationship between the wavelength lambda of a material particle, its linear momentum p and planck constant h. lambda=h/p=h/(mv) The de Broglie relation implies that the wavelength of a particle should decrease as its velocity increases. It also implies that for a given velocity heavier particles should have shorter wavelength than fighter particles.The wave associated with particles in motion are called matter waves or de Broglie waves.These waves differ from the electromagnetic waves as they (I)have lower velocities (II)have no electrical and magnetic fields and (III)are not emitted by the particle under consideration The experiment confirmation of the de Broglies relation was obtained when Davisson and Germer, in 1927, observed that a beam of electronsis diffrated by a nickel crystal.As diffraction is a characteristic property of waves hence the beam of electron behaves as a wave, as proposed by de broglie. Using Bohr's theory, the transition, so that the electrons de-Broglie wavelength becomes 3 times of its orginial value in He^+ ion will be
Answer» `2to6` `2to4` `1to4` `1to6` SOLUTION :`lambda=h/(MV)" " lambdaimplies 3 "TIMES" " " Vimplies 1/3 "times" vprop z/n` so, TRANSITION will be `2to6`

Discussion

30765.	The freezing temperature of pure benzene is 5.40^(@)C. When 1.15g of naphthalene is dissolved in 100g of benzene, the resulting solution has a freezing point of 4.95^(@)C. The molal f.p. depression constant for benzene is 5.12, what is the molecular weight of naphthalene?
Answer» Solution :MOLALITY `=(DletaT_(f))/(K_(f))=(5.40-4.95)/(5.12)=0.88` mole per `1000g` `:.1000g` of the solvent contains `11.5g` of naphthalene `:.` mol.wt.of naphthalene `=("weight in grams"//1000g)/("no.of moles"//1000g)` ………………..(RULE 1, Chapter 1) `=(11.5)/(0.088)=130`

Discussion

30766.	Thefreezingpoints of equimolarsolutionsofglucoseKNO_(3) and AlCl_(3) are inorderof_____.
Answer» `AlCl_(3)LT KNO_(3)lt` GLUCOSE Glucose `lt KNO_(3)lt AlCl_(3)` Glulcose `lt AlCl_(3)lt KNO_(3)` `AlCl_(3)lt` Glucose `lt KNO_(3)` ANSWER :A

Discussion

30767.	The freezing points of a 0.05 molal solution of a non-electrolyte in water is: (K, = 1.86 K molality^-1)
Answer» `-1.86^@C` `-0.093^@C` `-0.93^@C` `0.093^@C` ANSWER :C

Discussion

30768.	The freezing point of water is depressed by 0.37^@C in a 0.01 molal NaCl solution. The freezing point of 0.02 molal solution of urea is depressed by
Answer» `0.37^@C` `0.74^@C` `0.185^(@)C` `0^(@)C` Solution :The depression in freezing point is proprtional to motal concentration of the solute IE. `Delta T_f PROP m` `DeltaT_f = K_fmi or K_f = (DeltaT_f)/(ixxm) =` constant so, `(DeltaT_(f_(NACL)))/(i_(NaCl XX m_(NaCl)))=(Delta T_(furea))/(m_(Urea)xxi_(Urea))=` constant `(0.37)/(2xx001) = (DeltaT_(fUrea))/(0.02 xx 1)rArr Delta T_(fUrea) = (0.37 xx 0.02)/(0.02) =0.37^@C`

Discussion

30769.	The freezing point of water is depressed by 0.37^(@)C in a 0.01 molal NaCl solution. The freezing point of 0.02 molal solution of urea is depressed by
Answer» `0.37^(@)C` `0.74^(@)C` `0.185^(@)C` `0^(@)C` Solution :The depression in FREEZING point is proportional to molal concentration of the SOLUTE i.e. `Delta T_(f) prop m` `Delta T_(f)=K_(f)mi` or `K_(f)=(Delta T_(f))/(ixxm)` = constant so, `(Delta T_(f_(NaCl)))/(i_(NaCl)xxm_(NaCl))=(Delta T_(f_("UREA")))/(m_("Urea")xxi_("Urea"))` = constant `(0.37)/(2xx0.01)=(Delta T_(f_("Urea")))/(0.02xx1)rArr Delta T_(f_("Urea"))=(0.37xx0.02)/(0.02)=0.37^(@)C`.

