94420 + Interview Questions in Chemistry in Class 12

1.	Calculate the cell potential for Ag_((S))\|Ag^(+)(0.01M)\|\|Ag^(+)(0.1M)\|Ag_((S)) at 298K.
Answer» ANSWER :`0.0592V`

Discussion

2.	Calculate the cell e.m.f. at 25^(@)C for the cell: Mg(s)\|Mg^(2+)(0.01M)\|\|Sn^(2+)(0.1M)\|Sn(s) Given E_(Mg^(2+)//Mg)^(@)=-2.34V,E_(Sn^(2+)//Sn)^(@)=-0.136V,1F=96,500" C "mol^(-1) Calculate the maximum work that can be accomplished by the operation of this cell.
Answer» Solution :`MG+SN^(2+)toMg^(2+)+Sn,E_(cell)^(@)=-0.136-(-2.34)=2.204V` `E_(cell)=E_(cell)^(@)-(0.0591)/(2)"log"([Mg^(2+)])/([Sn^(2+_)]), thereforeE_(cell)=2.204-(0.0591)/(2)"log"(0.01)/(0.1)=2.500V` `w_(max)=-DeltaG^(@)=nFE_(cell)^(@)=2xx96500xx2.204=425372J` (Note that `w_(max)` is RELATED to `E_(cell)^(@)` and not `E_(cell)`).

Discussion

3.	Calculate the capillary depression of Hg in a tube of diameter 1.0 mm. Assume that the contact angle is zero. The density of Hg is 13.6 xx 10^3 kg m^(-3) and the surface tension of Hg is 0.460 Nm^(-1).
Answer» ANSWER :`1.38 XX 10^(-2)m`

Discussion

4.	Calculate the buffer capacity of 1L solution of : (i)0.1MCH_(3)COOH and 0.1MCH_(3)COONa (ii) 0.2MCH_(3)COOH and 0.2MCH_(3)COONa Given:pK_(a)(CH_(3)COOH)=4.74 Which will be a better buffer?
Answer» SOLUTION :Buffer capcaity =`(2.303(a+x)(b-x))/(a+b)~~(2.303ab)/(a+b)xltlta,b` Buffer capacity =`(0.1xx0.1xx2.303)/(0.1+0.1)=0.11515` Buffer capacity `=(0.2xx0.2xx2.303)/(0.2+0.2)=0.2303` Second buffer solution (having GREATER buffer CAPCITY )can be called BUTTER buffer.

Discussion

5.	Calculate the bond enthalpy of Xe-F bond as given in the equation, XeF_(4)(g)toXe^(+)(g)+F^(-)(g)+F_(2)(g)+F_(g),Delta_(r)H=292" kcal "mol^(-1) Ionisation energy of Xe=279 kcal/mol bond energy (F-F)=38 kcal/mol Electron affinity of F=85 kcal/mol
Answer» 8.5 kcal/MOL 34 kcal/mol 24 kcal/mol none of these Solution :`Delta_(r)H=`Heat supplied-heat EVOLVED `292=(4x+279)-(38+85)impliesx=34` kcal `mol^(-1)`

Discussion

6.	Calculate the bond energy of C-H bond from the following data : (a) C (s) + 2 H_(2) (g) to CH_(4) (g) , Delta H = -74.8 KJ (b) H_(2) (g) to 2 H (g) , Delta H = 435.4 KJ (c) C(s) to C(g) , Delta H = 718.4 KJ
Answer» 316.0 KJ/mol 416 KJ/mol 516 KJ/mol 616.0 KJ/mol Answer :B

Discussion

7.	Calculate the bond energy of C-=C in C_(2)H_(2) from the following data : (i)C_(2)H_(2)(g)+2(1)/(2)O_(2)(g) to 2CO_(2)(g)+H_(2)O,DeltaH=-310"kcal" (ii)C(s)+O_(2)(g) to CO_(2)(g),DeltaH=-94"kcal" (iii) H_(2)(g)+(1)/(2)O_(2)(g) to H_(2)O(g),DeltaH=-68 "kcal" Bond energy of C-H bonds =99 kcal Heat of atomisation of C=171 kcal Heat of atomisation of H=52 kcal
Answer» SOLUTION :Let us first calculate heat of formation of `C_(2)H_(6)` from which we can calculate bond energy of `C-=C` bond. `2C(s)+H_(2) to C_(2)H_(2)(g)` , `DeltaH=?` Applying the inspection method , i.e., `[2xx` Eqn. (ii) `+` Eqn. (iii) - Eqn. (i)] we get `2C(s)+2O_(2)(g)+H_(2)(g)+(1)/(2)O_(2)(g)-C_(2)H_(2)(g)-2(1)/(2)O_(2) to 2CO_(2)(g)+H_(2)O(g)-2CO_(2)(g)-H_(2)O(g), DeltaH=2xx(-94)+(-68)-(-310)kcal` or `2C(s)+H_(2)(g) to `C_(2)H_(2)(g) `, `DeltaH=54` kcal Now heat CHANGES for reactants Heat of atomisation of 2 moles of `C=2xx171` kcal Heat of atomisation of 2 moles of `H=2xx52` kcal And heat change for the product `(H-C-=C-H)` Heat of formation of 2 moles of C-H bonds `=-2xx99` kcal Heat of formation of 1 MOLE of `C-=C` bonds `=x` (say) Summing up, we get heat of formation of `C_(2)H_(2)`, `2xx171+2xx52-2xx99+x=54` `x=-194` kcal Hence bond energy of `C-=C` bond in `C_(2)H_(2)` is `+194` kcal

