94420 + Interview Questions in Chemistry in Class 12 Page 617 InterviewSolution

30801.	The freezing point of a solution containing 0.2 g of acetic acid in 20.0 g of benzeneis lowered by 0.45^@C . Calculate the degree of association of acetic acid in benzene. K_f(C_6H_6) = 5.12 K "mol"^(-1) kg
Answer» SOLUTION :ACETIC ACID EXISTS as DIMER in benzene 0.9453

Discussion

30802.	The freezing point of solution containing 0.2 g of acetic acid in 20.0 g of benzene is lowered by 0.45^(@)C. Calculate the degree of association of acetic acid in benzene. (K_(f)=5.12 K^(@) mol^(-1) kg^(-1))
Answer» Solution :Setp I. Calculation of observed molar mass of acetic acid `M_(B)=(K_(F)xxW_(B))/(DeltaT_(f)xxW_(A)), W_(A)=0.2 g, W_(A)=0.02 kg, DeltaT_(f)=0.45^(@)C=0.45 K` `K_(f)=5.12" K kg mol"^(-1), M_(B)=((5.12" K kg mol"^(-1))xx(0.2 g))/((0.45 K)xx(0.2 kg ))=113.8" g mol"^(-1)` Step II. Calculation of Van't Hoff factor (i) `i=("Normal molar mass")/("Observed molar mass")=(60" g mol"^(-1))/((113.8" g mol"^(-1)))=0.527` `"Step III. Calculation of DEGREE of ASSOCIATION "(alpha)` `"For association "alpha=(i-1)/(1/n-1)=(0.527-1)/(1/2-1)=((-0.473))/((-0.50))=0.946= 0.946xx100=94.6%`

Discussion

30803.	The freezing point of a 1.00 m aqueous solution ofHF is found to be -1.91 ^(@)C. The freezing point constant of water, K1is 1 .86 K kg mol^(-1) The percentage dissociation ofHF at this concentration is
Answer» `30%` `10%` `5.2%` `2.7%` Solution :`DELTA T_(t)=K_(f) xx m xx i` `I = (Delta T_(f))/(K _(f) xxm ) = (1.91)/(1.86xx1) =1.02` For `HI HARR H ^(+) + I ^(-)` `1- alpha + alpha + alpha =i= 1.027` `1+ alpha =1.02` `alpha = 0.02 or 2.7%`

Discussion

30804.	The freezing point of a solution containing 0.2 g of acetic acid in 20.0 g benzene is lowered by 0.45^@C. The degree of association of acetic acid in benzene is (Assume acetic acid dimerises in benzene and K_f for benzene = 5.12 K kg "mol"^(-1) ) ?
Answer» `94.6%` `54.9%` `78.2%` `100%` Solution :Given : `w_2=0.2 G , w_1=20 g , DeltaT_f =0.45^@C` `DeltaT_f = (1000xxK_fxxw_2)/(w_1xxM) rArr 0.45=(1000xx5.12xx0.2)/(20xxM)` `THEREFORE M_"(observed)"`=113.78 (ACETIC acid ) As acetic acid dimerises in benzene , so, `{:(,2CH_3COOH hArr, (CH_3COOH)_2),("Before association",1,0),("After association", 1-alpha,alpha//2):}` (where `alpha` is degree of association) Molecular weight of acetic acid =60 `i="Normal molecular mass"/"Observed molecular mass"` `therefore M_"(normal)"/M_"(observed)"=1-alpha + alpha/2` `60/113.78 = 1-alpha + alpha/2 therefore alpha` = 0.946 or 94.6%

Discussion

30805.	The freezing point of a 0.08 molal solution of NaHSO^(4) is -0.372^(@)C. Calculate the dissociation constant for the reaction. K_(f) for water =1.86 K m^(-1)
Answer» 0.04 0.02 0.01 0.2 Answer :A

Discussion

30806.	The freezing point of a 0.01 M aqueous glucose solution at 1 atmosphere is -0.18^(@)C. To it, an addition of equal volume of 0.002 M glucose solution will , produced a solution with freezingpoint of nearly
Answer» `-0.036^(@)C` `-0.108^(@)C` `-0.216^(@)C` `-0.422^(@)C` Solution :`DELTA T_(f)=(K_(f)xx1000xx W)/(m XX W)=-0.108^(@)C`