Discussion

30770.	The freezing point of the solution M is
Answer» 268.7 K 268.5 K 234.2 K 150.9 K Answer :D

Discussion

30771.	The freezing point of the solution M is -
Answer» 268.7 K 268.5 K 234.2 K 150.9 K SOLUTION :`"MOLARITY "m=(x_Bxx1000)/((1-x_B)m_A)=(0.1xx1000)/(0.9xx46)=2.415` `DeltaT=K_"F"xxm=2xx2.415=4.83` Freezing POINT of solution = 155.7 – 4.83 = 150.9 K.

Discussion

30772.	The freezing point of the aqueous solution of urea is -0.6^(@)C. How much urea should be added to 3 kg water to get such solution ? (K_(f)=1.5^(@)"C Kg. mol"^(-1))
Answer» 72 gram 6.0 gram 3.6 gram 2.4 gram Solution :`Delta T = K_(F).m=K_(f)xx ((1000xx W)/(M xx W_(o)))` `W=(Delta TXX M xx W_(o))/(K_(f)xx1000)` `therefore W=(0.6xx60xx3000)/(1.5xx1000)=72` gram.

Discussion

30773.	The freezing point of one molal NaCl solution assuming NaCl to be 100% dissociated in water is (molal depression constant = 1.86)
Answer» `-1.86^(@)C` `-3.72^(@)C` `+1.86^(@)C` `+3.72^(@)C` SOLUTION :For NaCli=2 `DELTA T_(F)=2K_(f)m=2xx1.86xx1=3.72` `T_(s)=T-Delta T_(f)=0-3.72=-3.72^(@)C`

Discussion

30774.	The freezing point of equimolar aqueoussolutions will be highest for
Answer» `CA(NO_3)_2` `LA(NO_3)_3` `C_6H_12O_6` ANSWER :D

Discussion

30775.	The freezing point of equimolal aqueous solution will be highest for
Answer» `C_(6)H_(5) CH_(3)^(+) CL^(-)` `Ca (NO_(3))_(2)` `L a (NO_(3)) _(2)` `C_(6) H_(12)O_(6)` Solution :Glucose is non electrolyte hence depression in freezing point will be MINIMUM, hence freezing point will be highest.

Discussion

30776.	The freezing point of cyclogexane is 279.65 K. A solution of 14.75 g of a solute in 500 g of cyclohexane has a freezing point of 277.33 K. Calculate the molar mass of the solute. (Given K_(f)=20.2 K kg mol^(-1))
Answer» Solution :`DeltaT_(f)=T_(f)^(@)-T_(f)=279.65 K-277.33=2.33 K` `K_(f)=20.2" K KG mol"^(-1)=20.2Km^(-1)` `"Molality, m"=(DeltaT_(f))/K_(f)=(2.23K)/(20.2Km^(-1))=0.115 m` Mass of SOLVENT = 500 g = 0.5 kg `"Moles of solute" = "molality"xx"mass of solvent in kg" =0.115" kg mol"^(-1)xx0.5" kg "=0.057" moles"` `"Molar mass of solute"=("Mass of sovent")/("Moles of solute")=(14.75 g)/(0.057" moles")=258.77" g mol"^(-1)`.