Discussion

8.	Calculate the bolling point of one mole aqueous solution (density 1.06 g cm^(-3)) of KBr. (Given : K_(b), for H_(2)O =0.52K kg mol^(-1), Atomic mass : K =39 , Br = 80 ]
Answer» Solution :One molar aqueous solution BOILING POINT = 7 d = 1.06 g/ml assuming VOLUME 1 L of KBr. Weight of KBr = 1.06 g `DeltaK_(b)=0.52` K kg /mol . `DeltaT_(b)=iK_(b)` m weight of solvent 1 kg `DeltaT_(b) = T_(2)-T_(1)` where `T_(1)= 373 K` `DeltaT_(b) = iK_(b) xx(W_(B)xx1000)/(M_(B)W_(A))` `DeltaT_(b)=2xx0.52 xx(1.06xx1000)/(119xx1)` `DeltaT_(b)=(1.1024xx1000)/(119)` =9.26 `T_(2)-373=9.26` `=382.26 K` .

Discussion

9.	Calculate the boiling point sollution when 2 g of Na_(2)SO_(4) (M=142 g mol^(-1)) was dissoved in 50 g of water assuming that NaSO_(4) undergoes complete ionization (K_(b) for water = 0.52 K kg mol^(-1))
Answer» Solution :Step I. Calculation of elecation in b.p. TEMPERATURE `Na_(2)SO_(4)" DISSOCIATES in aqueous solution as ":` `Na_(2)SO_(4)(s)overset(("aq"))to2Na^(+)("aq")+SO_(4)^(2-)("aq")` `i=3, W_(B)==2g, M_(B)=142"g MOL"^(-1), W_(A)=50g=0.050 kg` `K_(b)=0.52" K kg mol"^(-1)` `DeltaT_(b)=ixxK_(b)xxm=(xx(0.52" K kg mol"^(-1))xx(2g))/((142" g mol"^(-1))xx(0.050" kg "))=0.44 K` Step II. `"Calculation of boilling point of solution" (T_(b))` `T_(b)=T_(b)^(@)+DeltaT_(b)=373 K+0.44 K=373.44 K`

Discussion

10.	Calculate the boiling point of solution when 4 g of MgSO_4 (M = 120 g "mol"^(-1))was dissolved in 100 g of water, assuming MgSO_4undergoes complete ionisation.
Answer» SOLUTION :Apply the relation ` Delta T_b = i xx K_b xx m` For `MgSO_4 , i = 2` Molarity of solution = ` (4//120)/(100) xx 1000 = 4/120 xx 10 = 1/3` Substituting the values in EQUATION (i), we have ` Delta T_b = 2 xx 0.52 xx 1/3 = 0.347` Boiling point of the solution = `100 + 0.347 = 100.347^@C = 373.347 K `

Discussion

11.	Calculate the boiling point of bromine from the following data : DeltaH^(0) and DeltaS^(0) values of Br_(2)(l) to Br_(2)(g) are 30.91 kJ/"mole" and 93.2 J/"mol". K respectively. Assume that DeltaH and DeltaS do not vary with temperature.
Answer» Solution :Consider the process : `Br_(2)(L) to Br_(2)(G)` The b.p. of a LIQUID is the temperature at which the liquid and the pure gas coexist at equilibrium at 1 atm `:. DeltaG=0` As it is given that `DELTAH` and `DeltaS` do not change with temperture `DeltaH=DeltaH^(@)=30.91kJ` `DeltaS=DeltaS^(@)=93.2J//K=0.0932kJ//K` We have, `DeltaG=DeltaH-TDeltaS=0` `:.T=(DeltaH)/(DeltaS)=(30.91)/(0.0932)=331.6K` This is the temperature at which the SYSTEM is in equilibrium that is the b.p. of bromine.