Discussion

30807.	The freezing point of 5% W/W solution of sucrose is 271 K. The freezing point of pure water is 273.15 K. What will be the freezing point of 5% W/W Glucose solution ?
Answer» 271.2 K 273.15 K 269.07 K 277.23 K SOLUTION :`DELTA T_(F)=K_(f)XX m` `2.15=(1000xx5)/(95xx342)xx K_(f)` `K_(f)=13.9707` `Delta T_(f)=(13.97xx1000xx5)/(95xx180)` `=4.084=269.07 K`

Discussion

30808.	The freezing point of a 0.05 molal solution of a non-eletrolyte in water is ( K_(f)=1.86Km^(-1))
Answer» `-1.86^(@)C` `-0.93^(@)C` `-0.093^(@)C` `0.93^(@)C` SOLUTION :`DeltaT_(f)=K_(f)xx m = 1.86xx0.05=0.093` FREEZING POINT `=0-0.093= - 0.093^(@)C`

Discussion

30809.	The freezing point of a 0.05 m BaCl_(2) in water ( 100% ionisation) is about (K_(f)=1.86 Km^(-1)) :
Answer» `-0.279^(@)C` `-0.558^(@)C` `-0.093^(@)C` `-0.186^(@)C` Solution :`BaCl_(2) hArr Ba^(2)+2Cl^(-), i=3` `DeltaT_(F)=iK_(f) xx m` `DeltaT_(f)=3xx1.86 xx 0.05=0.279^(@)C`

Discussion

30810.	The freezing point of 1 molar NaCl solution assuming NaCl to be 100% dissociated in water is: (K_f = 1.86 K molality^-1)
Answer» `-1.86^@C` `-3.72^@C` `+1.86^@C` `+3.72^@C` ANSWER :B

Discussion

30811.	The freezing point of 1 molal CaCl_(2) aq. Solution is -3.62^(@)C, (K_(f)=1.86 unit). The percentage ionisation of CaCl_(2) is
Answer» `94.6%` `23.65%` `47.3%` `69.73%` ANSWER :C

Discussion

30812.	The freezing point of 1% aqueous solution of calcium nitrate will be:
Answer» `0^@C` Above `0^@C` `1^@C` Below `0^@C` ANSWER :D

Discussion

30813.	The freezing point of 0.1 M solution of glucose is -1.86^(@) C. If an equal volume of 0.3 M glucose solution is added, the freezing point of the mixture will be (assume M=m).
Answer» `-7.44^(@)` C `-5.58^(@)` C `-3.72^(@)` C `-2.79^(@)` C ANSWER :C

Discussion

30814.	The freezing point of 0.2 molal solution of acetic acid in benzene is 277.65 . Freezing point of pure benzene is 278.4 K and heat of fusion of benzene is 10.04 Kj//"mol". If molarity of solution is equal tomolality , and acetic acid is found in equilibrium with its dimmer, then answer the following based on this data The degree of association of acetic acid should be
Answer» `3 KXX "molality"^(-1)` `2Kxx"molality"^(-1)` `5Kxx"molality"^(-1)` `10Kxx"molality"^(-1)` Answer :C

Discussion

30815.	The freezing point of 0.08m NaHSO_(4) is -0.345^(@)C. Calculate the percentage of HSO_(4)^(-1) that transfer a proton to water to form SO_(4)^(-2) ion. K_(f) of H_(2)O=1.86 k "mol"^(-1)kg.
Answer» Solution :`NaHSO_(4)+H_(2)O(aq)toNa^(+)+H_(3)O^(+)+SO_(4)^(2-)` `t=0"" 1""0""0""0` `"at eq. " ""1-ALPHA""alpha""alpha""alpha` `i=(1+2 alpha)/1` `DeltaT_(f)=ixxK_(f)xxm` `0.345=(1+2alpha)xx1.86xx0.08` So `alpha=0.6591` % dissociation `=65.92%`

Discussion

30816.	The freezing point of 0.08 molal aq NaHSO_(4) solution is -0.372. Calculate alpha (degree of dessociation) of HSO_(4)^(-) given alpha (degree of dissociation) of NaHSO_(4) is 100% & K_(f) for H_(2)O=1.86 (K)/(m). Give answer after multiplying by 10.
Answer» ANSWER :5

