94420 + Interview Questions in Chemistry in Class 12

1.	Calculate the molarity of a solution containing 14 g of KOH in 750 ml of solution.
Answer» Solution :MOL wt. of KOH = 56 Molarity `= (14)/(750 )xx(1000 )/(56)` `=0.333`

Discussion

2.	Calculate the molarity of 9.8% (W/W) solutionof H_(2)SO_(4) if the density of solution is 1.02 g mL^(-1) (Molar mas of H_(2)SO_(4)=98 g mol^(-1)).
Answer» SOLUTION :`"MOLAR mass of solute" (H_(2)SO_(4))=98" g mol"^(-1)` Mass of solution = 100 g Mass of solute = 9.8 g `"Density of the solution" = 1.02 " g mL"^(-1)` `"Volume of the solution"=("Mass of solution")/("Density of solution")` `=((100g))/((1/02"g mL"^(-1))=98.04 mL = 0.098 L.` `"Molarity of solution (M)"=("Mass os solute/Molar mass of solute")/("Volume of solution in litres")` `((9.8g)//(98" g mol"^(-1)))/((200//1000)L)=0.5 mol L^(-1)=1.02 M.`

Discussion

3.	Calculate the molarity and normality of a solution containing 5 g of NaOH in 450 mL Solution.
Answer» SOLUTION :0.278 each

Discussion

4.	Calculate the molarity and molarity of a solution of ethanol in water if the molre fraction of ethanol is 0.05 and the density of solution is 0.997 g/"cc"
Answer» Solution :SINCE mole fraction of `C_(2)H_(5)OH=("MOLES of " C_(2)H_(5)OH)/("moles of" C_(2)H_(5)OH+"moles of " H_(2)O)` `=0.05=(5)/(100)` `:.100` moles of solution contain 5 moles of `C_(2)H_(5)OH` Weight of `C_(2)H_(5)OH=` moles `xx`mol.wt. `=5xx46=230g` Weight OG `H_(2)O=95xx18=1710g` `:.` weight of solution `=230+1710=1940g` Volume of solution `=(1940)/(0.997)"cc"=1945.8mL=1.9458` litres `:.` molality `=(5)/(1710)xx1000=2.92m` Molarity `=(5)/(1.9458)=2.57M`

Discussion

5.	Calculate the molarity and molality of20 per cent aqueous ethanol (C_(2)H_(5)OH) solution by volume (density of the solution = "0.960 g per cm"^(3)).Assume the solution to be ideal.
Answer» Solution :ASSUMING the solution to be IDEAL, `"100 cm"^(3)" of alcohol "+80 cm^(3)" of WATER, 100 cm"^(3)" of solution "=100xx0.960 =96g` `therefore"96 g of solution = 20 cm"^(3)" of alcohol "+"80 g of water"because"20 cm"^(3)" of alcohol "=96-80 = 16 g.`

Discussion

6.	Calculate the molar volume of gas at STP -
Answer» 22.8 ml 25.8 ml 22.4 ml 11.2 ml Solution :At S.T.P P, P=1 , R = 0.0821 L ATM `K^(-1) MOL^(-1)` , n=1 mol, T= 273 K, PV = NRT

Discussion

7.	Calculate the molar solubility of Fe(OH)_(3) in a buffer solution that is 0.1M in NH_(4)OH and 0.1M in NH_(4)Cl (K_(b) of NH_(4)OH=1.8xx10^(-5),K_(sp) of Fe(OH)_(3)=2.6xx10^(-39))
Answer» `4.46xx10^(-22)M` `8.92xx10^(-25)M` `2.23xx10^(-24)M` `4.46xx10^(-25)M` SOLUTION :`N//A`

Discussion

8.	Calculate the molar mass of water if it contains 50% heavy water (D_(2)O).
Answer» Solution :As water contains `50%D_(2)O`, this means that it contains `(1)/(2)` MOLE of `H_(2)O` and `(1)/(2)` mole of `D_(2)O`. Mass of `(1)/(2)" mole of "H_(2)O=(1)/(2)xx18="9 g. Mass of "(1)/(2)" mole of "(1)/(2)" mole of "D_(2)O=(1)/(2)(2xx2+16)="10 g. Hecne, molar mass of the GIVNE sample of water"=9+10="19 g mol"^(-1)`.

Discussion

9.	Calculate the molar mass of a substance 1.3 g of which when dissolved in 169 g of water gave a solution boiling at 100.025^(@)C at a pressure of one atmosphere (K_(b) for water = "0.52 K m"^(-1))
Answer» SOLUTION :`"160 G MOL"^(-1)`

Discussion

10.	Calculate the molar ionic conductance of Al^(3+) ions at inifinite dilution, given that the molar conductance of Al_(2)(SO_(4))_(3) and molar ionic conductance of SO_(4)^(2-) ions at infinite dilution are 858" S "cm^(2)mol^(1) and 160" S "cm^(2)mol^(-1) respectively.
Answer» Solution :`wedge_(m)^(@)[Al_(2)(SO_(4))_(3)]=2lamda_(m)^(@)(Al^(3+))+3lamda_(m)^(@)(SO_(4)^(2-))`

Discussion

11.	Calculate the molar conductance of 0.025 M aqueous solution of calcium chloride at 25^@C . The specific conductance of calcium chloride is 12.04 xx 10^(-2) Sm^(-1).
Answer» Solution :Molar CONDUCTANCE = `Lambda_m = ((SM^(-1)) xx 10^(-3))/(M) mol^(-1) m^3` `= ((12.04 xx 10^(-2) Sm^(-1)) xx 10^(-3))/(0.025) mol^(-1) m^(3) = 481.10^(-2) Sm^(2) mol^(1)` .