Discussion

12.	Calculate the boiling point of a solution prepared by adding 15.00 g of NaCl to 250.0 g of water. (K_(b), for water 0.512 K. kg mol^(-1), molar mus of NaCl = 58.44g)
Answer» SOLUTION :`DeltaT_(B)=K_(b)xxm` `DeltaT_(b)=K_(b)XX(W_(B))/(M_(B))xx(1000)/(W_(A)" in g")` `T_(b)-T_(b)^(0)= (0.512xx15xx1000)/(58.44) ` = 0.5256 K `T_(b) = T_(b)^(0)+DeltaT_(B) = 0.53K` `T_(b)= 373 +0.5256= 373.53K`

Discussion

13.	Calculate the boiling point of a solution prepared by adding 15.00 g of NaCl to 250.0 g of water (K_(b) for water = "0.512 K kg mol"^(-1) , Molar mass of NaCl = 58.44 g)
Answer» Solution :`DeltaT_(B)=iK_(b)m=2xx0.512xx(15)/(58.44)XX(1)/(250)xx1000=1.05, T_(b)=100+1.05^(@), T_(b)=100+1.05^(@)C=101.05^(@)C`

Discussion

14.	Calculate the boiling point of a solution prepared by adding 15.0 g of NaCl of 250.0 g of water (K_(b) for water ="0.512 K kg mol"^(-1), Molar mass of NaCl = 58.44 g)
Answer» SOLUTION :(B) `101.05^(@)C(i=2)`

Discussion

15.	Calculate the boiling point of a solution containing 25 g urea (NH_(2)CONH_(2)) and 25 g thiourea (NH_(2)CSNH_(2)) in 500 g chloroform, CHCl_(3). The boiling point of pure chloroform is 61.2^(@)C and K_(b)=3.63 K m^(-1).
Answer» SOLUTION :25 G urea `=(25)/(60)` mole = 0.417 mole, 25 g THIOUREA `= (25)/(76)` mole = 0.329 mole TOTAL moles `= 0.417+0.329=0.746` Mass of solvent `= 500 g = 0.500 kgtherefore "Molality"=(0.746 MOL)/(0.500 kg)=1.492 m`, `Delta T_(b)=K_(b)xx m=3.63xx1.492 = 5.416^(@)` Boiling point of solution `= 61.2+5.416^(@)C=66.616^(@)C`.

Discussion

16.	Calculate the boiling point of a solution containing 0.61 g of benzoic acid in 50 g of carbon disulphide assuming 84% dimerisation of the acid. The boiling point and K_(b) of CS_(2) are 46.2^(@)C and "2.3 K kg mol"^(-1) respectively.
Answer» SOLUTION :`i=1-(alpha)/(2)=1-(0.84)/(2)=1-0.42=0.58` `DeltaT_(b)=iK_(b)m=0.58xx2.3xx(0.61)/(122)xx(1000)/(50)=0.1334^(@)."HENCE "T_(b)=46..2+0.133=46.333^(@)C`.

Discussion

17.	Calculate the boiling point of a solution containing 0.456 g of camphor (mol. Mass = 152) dissolved in 31.4 g of acetone (b.p. = 56.30^(@)C), if the molecular elevation constant per 100 g of acetone is 17.2^(@)C.
Answer» SOLUTION :Here, we have `w_(2)=0.456 g, M_(2)=152, w_(1)=31.4 g, T_(0)=56.30^(@)C, K_(b)=17.2^(@)C//100 g` `Delta T_(b)=(100 K_(b).w_(2))/(w_(1)M_(2))=(100xx17.2xx0.456)/(31.4xx152)=0.16^(@)C` `therefore` BOILING point of solution `(T_(b))=T_(b)^(@)+Delta T_(b)=56.30+0.16 = 56.46^(@)C`

Discussion

18.	Calculate the boiling point of a solution containing 1.8 g of a non-volatile solute dissolved in 90 g of benzene. The boiling point of pure benzene is 353.23 K, (K_(b)=2.53 K kg mol^(-1) , density of water= 1 g mol^(-1)).
Answer» Solution :`W_(B)=1.8g, W_(A)=90g=0.09 kg, T_(b)^(@)=353.23 K` `K_(b)=2.053" K kg mol"^(-1), M_(B)=58"g mol"^(-1),T_(b)=?` `DeltaT_(b)=(K_(b)xxW_(B))/(DeltaT_(b)xxW_(A))=((2.53" K kg mol"^(-1))XX(1.8g))/((58" g mol"^(-1))xx(0.09 kg))=0.872 K` `T_(b)=T_(b)^(@)+DeltaT_(b)=353.23+0.872=354.102 K.`