Discussion

30777.	The freezing point of equimolal aqueous solution will be highest for:
Answer» `C_6H_5NH_3Cl` `CA(NO_3)_2` `LA(NO_3)_3` `C_6H_12O_6` ANSWER :D

Discussion

30778.	The freezing point of benzene decreases by 0.45^(@)C when 0.2 g of acetic acidis added to 20 g of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be ……. (K_(f) for benzene = 5.12 K kg mol^(-1))
Answer» `64.6%` `80.4%` `74.6%` `94.6%` Solution :`Delta T_(f)=0.45` `m=(((0.2)/(60))xx1000)/(20)=(1)/(6)` `K_(f)=5.12` K kg / mol `i=1+((1)/(n)-1)x "" (n=2)` `= 1-(x)/(2)` `Now, Delta T_(f)=iK_(f)m` `0.45=(1-(x)/(2))(5.12)((1)/(6))` x = 0.94 `THEREFORE` PERCENTAGE of `~~ 94%`

Discussion

30779.	The freezing point of benzene decreases by 2.12 K when 2.5 g of benzoic acid (C_6H_3COOH)is dissolved in 25 g of benzene. If benzoic acid forms a dimer in benzene, calculate the van't Hoff factor and the percentage association of benzoic acid. [K_f for benzene = 5.12 K kg "mol"^(-1) ]
Answer» Solution :Apply the relation : `Delta T_f= i xx K_f xx m ` where i is van.t Hoff FACTOR. `K_f` = 5.12, molalily (m) = ` (2.5)/(122) xx 1000/25` Substituting the values in the equation above, we have `2.12 = i xx 5.12 xx (2.5)/(12) xx 1000/25` `i = (2.12 xx 122 xx 25)/(5.12 xx 2.5 xx 1000) = 0.505` Two molecules of `C_6H_5COOH `associate to FORM one MOLECULE of `(C_6H_5COOH)_2`. For association i = `1- alha/2`where a is DEGREE of association. Substituting the values, we have `0.505 = 1 - alpha/2` `alpha/2 = 1- 0.505 " or" alpha/2 = 0.495 " or " alpha = 0.99` Percentage association of benzoic ACID = 99

Discussion

30780.	The freezing point of benzene decreases by 0.45^(@)C when 0.2 g of acetic acid is added to 20 g of benzene. IF acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be (K_(f) "for benzene" = 5.12 K kg mol^(-1))
Answer» 0.746 0.946 0.646 0.804 Solution :`"MOLALITY (m)"=("No.of moles of "CH_(3)COOH)/("MASS of benzene in kg")` `((0.2))/((60 g MOL^(-1))xx(0.02kg))` `=0.167 mol kg^(-1)=0.167m` `DeltaT_(f)=ixxK_(f)xxm or i=(DeltaT_(f))/(K_(f)xxm)` `i=((0.45K))/((5.12 "K kg mol"^(-1))xx(0.167"mol kg "^(-1)))=0.526` `alpha=((i-1))/((n-1))=((0-526-1))/((0.5-1))` `=((-0474))/((-0.5))=0.948 %`

Discussion

30781.	The freezing point of benzene decreases by 0.45^(@)C when 0.2 g of acetic acid is added to 20 g of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be (K_(f) for benzene = 5.12 K kg "mol"^(-1))
Answer» `76.6%` `94.6%` `64.6%` `80.4%` Solution :`DeltaT_(f"(observed)")=0.45^(@)` `DeltaT_(f"(calculated)")=(1000K_(f)w_(2))/(w_(1)xxM_(2))` `=(1000xx5.12xx0.2)/(20xx60)(M_(2)(CH_(3)COOH)=60)` `= 0.853^(@)` `therefore""=(DeltaT_(f"(observed)"))/(DeltaT_(f"(calculated)"))=(0.45)/(0.853)=0.527` `underset(1-alpha)(2CH_(3)COOH)hArr underset((alpha)/(2))((CH_(3)COOH)_(2))` Total moles `=1-alpha+(alpha)/(2)=1-(alpha)/(2)` `i=1-(alpha)/(2)=0.527 or (alpha)/(2)=0.473 or alpha=0.946` `%" association "=94.6%`