Discussion

30817.	The freezing point of 0.01 M KCl solution is -2^(@)C. If BaCl_(2) is completely ionized, what is the freezing point of 0.01 M BaCl_(2) solution ?
Answer» `-3^(@)C` `+3^(@)C` `-2^(@)C` `-4^(@)C` SOLUTION :`Delta T_(F)=K_(f)xx m xx i` `2=K_(f)xx0.01xx2` `K_(f)=100^(@)C"kgmol"^(-1)` `Delta T_(f)=K_(f)xx m xx i` `=100xx0.01xx3` `= 3^(@)C` `T_(f)=-3^(@)C`

Discussion

30818.	The freezing point of 0.05 m solution of glucose in water is (K_f = 1.86°C m^(-1))
Answer» 0.093°C -1.86°C -0.093 °C -0.93°C Answer :C

Discussion

30819.	The freezing point (in .^(@)C) of a solution containing 0.1 g of K_(3)[Fe(CN)_(6)] (Mol. Wt. 329) in 100 g of water (K_(f)=1.86 "K Kg mol"^(-1)) is
Answer» `-2.3xx10^(-2)` `-5.7xx10^(-2)` `-5.7xx10^(-3)` `-1.2xx10^(-2)` SOLUTION :`Delta T_(f)=ixx K_(f)xx m` `=4xx1.86xx(0.1)/(329xx0.1)=2.3xx10^(-2)` `rArr T_(f)=0-2.3xx10^(-2)=-2.3xx10^(-2) .^(@)C`. `rArr V = 6 ML`.

Discussion

30820.	The freezing point (in .^(@)C) of a solution containing 0.1 g of K_(3)[Fe(CN)_(6)[ (Mol. Wt. 329) in 100 g of water (K_(f)="1.86 K kg mol"^(-1)) is
Answer» `-2.3xx10^(-2)` `-5.7xx10^(-2)` `-5.7xx10^(-3)` `-1.2xx10^(-2)` SOLUTION :`K_(3)[Fe(CN)_(6)]HARR 3K^(+)+[Fe(CN)_(6)]^(3-)` `therefore I = 4` `DeltaT_(F)=iK_(f)m=ixxK_(f)xx(w_(2))/(M)xx(1)/(w_(1))xx1000` `=4xx1.86xx(0.1)/(329)xx(1)/(100)xx1000` `=0.023 = 2.3xx10^(-2).^(@)C or K` `T_(f)=0-2.3xx10^(-2).^(@)C=-2.3xx10^(-2).^(@)C`

Discussion

30821.	Calculate the depression in the freezing point of a solution containing 0.1 g of K_(3)[Fe(CN)_(6)] in 100 g of H_(2)O ? (molecular weigth of K_(3)[Fe(CN)_(6)]=329, K_(f)=1.86K " kg mol"^(-1))
Answer» `-2.72xx^(-2)` `-5.7xx10^(-2)` `-5.7xx10^(-3)` `-1.2xx10^(-2)` ANSWER :A

Discussion

30822.	The freezing point depression of 0.1 molal solution acid in benzene is 0.256 K. For benzene K_(f) is 5.12 K kg mol^(-1) Calculate the value of Van't Hoff factor for enzoic acid in benzene. What conclusion can you drow about the molecular state of benzonic acid in cenzene ?
Answer» SOLUTION :Step I. Calculation of Van't Hoff factor. `"Freezing point depression "(DeltaT_(f))=ixxK_(f)xxm` `DeltaT_(f)=0.256 K, K_(f)=5.12" K kg mol"^(-1),m=0.1m=(0.1" mol kg"^(-1))` `i=(DeltaT_(f))/(K_(f)xxm)=((0.256 K))/((5.12" K kg mol"^(-1))xx(0.1"mol"^(-1)kg))=0.5` Step II. Predicting the MOLECULAR state of benzoic acid. `i=("Normal MOLAR mass")/(Observed molar mass")=0.=1/2.` This shows that the observed molar mass of benzoni acid is double the normal molar mass or benzoic axists as a dimer in benzene.