Discussion

19.	Calculatethe boiling point of a one molaraqueous solution (density 1.04g"mL "^(1))ofpotassium chlorideK_(b)for water = 0.52K "mol "^(-1). Atomicmasses : K = 39, Cl = 35.5 )
Answer» Solution :Concentrationof solution = 1M ,Density of solution = 1.04 g `" mL"^(-1)` LET us FIRST calculate the motality of the solution. mount of solute (KCI) = 1mol = 74.5 g , Volumeof solution= 1:L = 1000 mL Mass of the solution ` = 1000 xx 1.04 g = 1040 g` `:. ` Mass of solvent=` 1040 - 74.5` g = 965 . 5 g = 0.9655 kg `"Molality of the solution"= ("No. of moles of the solute ")/("Mass of the solvent in kg")= ("1mol")/(0.9655 "kg") = 1.0357 "molkg"^(-1) = 1.0357 m` KCl DISSOCIATES as : ` KCl to K^(+)Cl^(-)` ` :. ` Numberof particles after dissociation=2`:. `van't Hoff factor , i = 2 Now` Delta T _(b) = i xx K_(b) xx m = 2 xx 0.52 xx 1.0357 = 1.078^(@) C` `:. ` BOILING point of the solution = ` 100 + 1.078 = 101.078 ^(@)C`

Discussion

20.	Calculate the boiling point elevation for a solution prepared by adding 10 g of CaCI_(2) to 200 g of water. (K_(b) for water = 0.512 K kg mol, Molar mass of CaCI_(2) = 100 g mol^(-1)).
Answer» SOLUTION :`CaCI_(2)" DISSOCIATES in aqueous solution"` `CaCI_(2)(s)OVERSET(("aq"))toCa^(2+)("aq")+2CI^(-)("aq")` `i=3, W_(B)=10 g, W_(A)=200 g= 0.2, K_(b)=0.512" K KG mol"^(-1)` `M_(B)=110" g mol"^(-1)` `DeltaT_(b)=ixxK_(b)m=(ixxK_(b)xxW_(B))/(M_(B)xxW_(A))` `DeltaT_(b)=((3)xx(0.512" K kg mol"^(-1))xx(10g))/((110" g mol"^(-1))xx(0.2 kg ))=0.69 K`

Discussion

21.	Calculate the Avogadro constant from the following data: Density of solid NaCl = 2.165 g/cc. Distance between centres of adjacent Na^(+)and Cl^(-) = 0.2819 nm. Also, calculate the edge length of a cube containing 1 mole of NaCl and the number of ions (Na^+ Cl^(-)) along one edge of the cube.
Answer» ANSWER :`6.02 XX 10^(23), 3.0 cm, 1.064 xx 10^(8)`

Discussion

22.	Calculate the binding energy for ._(1)H^(2) atom. The mass of ._(1)H^(2) atom is 2.014102 amu where 1n and 1p have their weights 2.016490 amu. Neglect mass of electron.
Answer» ANSWER :2.2232 MEV

Discussion

23.	Calculate the average volume available to a molecule in a sample of nitrogen gas at STP. What is the average distance between neighbouring molecules if nitrogen molecules are spherical in nature ?
Answer» SOLUTION :`6.023 xx10^(23)` molecules of `N_(2)` occupy `22400 cm^(3)` `THEREFORE` one molecules of `N_(2)` occupies `=(22400)/(6.023xx10^(23))cm^(3)` or volume of one molecule of `N_(2)=3.72xx10^(-20) cm^(3)` Also, the average distance between TWO molecules =2r and `(4)/(3) pi r^(3)=3.72xx10^(-20)` `therefore r^(3) =(3.72xx10^(-20)xx3xx7)/(4xx22)` `therefore r=20.7xx10^(-8) cm` Thus, average distance `=2xx20.7xx10^(-8)` `=41.4xx10^(-8) cm`

Discussion

24.	Calculate the average atomic mass of hydrogen using the following data : {:("Isotops",%"Natural abundance","Molar mass"),(.^(1)H,"99.985","1"),(.^(2)H,"0.015","2"):}
Answer» <P> Solution :`"Average atomic mass "= (Sigmap_(i)A_(i))/(100)""(p_(i)="PERCENT abundance, "A_(i)="atomic mass of ISOTOPE ")` `=(99.985xx1+0.015xx2)/(100)=(99.985+0.030)/(100)=(100.015)/(100)=1.00015u.`