Discussion

30782.	The freezing point of benzene decreaes by 0.45^(@)C when 0.2g of acetic acid is added to 20g of benzene. If acetic acid associates to form a dimer in benzene percentage association of acetic acid in benzene will be (K_(f) for benzene =5.12K kg mol^(-1))
Answer» `64.6%` `80.4%` `74.6%` `94.6%` ANSWER :D

Discussion

30783.	The freezing point of aqueous solution that contains 5% by mass urea, 1.0% by mass KCi and 10% by mass of glucose is (K_(f)H_(2)O)=1.86 k kg/mol
Answer» `290.2K` `285.5 K` `269. 93 K` `250 K` ANSWER :C

Discussion

30784.	The freezing point of aqueous solution that contains 5% by mass urea, 1.0% by mass KCl and 10% by mass of glucose is: (K_fH_2O = 1.86K molality^-1):
Answer» 290.2 K 285.5 K 269.93 K 250 K Answer :C

Discussion

30785.	Thefreezing pointof anaqueoussolutionsis 272 .93 K . Calculatethemolalityof the solutionif molal depressionconstantforwateris 1.86 Kg mol^(-1)
Answer» Solution :Given :Forpurewater `T_(o) = 273 K` Forsolution`T_(R)= 273 .93 K` `K_(r)= 1.86 K kg MOL^(-1)` Molalityof the solution`= m= ?` Depression in thefreezingpoint `= Delta T_(f)= T_(o)- T_(f)= 273- 272 .93= 0.07 K` `Delta T_(f)= K_(f)xx M` `:. m = (Delta T_(f))/(K_(f)) = (0.07)/(1.86)= 0.0376 "mol" kg ^(-1)`

Discussion

30786.	The freezing point of an aqueous solution of KNC containing 0.1892 "mol" kg^(-1) was 0.74^(@)C. On adding 0.95 mol of Hg (CN)_(2) the freezing point of solution was -0.53^(@)C. Assuming that complex is formed to the equation Hg(CN)_(2)+mCN^(-) to [Hg(CN)_(m+2)]^(n-) Find. m
Answer» ANSWER :`m=2`

Discussion

30787.	The freezing point of an aqueous solution of non-electrolyte having an osmotic pressure of 2.0 atm at 300 K will be, (K_(f) "for " H_(2)O = 1.86K mol^(-1)g) and R = 0.0821 litre atm K^(-1) mol^(-1) . Assume molarity and molarity to be same :
Answer» `-0.151^(@)C` `-1.151^(@)C` `-3.151^(@)C` `-2.151^(@)C` ANSWER :A

Discussion

30788.	The freezing point of an aqueous solution of KCN containing 0.1892 mole/kg H_(2)O was -0.704^(@)C. On adding 0.095 mole of Hg(CN)_(2), the freezing point of the solution was -0.53^(@)C. Assuming that the complex is formed according to the equation Hg(CN)_(2)+mCN^(-) to Hg(CN)_(m+2)^(m-) and also Hg(CN)_(2) is the limiting reactant, calculate m .
Answer» Solution :`K_(F)=(DeltaT_(f))/(m)=(0.704)/(2xx0.1892)` (KCN DISSOCIATES COMPLETELY) `=1.86` TOTAL molality after the addition of `Hg(CN)_(2)` `=` molality of `K^(+)+` molality of `CN^(-)``+` molality of Hg `(CN)_(m+2)^(m-)` `=0.1892+(0.1892-0.095m)+0.095` `=(0.4734-0.095m)` Now, `K_(f)=(DeltaT_(f))/(m)` `1.86=(0.53)/(0.4734-0.095m)` `m=1.984` or `m=2`