Discussion

30823.	The freezing point depression of 0.1 molal solution of acetic acid in benzene is 0.256 K , K_(f) for benzene is 5.12 K Kg mol^(-1). What conclusion can you draw about the molecular state of acetic acid in benzence ?
Answer» Acetic acid is DOUBLY ASSOCIATED Benzene is doubly associated Both are equally associated None of the above. Answer :A

Discussion

30824.	The freezing point depression of 0.1 m NaCl solution is 0.372^(@)C. What conclusion would you draw about its molecular state ? K_(f) for water is "1.86 K kg mol"^(-1).
Answer» Solution :`DeltaT_(f)"(obs.)"=0.372^(@)C, DeltaT_(f)"(cal.)"=1.86xx0.1=0.186^(@)C` As `DeltaT_(f)"(obs.)" gt DeltaT_(f)"(cal.)"` this MEANS NaCl is dissociated in solution. `"Further,i"=(DeltaT_(f)"(obs.)")/(DeltaT_(f)"(cal.)")=(0.372)/(0.186)=2."Buti"=("No. of particles after dissociation")/("No. of molecules taken")` This means that each NaCl MOLECULE ionizes to give two particles, i.e., `Na^(+) and Cl^(-)` ions as follows: `NaCl overset(+AQ)rarrNa^(+)+Cl^(-)`

Discussion

30825.	The freezing point depression constant for water is -1.86cm^(-1). If "5.00 g "Na_(2)SO_(4) is dissolved in "45.0 g "H_(2)O, the freezing point is changed by -3.82^(@)C. Calculate the van't Hoff factor for Na_(2)SO_(4)
Answer» 0.381 2.05 2.63 3.11 Solution :`DeltaT_(f)=i xxK_(f)xxm =i xxK_(f)xx(w_(2))/(M_(2))xx(1)/(w_(1))xx1000` `3.82=ixx1.86xx(5)/(142)xx(1)/(45)xx1000` `""("Molar mass of "Na_(2)SO_(4)=142)` `"or"i=2.62`

Discussion

30826.	The freezing point depression of 0.1 m NaCI solution is 0.372^(@)C. What conlusion would you draw about the state in aqueous soluion ? (K_(f) "for water"=1.86 "K kg mol"^(-1)).
Answer» Solution :Step. Calculation of Van't HAFF factor. `"FREEZING point DEPRESSION"(DeltaT_(f))=ixxK_(f)xxm` `DeltaT_(f)=0.372^(@)C or 0.372 K, K_(f)=1.86" K kg MOL"^(-1), m=0.1" m"or 0.1" mol kg"^(-1)` `i=(DeltaT_(f))/(K_(f)xxm)=((0.372 K))/((1.86" K kg mol"^(-1))xx(0.1" mol"^(-1)kg))=2.` Step II.`"Calculation of degree of DISSOCIATION"(alpah).` `NaCI to Na^(+)+CI^(-)` `therefore""alpha=(i-1)/(n-1)=(2-1)/(2-1)1.` `"Thus, degree of dissociation of "alpha=1 "or it is 100% dissociated in dqueous soltion.`

Discussion

30827.	The freezing point depression of 0.1 M molal solution of acetic acid in benzene is 0.256 K. K_(f) for benzene is "5.12 K kg mol"^(-1). What conclusion can you draw about the molecular state of acetic acid in benzene?
Answer» Solution :Here, we are given `m="0.1 mol kg"^(-1),""K_(f)="5.12 K kg mol"^(-1)` `therefore"Theroretically calculated value of "DeltaT_(f)" will be "DeltaT_(f)=K_(f) xx m ="5.12 K kg mol"^(-1)xx"0.1 mol kg"^(-1)=0.512K` Observed (experimental) value of `DeltaT_(f)=0.256K` (Given) Thus, the observed value is HALF of the theoretical value. Since the observed value depends upon the number of particles actually present, this menas that the number of particles actually present in half of the theoretical value. In other words, the molecules of acetic acid are DOUBLY associated in benzene, i.e., they exist as dimers, `(CH_(3)COOH)_(2)`. Alternatively, van't HOFF factor, `i=("Observed colligative property")/("Calculated colligative property")=(0.256)/(0.512)=(1)/(2)` `"Also,i"=("Calculated mol. mass")/("Observed mol. mass")""therefore"Observed mol. mass "=("Calculated mol. mass")/("i")` But calculated molar mass of `CH_(3)COOH="60 g mol"^(-1) ""therefore"Observed molar mass "=(60)/(1//2)="120 g mol"^(-1)` Thus, the observed molar mass is double of the theoretical value. Hence, acetic acid EXISTS as doubly associated, i.e., as `(CH_(3)COOH)_(2)`.