Discussion

25.	Calculate the atomicity of mercury molecules from the following data : (a) 10.0 g of mercury combine with 0.8 g of oxygen to form an oxide. (b) 500 mL of mercury vapour at S.T.P. weigh = 4.465g (c) Specific heat of mercury is 0.033.
Answer» Solution :Calculation of equivalent mass `"0.89 g of oxygen combine with Hg = 10.0 g"` `therefore"8 g of oxygen will combine with Hg "=(10)/(0.8)xx8=100g` `therefore"Equivalent mass of mercury"=100` Calculation of molar mass `"500 ML of mercury VAPOUR at S.T.P. weigh = 4.465 g"` `therefore"22400 mL of mercury vapour at S.T.P. will weigh"=(4.465)/(500)xx22400g=200g` `therefore"Molar mass ofmercury = 200 g mol"^(-1)` Calculation of valency. By Dulong and Petit's law. `"Approx. atomic mass of mercury "=(6.4)/("Sp. Heat")=(6.4)/(0.033)=193.9` `therefore"Valency of mercury"=("Approx. atomic mass")/("Valency")=(193.9)/(100)~~2"(as valency is a whole no.)"` `therefore"Actual atomic mass "="Eq. mass"XX"Valency"=100xx2=200` `"Calculation of atomicity.Atomicity"=("Mol. mass")/("At. mass")=(200)/(200)=1` Thus, mercury molecules are monoatomic.

Discussion

26.	Calculate the atomic mass (average) of chlorine using the following data : {:(,""%"Natural Adundance","Molar mass"),(.^(35)Cl,"75.77","34.9689"),(.^(37)Cl,"24.23","36.9659"):}
Answer» SOLUTION :`"FRACTIONAL abundance of ".^(35)Cl=0.7577," Molar MASS "=34.9689` `"Fractional abundance of ".^(37)Cl=" 0.2423. Molar mass "=36.9659` `therefore"Average atomic mass "=(0.7577)+(34.9689 AMU)+(0.2423)(36.9659amu)` `=26.4959+8.9568= 34.4527`

Discussion

27.	Calculate the angle at which (a) first order reflection and (b) second order reflection will occur in an X-ray spectrometer when X-ray of wavelength 1.54A^@ are diffracted by the atoms of a crystal, given that the interplanar distance is 4.04A^@.
Answer» SOLUTION :`10^@59. , 22^@24`

Discussion

28.	Calculate the amount of (NH_(4))SO_(4) in grams which must be added to 500 ml of 0.200 M NH_(3) to yield a solution with pH = 9.35 (K_(b) for NH_(3) = 1.78 xx 10^(-5))
Answer» 10.56 gm 15 gm 12.74 gm 16.25 gm Answer :A

Discussion

29.	Calculate the amount of limen, Ca(OH)_(2), required to remove hardness of 50,000 litres of well water which has been found to contain 1.62 g of calcium bicarbonate per 10 litre.
Answer» Solution :(i) Calculation of total `Ca(HCO_(3))_(2)` present 10 L of water contains `Ca(HCO_(3))_(2)=1.62 G` `therefore 50, 000 L` of water will contain `Ca(HCO_(3))_(2)=(1.62)/(10)xx50000 g=8100 g` (ii) Calculation of lime REQUIRED. The BALANCED EQUATION for the reaction involved is : `{:(""Ca(HCO_(3))_(2),+,""Ca(OH)_(2),rarr,2CaCO_(3)+2H_(2)O),("1 mole",,"1 mole",,),(40+(1+12+48)xx2=162 g,,40+(16+1)xx2=74 g,,):}` `162 g Ca(HCO_(3))_(2)` required lime = 74 g `therefore 8100g Ca(HCO_(3))_(2)` will required lime `=(74)/(162)xx8100 g=3700 g = 3.7 kg`

Discussion

30.	Calculate the amount of NaCl which must be added to 100 g of water so that the freezing point is depressed by 2 K. For water, K_(f)=1.86K//m.
Answer» Solution :`NaCl rarrNa^(+)+Cl^(-),i=2, DeltaT_(F)=eK_(f)m, i.e., 2=2xx1.86xxm or m=0.538` `therefore"AMOUNT of NaCl to bedissolved in 100 G"=(0.538xx100)/(1000)xx58.5g=3.147g.`

Discussion

31.	Calculate the amount of KCl which must be added to 1 kg of water so that the freezing point is depressed by 2 K. (K_(f) " for water = " 1.86 " K kg "mol^(-1))
Answer» Solution :SINCE one mole of KCI gives 2 mole PARTICLES, the value of I = 2 `DELTA T_(f) = 2 K ` `K_(f) = 1.86 " kg mol"^(-1)` Applying equation, `T_(f) = iK_(f) `m `m = (Delta T_(f))/(iK_(f)) = (2)/(2 xx 1.86)` Therefore, 0.54 mole of KCI should be added to one kg of water. MOLAR mass of KCI = 39 + 35.5 = 74.5 g AMOUNT of KCl = 0.54 `xx 74.5 `g = 40.23 g