Discussion

30789.	Calculate the freezing point of an aqueous soltuion of non-electrolyte having an osmotic pressure 2.0 atm at 300 K. (K'_(f) = 1.86 K mol^(-1) kg and S = 0.0821 litre atm K^(-1) mol^(-1))
Answer» `-0.15^(@)C` `+0.15^(@)C` `-0.51^(@)C` `-0.17^(@)C` SOLUTION :`PI = CRT` `:. C = (pi)/(RT)` `=(2.0 atm)/((0.0821 L atm K^(-1) MOL^(-1))xx(300K))` `=0.0812 mol L^(-1) =0.0812 M ~~0.0812 m` `( :'` solution is very dilute) `DELTA T_(f) = K_(f)m = (1.86 Km^(-1)) xx (0.0812m)` `=0.15K = 0.15^(@)C` `:.` Freezing point `= (0.00 - 0.15)^(@)C =- 0.15^(@)C`.

Discussion

30790.	The freezing point of an aqueous solution of KCN containing 0.1 mol kg^(-1) was -0.38^(@) C. On adding 0.005 moles of Hg(CN)_(2) per kg of solvent, the freezing point of the solution remined as -0.38^(@) C. Assuming following reaction to be occured to 100% extent and none of the Hg(CN)_(2) remaining. Hg(CN)_(2)(aq)+x(CN)^(-)toHg(CN)_(x+2)^(x-)(aq) calculate the value of 'x'.
Answer» ANSWER :1

Discussion

30791.	The freezing point of a solution that contains 10 g urea in 100 g water is (K_1for H_2O = 1.86°C m^(-1))
Answer» -5.1°C -2.5°C -3.6°C -3.1°C ANSWER :D

Discussion

30792.	The freezing point of a solution prepared from 1.25 gm of a non-electrolyte and 20 gm of water is 271.9 K. If molar depression constant is 1.86 Kg K "mole"^(-1), then molar mass of the solute will be
Answer» `105.7` `106.7` `115.3` `93.9` SOLUTION :MOLAR mass `=(K_(f)xx1000xx w)/(DELTA T_(f)xx W)=(1.86xx1000xx1.25)/(20xx1.1)` `=105.68=105.7`.

Discussion

30793.	The freezing pointof a solution of acetic acid (mole fraction0.02) in benzene is 277.K. Acetic acid exists party as a dimmer, 2A hArr A_(2). Calculate equilibrium constant for dimerization. Freezing point of benzene is 278.4 K and K_(t) for benzen is 5.
Answer» Solution :LET acetic acid =A Benzene =B ASSUME `alpha` part of A forms dimer `{:(2A ,hArr, A_(2)),(1,0,"Intial moles"),(1-alpha, alpha//2 ,"moles after at eqm"):}` `:.I=((1-alpha)+alpha//2)/(1)=1-alpha//2` Mol. Fraction of `A=X_(A)=0.02` Mol. fraction of `B=X_(B)=0.98` Molality of A in B `=(X_(A))/(m_(1))XX(1000)/(X_(B))=(0.02)/(78)xx(1000)/(0.98)= 0.262 "mol" kg^(-1) (m_(1)="mol"` wt of solvent) Since `DeltaT_(t)=K_(1)xxlxx"molality"` `278.4-277.4=5xxixx0.262` or `1=5xxixx0.262` `i=(1)/(5)xx0.262=0.763` `1-alpha//2=0.763 rArr alpha=0.48` Hencethe molality of A after dimer is formed `(1-alpha)xx` initial molality `=0.52xx0.262` Molality of `A_(2)` after dimer formed `=(alpha)/(@)xx"molality" =(0.48)/(2)xx 0.262` `=0.24xx0.26=0.06288` The equilibrium constant `K_(eq)=([A_(2)])/([A]^(2))=(0.06288)/((0.13624)^(2))=3.39`

Discussion

30794.	The freezing point of a solution prepared from 1.25 g of non-electrolyte and 20 g of water is 271.9 K. If molar depression constant is 1.86 K mol^-1 then molar mass of the solute will be:
Answer» 105.7 106.7 115.3 93.9 Answer :A