Discussion

30828.	The freezing point depression constant for water is -1.86^@C m^(-1). If 5.00g Na_2SO_4 is dissolved in 45.0 g H_2O the freezing point is changed by -3.82^@C, calculate the van't Hoff factor for Na_2SO_4.
Answer» 0.381 2.63 2.05 3.11 Answer :B

Discussion

30829.	The freezing point depression constant for water is- 1.86 cm^(-1) . If 5.00 g Na_2SO_4is dissolved in 45.0 g H_2O , the freezing point is changed by-3.82^@C . Calculate the van't Hoff factor for Na_2SO_4 .
Answer» 2.05 2.63 3.11 0.381 Answer :B

Discussion

30830.	the freeof burning liquidparaaffincannotbeextingushed by throwingoverthisExpainby givingreaason .
Answer» SOLUTION :Theliquidparaffinis ligtherthanwater.itflowsoverwatersurfaceandcontinues BURNING .

Discussion

30831.	The free energy profile for the transformation of A into G is shown below: What is the least stable transition state:
Answer» E C Decreases with INCREASE of temperature B Solution :LEAST STABLE TRANSITION state has HIGHEST energy.

Discussion

30832.	The free energy of formation of NO is 78 kJ mol^(-1) at the temperature of an automobile engine (1000 K). What is the equilibrium constant for this reaction at 1000 K (1)/(2)N_(2)(g)+(1)/(2)O_(2)(g)iffNO(g)
Answer» `8.4xx10^(-5)` `7.1xx10^(-19)` `4.2xx10^(-10)` `1.7xx10^(-19)` ANSWER :A

Discussion

30833.	The free energy for a reaction having DeltaH=31400 cal. DeltaS=32 cal K^(-1)mol^(-1) at 1000^(@)C is
Answer» `-9336` CAL `-7386` cal `-1936` cal `+9336` cal Solution :`DeltaG=DeltaH-TDeltaS=31400-1273xx32` `=31400-40736=-9336` cal.

Discussion

30834.	The free energy profile for the transformation of A into G is shown below: The slowest step in the forward direction:
Answer» A to C C to E E to G C to F Solution :rate `prop(1/("ACTIVATION energy"))`

Discussion

30835.	The free energy for a reaction having DeltaH=31400cal,DeltaS=32" cal "K^(-1)mol^(-1) at 1000^(@)C is
Answer» `-9336cal` `-7386cal` `-1936cal` `+9336cal` SOLUTION :From Gibbs-Helmholtz EQUATION, we KNOW `DeltaG=DeltaH-T*DeltaS=31400-1273xx32` `=31400-40736=-9336cal`

Discussion

30836.	The free energy for a recersible reaction at equlibrium is
Answer» NEGATIVE Positive Zero EITHER positive or negative but not zero ANSWER :B

Discussion

30837.	The free energy change of a reaction is zero when:
Answer» The REACTANTS are INITIALLY MIXED A catalyst is added The system is at equilibrium The reactants are completely consumed ANSWER :C

Discussion

30838.	The free energy change for the following reactions are given below C_(2)H_(5)(g)+5/2 O_(2)(g) to 2CO_(2)(g)+H__(2)O (I), triangleG^(@)=-1234kJ C(s)+O_(2)(g) to CO_(2) (g), triangleG^(@)=-384kJ H_(2)(g)+1/2O_(2)(g) to H_(2)O(l), triangleG^(@)=-237kJ What is the standard free energy change for the reaction
Answer» `-208 KJ` `-2259kJ` `+2259kJ` `209kJ` ANSWER :D

Discussion

30839.	The free energy change for the following reactions are given below, C_(2)H_(2)(g)+(5)/(2)O_(2)(g)rarr2CO_(2)(g)+H_(2)O(l),DeltaG^(@)=-1234 kJC(s)+O_(2)(g)rarrCO_(2)(g)DeltaG^(@)=-394 kJH_(2)(g)+(1)/(2)O_(2)(g)rarrH_(2)O(l)DeltaG^(@)=-237 kJWhat is the standard free energy change for the reaction H_(2)(g)+2C(s)rarrC_(2)H_(2)(g)
Answer» `-209 kJ` `-2259 kJ` `+2259 kJ` 209 kJ Solution :By `2XX(ii)-(i)+(III)` `H_(2(g))+2C_((s))rarrC_(2)H_(2(g)), DeltaG^(@)=209 kJ`.