Discussion

32.	Calculate the amount of KCl which must be added to 1 kg of water so that the freezing point is depressed by 3 K. (K for water = 1.86 K "mol"^(-1)).
Answer» Solution :KCI is a strong ELECTROLYTE and is completely DISSOCIATED in water as : `KCIto K^+ +Cl^-` Now, 1 mole of KCI will give 2 mole IONS. THEREFORE, the value of Van.t Hoff factor (i) will be equal to 2. `Deltat=i k_f m ` `K_f =1.86 (K kg "mol"^(-1))` `DeltaT = 3 K,i=2` `m= (Delta T )/(i K _f ) = ( 3 )/( 2 xx 1.86 ) = 0.81` mole Thus, 0.81 mole or `0.81 xx74.5 g = 60-34 g` must be added to 1 kg of water.

Discussion

33.	Calculate the amount of ice that will separated out when a solution containing 50 g of ethylene glycol in 200 g of water is colled to -9.3^(@)C.""(K_(f)" for "H_(2)O="1.86 K kg mol"^(-1))
Answer» Solution :`DeltaT_(f)=(1000K_(f)w_(2))/(w_(1)M_(2))` LET us calculate the amount of WATER `(w_(1))`present when `DeltaT_(f)=9.3^(@)` `9.3=(1000xx1.86xx50)/(w_(1)xx62)""("MOL. mass of "(CH_(2)OH)_(2)=62)` `"or"w_(1)=161.29 g` `therefore"Water frozen to ice "=200-161.29=38.71g`

Discussion

34.	Calculate the amount of heat that must be supplied to raise the temperature of 2 kg of water from 25^(@)C to its boiling point at one atmospheric pressure. The average specific heat of water in the range 25 - 100^(@)C is 4.184 JK^(-1) g^(-1).
Answer» 628 KJ 418.4 kJ 209.2 kJ 108.6 kJ Solution :Heat required for HEATING = `m xx C_(s) xx Delta T` `m = 2000g, Delta T = 100 - 25 = 75^(@)`, `C_(s) = 4.184 JK^(-1) g^(-1)` Heat required = `2000 x 4.184 xx 75` `= 6.28 xx 10^(5) J` or= 628 kJ

Discussion

35.	Calculate the amount of heat evolved during the complete combustion of 100 ml of liquid benzene from the following data. Predict your answer as (Delta H)/(100) ( in KJ/mol). (i) 18 gm of graphite on complete combustion evolve 585 KJ heat (ii) 15540 KJ heat is required to dissociate all the molecules of 1 litre water into H_(2) and O_(2). (iii) The heat of formation of liquid benzene is 48 kJ/mol (iv) Density of C_(6)H_(6)(l)=0.87 gm//ml
Answer» Solution :`(i)""C(s) + O_(2)(G) rarr CO_(2)(g) , DeltaH_(f)^(0) = -390KJ//mol, "" DeltaH_(f)^(0) = (585)/(18)xx12` `(ii) "" H_(2)(g)+(1)/(2)O_(2)(g) rarr H_(2)(l) , DeltaH_(f)^(0) =-280 KJ//mol ,"" Delta H_(f)^(0) = (15540)/(55.5)` `(III)"" C_(6)H_(6)(l) + (15)/(2)O_(2)(g) rarr 6CO_(2)(g) + 3H_(2)O(l) ,"" DeltaH_(f)^(0) = -48` `THEREFORE""DeltaH^(0) = [6(-390) + 3(-280)] - 48 = -3228 KJ//mol` `""` Mass of benzene is `= 0.87 xx 100 = 87g` `therefore""` Heat evolved from `87` gm benzene `=3600KJ` `""` Hence , `(DeltaH)/(100)=36KJ.`

Discussion

36.	Calculate the amount of energy released in ergs, calories and in joules when 0.001 kg of mass disappears. [Given, Velocity of light =3xx10^(8)m s^(-1)]
Answer» SOLUTION :According to Einstein equation , `E=mc^(2)` `m=0.001kg=1XX10^(-3)kg, c=3XX10^(8)ms^(-1)` `E=(1xx10^(-3))(3xx10^(8))^(2)=9xx10^(13)J` 1J`=10^(7)"erg"=0.24` cal `9xx10^(13)J=9xx10^(13)xx10^(7)"erg"=9xx10^(13)xx0.24` cal `=9xx10^(20)` erg `=2.16xx10^(13)` cal.