Discussion

30795.	The freezing point of a solution containing 50cm^(3) of ethylene glycol in 50 g water is found to be -34^(@)C. Assuming ideal behaviour, calculate the density of ethylene glycol, (K_(f)" for water "="1.86 K kg mol"^(-1))
Answer» SOLUTION :`w_(2)=(M_(2)xxDeltaT_(F)xxw_(1))/(1000xxK_(f))=("62 g MOL"^(-1)xx34Kxx50g)/("1000 g kg"^(-1)xx"1.86 K kg mol"^(-1))="56.67 g"` `d=(w_(2))/(V)=("56.67 g")/("50 cm"^(3))="1.13 g cm"^(-3)`

Discussion

30796.	The freezing point of a solution containing 5g of benzoic acid (M=122 g mol^(-1)) in 35g of benzene is depressed by 2.94 K. What is the percentage association of benzoic acid if it forms a dimer in solution.[K_(f) for benzene = 4.9 K kg mol^(-1) ]
Answer» Solution :NUMBER of moles of BENZOIC acid `=(5)/(122)` Molality of benzoic acid solution `= (5)/(122) xx (1000)/(35) =1.17` Apply the relation `Delta T_(f) =i K_(f) m`, where i is van.t Hofff factor `2.94 K=ixx4.8 Kg mol^(-1) xx 1.17 mol kg^(-1)`, or `i=(2.94)/(4.9 xx 1.17)""....(i)` `(C_(6) H_(5) COOH)_(2) to (C_(6) H_(5) COOH)_(2)` Total number of moles `=1-x (x//2) =1 - (x//2)` `i= (1-(x)/(2))/(1)""...(II)` From (i) and (ii), we have `1-(x)/(2) = (2.94)/(4.9 xx 1.17)=0.5128` or `(x)/(2) =1-0.5128=0.4872` or `x=0.9744` or `97.44%`

Discussion

30797.	The freezing point of a solution containing 4.8 g of a compound in 60 g of benzene is 4.48. What is the molar mass of the compound (K_(f)=5.1 km^(-1)), (freezing point of benzene = 5.5^(@)C)
Answer» 100 200 300 400 Solution :`m=(K_(f)xx1000xxw)/(W XX Delta T_(f))=(5.1xx1000xx4.8)/(60xx1.02)=400`.

Discussion

30798.	The freezing point of a solution containing 28.335 cm^(3) of ethylene glycol in 50 gm water is found to be -34^(@)C. Assuming ideal behaviour, calculate the density of ethylene glycol in g//ml.K_(f) for water 1.86 k Kg "mol"^(-1)
Answer» ANSWER :B

Discussion

30799.	The freezing point of a solution containing 4.8 g of a compound in 60 g of benzene is 4.48. What is the molar mass of the compound ? ( K_(f)=5.1 Km^(-1), freezing point of benzene =5.5^(@)C )
Answer» 200 40 400 256 Solution :`M_(B)=(k_(f)xx1000xxw_(B))/(w_(A) xx DELTA T_(f))=(5.1xx1000xx4.8)/(60xx1.02)=400`

Discussion

30800.	The freezing point of a solution containing 18 g of a non-volatile solute dissolved in 200 g of water decreases by 0.93. Find the molecular mass of solute. K_(f) for water is 1.86 K/m.
Answer» SOLUTION :`W_(B)=18g, W_(A)=0.2"kg", DeltaT_(b)=0.93 K, K_(f)=1.86" K kg mol"^(-1), M_(B)=?` `M_(B)=(K_(f)xxW_(B))/(DeltaT_(f)xxW_(A))=((1.86" K kg mol"^(-1))xx(18 g))/((0.93 K)xx(0.2" kg"))=180" g mol"^(-1)`.

Discussion

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