Discussion

30840.	The free energy change for the following reaction are given below: C_(2)H_(2)(g)+(5)/(2)O_(2)(g) to 2CO_(2)(g)+H_(2)O(l),DeltaG^(@)=-1234J C(s)+O_(2)(g)toCO_(2)(g),DeltaG^(@)=-394kJ H_(2)(g)+(1)/(2)O_(2)(g)toH_(2)O(l),DeltaG^(@)=-237kJ what is the standard free energy change for the reaction? H_(2)(g)+2C(s)toC_(2)H_(2)(g)
Answer» `-209kJ` `-2259kJ` `+2259kJ` `209kJ` SOLUTION :`C_(2)H_(2)(g)+(5)/(2)O_(2)(g) to 2CO_(2)(g)+H_(2)O(l),DeltaG^(@)=-1234kJ`. . (i) `C(s)+O_(2)(g) to CO_(2)(g),DeltaG^(@)=-394KJ` . . (ii) `H_(2)(g)+(1)/(2)O_(2)(g) to H_(2)O(l),DeltaG^(@)=-237kJ` . . (iii) EQS. (ii)+(iii)-(i). `2C(s)+H_(2)(g)toC_(2)H_(2)(g)` `DeltaG^(@)=2(-394)+(-237)-(-1234)=209kJ`.

Discussion

30841.	The free energy change for a reversible reaction at equilibrium
Answer» Large POSITIVE Small negative Small positive 0 Solution :`DeltaG` at EQUILIBRIUM = 0.

Discussion

30842.	The free energy change for a reversible reaction at equilibrium is:
Answer» Zero Small positive Small negative Large positive Answer :A

Discussion

30843.	Thefree energiesof formation( Delta_fG )of MgO(s) and CO(g) at1273 Kand2273 Krepectivelyare givenbelow :Delta _f[ Mg O (s)] =-941 kJ mol^(-1)at1273 KDelta_fG [MgO (s)] =- 314 kJ mol ^(-1)at 2273 KDelta _fG [CO (g) ] =-439 kJ mol ^(-1)at1273 KDelta _fG [CO (g)]=-628 kJmol^(-1) at2273 KOnthebasis of abovepredictthe temperature at which carboncan beusedas areducingagentforMgO(s).
Answer» Solution :The EQUATIONFOR reductionofMgO by carbon MAYBE writtenas `MgO (s)+C (s)to M g O (s)+CO (g) ` Thefreeenergy`(Delta _ R G ) `oftheabove reaction attwodifferenttemperaturemay becalculatedas follows : ` ""Delta _ rG=Delta_f G [CO (g) ] - Delta _fG [MgO (s)] ` At 1273 K,`"" Delta _ rG=-439 -(-941) = +502 kJ mol ^(-1) ` At2273 K,` "" Delta_rG =-623-(-314) =-314 kJ mol^(-1) ` Since ` Delta_r G `is- veat2273 K,therefore , carboncan be usedasareducing AGENT forMgOat2273K.

Discussion

30844.	The free energy change for a reaction is zero when
Answer» The reactants are INITIALLY mixed A CATALYST is added The SYSTEM is at equilibrium The reactants are COMPLETELY consumed Answer :C

Discussion

30845.	What is the free energy change at equilibrium state ?
Answer» ANSWER :ZERO

Discussion

30846.	The free energy change Delta G = 0, when
Answer» the system is at equilibrium catalyst is added reactants are INITIALLY MIXED thoroughly the reactants are completely consumed Answer :A

Discussion

30847.	The fragrance of flowers is due to the presence of some steam volatile organic compounds called essential oils. These are generally insoluble in water at room temperature but are miscible with water vapour in vapour phase. A suitable method for the extraction of these oils from the flowers is
Answer» Distillation Crystallisation Distillation under REDUCED pressure Steam distillation Answer :D

Discussion

30848.	The fraction of total volume occupies by the atoms present in a simple cube is
Answer» `( pi )/(3 sqrt( 2))` `( pi )/( 4 sqrt( 2))` `( pi )/( 4)` ` ( pi )/( 6)` Solution :Packing fraction of a simple cube `= ( " Volume of one sphere ")/( "Volume of cubic unit cell")` Edge LENGTH `a=2r` or `r = a//2 ` `:.` Packing fraction `= (((4)/(3) pi (a)/(2))^(3))/(a^(3)) = ( pi )/( 6)`