Discussion

37.	Calculate the amount of chlorine required to completely react with 4.56 g of hydrogen (h_2) to yield hydrochloric acid (HCl) ? Also calculate the amount of HCl formed.
Answer» ANSWER :161.88 , 166.44 G

Discussion

38.	Calculate the amount of CaCl_2[molar mass = 111 g "mol"^(-1) ] which must be added to 500 g of water to lower the freezing point by 2 K, assuming CaCl_2is completely dissociated.
Answer» Solution :APPLY the relation (without taking dissociation into CONSIDERATION) ` Delta T_f = (K_f xx 1000 xx w_2)/(M_2 xx w_1)` `2K = (1.86 K KG "MOL"^(-1) xx 1000 g kg^(-1) xx w_2)/(111 g "mol"^(-1) xx 500 g) ` ` w_2 = (2 xx 111 xx 500)/(1.86 xx 1000) = 59.68 g` One molecule of `CaCl_2` dissociates into one `Ca^(2+)` ion and two `Cl^-` ions. `CaCl_2 to Ca^(2+) + 2Cl^(-)` PARTICLE dissociates into three particles. Therefore, the amount of `CaCl_2`needed = `(59.68)/(3) = 19.89 g`

Discussion

39.	Calculate the amount of carbon dioxide that could be produced when (i) 1 mole of carbon is burnt in air. (ii) 1 mole of carbon is burnt in 16 g of dioxygen. (iii) 2 moles of carbon are burnt in 16 g of dioxygen.
Answer» Solution :The balanced equation for the combustion of carbon in dioxygen/air is `underset("1 mole")(C(s))""+""underset("(32 g)")underset("1 mole")(O_(2)(g))""rarr""underset("(44 g)")underset("1 mole")(CO_(2)(g))` (i) In air, combustion is complete. THEREFORE, `CO_(2)` produced from the combustion of 1 mole of carbon = 44 g. (II) As only 16 g of dioxygen is available, it can combine only with 0.5 mole of carbon, i.e., dioxygen is the limiting REACTANT. Hence, `CO_(2)` produced = 22 g. (III) Here again, dioxygen is the limiting reactant. 16 g of dioxygen can combine only with 0.5 mole of carbon. `CO_(2)` produced again is equal to 22 g.

Discussion

40.	Calculate the amount of calcium oxide required when it reacts with 852 g ofP_4O_10
Answer» <P> Solution :`CaO + P_4O_10 to Ca_3 (PO_4)_2` APPLY POAC for CA and P atoms 1008 G

Discussion

41.	Calculate the amount of CaCl_(2) (van't Hoff factor I =247) dissolved in 25 L solution so that its osmotic pressure at 300 K is 075 atmosphere. Given : Molar mass of CaCl_(2) is 111 g."mol"^(-1), R=0082L." atm "K^(-1)"mol^(-1).
Answer» Solution :The osmostic pressue (calculated), `pi=(n)/(v)RT` . . .(i) where, V=volume `=25L,` T=Temperature = 300 K, n= dissolving moles of `CaCl_(2)` `=("Wt. of solute(g)")/("Molecular mass of solute (M)")` Thus equation (i) becomes `pi_("Cal")=(g)/(MV)RT` `:,""pi_("Cal")=(g)/(111"g mol"^(-1)xx25l)xx0082l" atm"` `K^(-1)"mol"^(-1)xx300K` `=(246)/(2775)atm=08=886xxg" atm. gm"` Van's HOFF factor (i) `=("observed osmotic pressure "(pi_(ob)))/("calculated osmotic pressure"(pi_(Cal)))` `:.""247=(075" atm")/("calculated osmotic pressure")` `:.""247=(075" atm")/(pi_"Cal")or,pi_("Cal")=03036` atm Thus, `03036=00886xx"g gm or",g=34271" gm"` Hence, the amount of `CaCl_(2)` dissolve in `25" l"` solution is `3*4271` gm.

Discussion

42.	Calculate the amount of benzoic acid (C_6H_3COOH)required for preparing 250 mL of 0.15 M solution in methanol
Answer» SOLUTION :0.15 M solution MEANS that 0.15 mole of benzoic acid is present in 1 L, i.e., 1000 mL of thesolution. Molar mass of benzoic acid `(C_6H_5COOH) = 72 + 5 + 12 + 32 + 1 = 122 g "mol"^(-1)` ` THEREFORE `0.15 mole of benzoic acid contains = 0.15 x 122 g = 18.3 g of benzoic acid Thus, 1000 mL of the solution contain = 18.3 g benzoic acid ` therefore `250 mL of the solution will contain =` (18.3)/(1000) XX 250 = 4.575 g `benzoic acid

Discussion

43.	Calculate the amount of benzoic acid (C_(6)H_(5)COOH) required for preparing 250 mL of 0.15 M solution in methanol.
Answer» Solution :0.15 M solution MEANS that 0.15 mole of benzoic acid is PRESENT in 1 L, i.e., 1000 ML of the solution. Molar MASS of benzoic acid `(C_(6)H_(5)COOH)=72+5+12+32+1="122 g MOL"^(-1)` `therefore"0.15 mole of benzoic acid "=0.15xx122 g=18.3g` `therefore"250 mL of the solution will contain benzoic acid "=(18.3)/(1000)xx250="4.575 g."`