Discussion

30849.	The fraction of total volume occupied by the atoms in a simple cubic is …………….. .
Answer» `((PI)/(4sqrt2))` `((pi)/(6))` `((pi)/(6))` `((pi)/(3sqrt2))` Solution :HINT : `(((4)/(3)pir^(3))/(a^(3)))=(((4)/(3)pi((a)/(2))^(3))/(a^(3)))=((pi)/(6))`

Discussion

30850.	The fraction of total volume occupied by the atoms present in a simple cube is
Answer» `(pi)/(6)` `(pi)/(3sqrt(2))` `(pi)/(4sqrt(2))` `(pi)/(4)` Solution :P.F.`=("vol. occupied by the atom in a unit CELL")/("vol. of unit cell")=((4)/(3)pir^(3))/(a^(3))` For SIMPLE unit `a=2r=((4)/(3)pir^(3))/(8r^(3))=(4pi)/(24)=(pi)/(6)`

Discussion

Explore topic-wise InterviewSolutions in .

The freezing point of a solution containing 0.2 g of acetic acid in 20.0 g of benzeneis lowered by 0.45^@C . Calculate the degree of association of acetic acid in benzene. K_f(C_6H_6) = 5.12 K "mol"^(-1) kg

The freezing point of solution containing 0.2 g of acetic acid in 20.0 g of benzene is lowered by 0.45^(@)C. Calculate the degree of association of acetic acid in benzene. (K_(f)=5.12 K^(@) mol^(-1) kg^(-1))

The freezing point of a 1.00 m aqueous solution ofHF is found to be -1.91 ^(@)C. The freezing point constant of water, K1is 1 .86 K kg mol^(-1) The percentage dissociation ofHF at this concentration is

The freezing point of a solution containing 0.2 g of acetic acid in 20.0 g benzene is lowered by 0.45^@C. The degree of association of acetic acid in benzene is (Assume acetic acid dimerises in benzene and K_f for benzene = 5.12 K kg "mol"^(-1) ) ?

The freezing point of a 0.08 molal solution of NaHSO^(4) is -0.372^(@)C. Calculate the dissociation constant for the reaction. K_(f) for water =1.86 K m^(-1)

The freezing point of a 0.01 M aqueous glucose solution at 1 atmosphere is -0.18^(@)C. To it, an addition of equal volume of 0.002 M glucose solution will , produced a solution with freezingpoint of nearly

The freezing point of 5% W/W solution of sucrose is 271 K. The freezing point of pure water is 273.15 K. What will be the freezing point of 5% W/W Glucose solution ?

The freezing point of a 0.05 molal solution of a non-eletrolyte in water is ( K_(f)=1.86Km^(-1))

The freezing point of a 0.05 m BaCl_(2) in water ( 100% ionisation) is about (K_(f)=1.86 Km^(-1)) :

The freezing point of 1 molar NaCl solution assuming NaCl to be 100% dissociated in water is: (K_f = 1.86 K molality^-1)

The freezing point of 1 molal CaCl_(2) aq. Solution is -3.62^(@)C, (K_(f)=1.86 unit). The percentage ionisation of CaCl_(2) is

The freezing point of 1% aqueous solution of calcium nitrate will be:

The freezing point of 0.1 M solution of glucose is -1.86^(@) C. If an equal volume of 0.3 M glucose solution is added, the freezing point of the mixture will be (assume M=m).

The freezing point of 0.08m NaHSO_(4) is -0.345^(@)C. Calculate the percentage of HSO_(4)^(-1) that transfer a proton to water to form SO_(4)^(-2) ion. K_(f) of H_(2)O=1.86 k "mol"^(-1)kg.

The freezing point of 0.08 molal aq NaHSO_(4) solution is -0.372. Calculate alpha (degree of dessociation) of HSO_(4)^(-) given alpha (degree of dissociation) of NaHSO_(4) is 100% & K_(f) for H_(2)O=1.86 (K)/(m). Give answer after multiplying by 10.

The freezing point of 0.01 M KCl solution is -2^(@)C. If BaCl_(2) is completely ionized, what is the freezing point of 0.01 M BaCl_(2) solution ?

The freezing point of 0.05 m solution of glucose in water is (K_f = 1.86°C m^(-1))

The freezing point (in .^(@)C) of a solution containing 0.1 g of K_(3)[Fe(CN)_(6)] (Mol. Wt. 329) in 100 g of water (K_(f)=1.86 "K Kg mol"^(-1)) is

The freezing point (in .^(@)C) of a solution containing 0.1 g of K_(3)[Fe(CN)_(6)[ (Mol. Wt. 329) in 100 g of water (K_(f)="1.86 K kg mol"^(-1)) is

Calculate the depression in the freezing point of a solution containing 0.1 g of K_(3)[Fe(CN)_(6)] in 100 g of H_(2)O ? (molecular weigth of K_(3)[Fe(CN)_(6)]=329, K_(f)=1.86K " kg mol"^(-1))

The freezing point depression of 0.1 molal solution acid in benzene is 0.256 K. For benzene K_(f) is 5.12 K kg mol^(-1) Calculate the value of Van't Hoff factor for enzoic acid in benzene. What conclusion can you drow about the molecular state of benzonic acid in cenzene ?

The freezing point depression of 0.1 molal solution of acetic acid in benzene is 0.256 K , K_(f) for benzene is 5.12 K Kg mol^(-1). What conclusion can you draw about the molecular state of acetic acid in benzence ?

The freezing point depression of 0.1 m NaCl solution is 0.372^(@)C. What conclusion would you draw about its molecular state ? K_(f) for water is "1.86 K kg mol"^(-1).

The freezing point depression constant for water is -1.86cm^(-1). If "5.00 g "Na_(2)SO_(4) is dissolved in "45.0 g "H_(2)O, the freezing point is changed by -3.82^(@)C. Calculate the van't Hoff factor for Na_(2)SO_(4)

The freezing point depression of 0.1 m NaCI solution is 0.372^(@)C. What conlusion would you draw about the state in aqueous soluion ? (K_(f) "for water"=1.86 "K kg mol"^(-1)).

The freezing point depression of 0.1 M molal solution of acetic acid in benzene is 0.256 K. K_(f) for benzene is "5.12 K kg mol"^(-1). What conclusion can you draw about the molecular state of acetic acid in benzene?

The freezing point depression constant for water is -1.86^@C m^(-1). If 5.00g Na_2SO_4 is dissolved in 45.0 g H_2O the freezing point is changed by -3.82^@C, calculate the van't Hoff factor for Na_2SO_4.

The freezing point depression constant for water is- 1.86 cm^(-1) . If 5.00 g Na_2SO_4is dissolved in 45.0 g H_2O , the freezing point is changed by-3.82^@C . Calculate the van't Hoff factor for Na_2SO_4 .

the freeof burning liquidparaaffincannotbeextingushed by throwingoverthisExpainby givingreaason .

The free energy profile for the transformation of A into G is shown below: What is the least stable transition state:

The free energy of formation of NO is 78 kJ mol^(-1) at the temperature of an automobile engine (1000 K). What is the equilibrium constant for this reaction at 1000 K (1)/(2)N_(2)(g)+(1)/(2)O_(2)(g)iffNO(g)

The free energy for a reaction having DeltaH=31400 cal. DeltaS=32 cal K^(-1)mol^(-1) at 1000^(@)C is

The free energy profile for the transformation of A into G is shown below: The slowest step in the forward direction:

The free energy for a reaction having DeltaH=31400cal,DeltaS=32" cal "K^(-1)mol^(-1) at 1000^(@)C is

The free energy for a recersible reaction at equlibrium is

The free energy change of a reaction is zero when:

The free energy change for a reversible reaction at equilibrium

The free energy change for a reversible reaction at equilibrium is:

The free energy change for a reaction is zero when

What is the free energy change at equilibrium state ?

The free energy change Delta G = 0, when

The fragrance of flowers is due to the presence of some steam volatile organic compounds called essential oils. These are generally insoluble in water at room temperature but are miscible with water vapour in vapour phase. A suitable method for the extraction of these oils from the flowers is

The fraction of total volume occupies by the atoms present in a simple cube is

The fraction of total volume occupied by the atoms in a simple cubic is …………….. .

The fraction of total volume occupied by the atoms present in a simple cube is