Discussion

44.	Calculate the amount of AgNO_3 which should be added to 60 ml of solution to perpare a concentration of 0.03 g ml^(-1)
Answer» 1.8 g 1.8 mg 0.018 g 0.018 mg Answer :A

Discussion

45.	Calculate the actual mass of one molecule of carbon dioxide (CO_(2))
Answer» SOLUTION :Molecular mass of `CO_(2)=44` amu 1 amu=`1.66xx10^(-24)G` So, the actual mass of `CO_(2)=44xx1.66xx10^(-24)` `=7.304xx10^(-23)g`

Discussion

46.	Calculate the Activation Energy (E_(a)) of a reaction that follows the equation ,k=(4.5xx10^(11)s^(-1)) e.
Answer» SOLUTION :`(E_(a))/(R)=28000 therefore E_(a)=232.79"kJ.mol"^(-1)`

Discussion

47.	Calculate the accelerating potential that must be imparted to a proton beam to give it an effective wavelength of 0.005nm.
Answer» Solution :Atomic weight of hydrogen is 1.008 amu. `therefore` weight of 1 MOLE of PROTON = 1.008g `therefore` weight of 1 proton `= (1.008)/(6.022 xx 10^(23))=0.167 xx 10^(-23)g` `=0.167 xx 10^(-26)kg` We have, `v= (h)/(m lamda)= ((6.63 xx 10^(-34)J.s))/((0.167 xx 10^(-26)kg) (0.005 xx 10^(-9) m))= 7.94 xx 10^(4) m//s` Kinetic energy `=(1)/(2) mv^(2) = (1)/(2) (0.167 xx 10^(-26)) (7.94 xx 10^(4))^(2)` `=5.26 xx 10^(-18)J` `=(5.26 xx 10^(-18))/(1.602 xx 10^(-19))eV` = 32.8eV As the magnitude of charge of a proton is the same as that of an electron, the potential required is equal in magnitude to the number of eV, i.e., 32.8 volts.

Discussion

48.	Calculate the accelerating potential that must be imparted to a photon to give it an effective wavelength of 0.005 nm (mass of proton =1.67xx10^(-27)kg)
Answer» ANSWER :`3.2255xx10^(-19)J`

Discussion

49.	Calculate sum of number of products formed in the reaction a,b and c .
Answer» SOLUTION :

Discussion

50.	Calculate sum of bond order between same bonded atoms in Q and R compounds.
Answer» SOLUTION : Bond order `[O-O]` in `H_(2)O_(2)=1.0` Bond order of `[O-O]` in `H_(2)O=2.0` Sum of bond order between same BONDED atoms in Q and R COMPOUNDS`=1+2.0=3.0`

Discussion

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Calculate the amount of calcium oxide required when it reacts with 852 g ofP_4O_10

Calculate the amount of CaCl_(2) (van't Hoff factor I =2*47) dissolved in 2*5 L solution so that its osmotic pressure at 300 K is 0*75 atmosphere. Given : Molar mass of CaCl_(2) is 111 g."mol"^(-1), R=0*082L." atm "K^(-1)"mol^(-1).

Calculate the amount of benzoic acid (C_6H_3COOH)required for preparing 250 mL of 0.15 M solution in methanol

Calculate the amount of benzoic acid (C_(6)H_(5)COOH) required for preparing 250 mL of 0.15 M solution in methanol.

Calculate the amount of AgNO_3 which should be added to 60 ml of solution to perpare a concentration of 0.03 g ml^(-1)

Calculate the actual mass of one molecule of carbon dioxide (CO_(2))

Calculate the Activation Energy (E_(a)) of a reaction that follows the equation ,k=(4.5xx10^(11)s^(-1)) e.

Calculate the accelerating potential that must be imparted to a proton beam to give it an effective wavelength of 0.005nm.

Calculate the accelerating potential that must be imparted to a photon to give it an effective wavelength of 0.005 nm (mass of proton =1.67xx10^(-27)kg)

Calculate sum of number of products formed in the reaction a,b and c .

Calculate sum of bond order between same bonded atoms in Q and R compounds.

Calculate the amount of CaCl_(2) (van't Hoff factor I =247) dissolved in 25 L solution so that its osmotic pressure at 300 K is 075 atmosphere. Given : Molar mass of CaCl_(2) is 111 g."mol"^(-1), R=0082L." atm "K^(-1)"mol^(-1).