Discussion

12.	Calculate the molar conductance of 0.01M aqueous KCl solution at 25^@C. The specific conductance of KCl at 25^@C is 14.114 xx 10^(-2) Sm^(-1).
Answer» SOLUTION :CONCENTRATION of KCl solution = 0.01M SPECIFIC conductance `(k) = 14.114 XX 10^(-2)Sm^(-1)` MOLAR conductance `(Lambda_m) = ?` `Lambda_, = (k xx 10^(-3))/(M) = (14.114 xx 10^(-2) xx 10^(-3))/(0.01) Sm^(-1) mol^(-1) m^(3)` `Lambda_m = 14.114 xx 10^(-5) xx 10^2 = 14.114 xx 10^(-3) Sm^2 mol^(-1)`.

Discussion

13.	Calculate the molar conductance of 0.01M aqueous KCl solution at 25^(@)C. The specific conductance of KCl at 25^(@)C is 14.114 times 10^(-2)Sm^(-1).
Answer» Solution :MOLAR conductance `wedge_(m)=(KAPPA times 10^(-3))/Mmol^(-1)m^(3)` Specific conductance `kappa=14.114 times 10^(-2)Sm^(-1)` Molar conductance `wedge_(m)=(14.114 times 10^(-2) times 10^(-3))/(0.01)` `""=14.114 times 10^(-3)Sm^(2)MOL^(-1)`.

Discussion

14.	Calculate the molar conductance at inifinite dilution for acetic acid, given. wedge_(m)^(oo)(HCl)=425Omega^(-1)cm^(2)mol^(-1),wedge_(m)^(oo)(NaCl)=188Omega^(-1)cm^(2)mol^(-1),wedge_(m)^(oo)(CH_(3)COONa)=96Omega^(-1)cm^(2)mol^(-1).
Answer» ANSWER :333` Omega^(-1)cm^(2)MOL^(-1)`

Discussion

15.	Calculate the molality of sulphuric acid solution acid solution with mole fraction of water is 0.85.
Answer» Solution :`"Mole fraction of WATER" =0.85, "Mole fraction of" H_(2)SO_(4)=1-0.85=0.15` `"Let"n_(B)"MOLES of " H_(2)SO_(4) "be dissolved in 1000 g of water to represent the molality of the solution"`. `THEREFORE"No. of moles of water"(n_(A))=(1000g)/((18"g mol"^(-1)))=55.55 mol` `"No. of moles of "H_(2)SO_(4)(n_(B))` `n_(B)/(n_(B)+n_(B))=0.15orn_(B)/(n_(B)+55.55)=0.15` `n_(B)=0.15""n_(B)+55.5xx0.15orn_(B)=(55.5xx0.15)/(0.85)=9.8`.

Discussion

16.	Calculate the molality of potassium carbonete solution formed by dissolving 2.5 g of it in one litre of solution (density of solution=0.85 g mL^(-1)).
Answer» Solution :`"MASS of 1000 ML (one litre ) of solution"=Vxxd=(1000mL)xx(0.85"g mL"^(-1))=850g` `"Mass of solvent (water)"=850g-2.5g)=847.5 g=0.8475 KG` `"Molar mass of"K_(2)CO_(3)=2xx39+12+3xx16=138" g mol"^(-1)` `"Molality of solution (m)"("Mass of" K_(2)CO_(3)//"Mola mass")/("Volume of solution in litres")` `"Molality of solution (m)"("Mass of" K_(2)CO_(3)//"Mola mass")/("Mass of solvent in kg")=((2.6g)//138" g mol"^(-1))/((0.8475kg))=0.021 m`.

Discussion

17.	Calculate the molalityof NaCl solutionwhoseelevation in boiling pointis equal to thedepressionin freezing point of 0.25 m sodiumcarbonate solutionin waterassumingcompletedissociationof salts. (k_(f) = 1.86 K m^(-1) , k_(b) = 0.52 K m^(-1))
Answer» SOLUTION :`NaClto NA^(+) + Cl^(-) (i=2)` `Na_(2)CO_(3)^(2-) (i=3)` `Delta T_(b) (NACL) = Delta T_(b) (Na_(2)CO_(3))` ` ixx K_(b) xx m = ixx k_(F) xx m` ` 2 xx 0.52 xx m= 3 xx 1.86 xx 0.25` ` m = 1.34` m

Discussion

18.	Calculate the molality of H_(2)SO_(4) if the density of 10% (w/w) aqueous solution of H_(2)SO_(4) is 1.84 "g cm"^(-3) ("Molar mass of "H_(2)SO_(4)="98 g mol"^(-1)).
Answer» Solution :10%` (w//w) H_(2)SO_(4)` solution MEANS 10 G `H_(2)SO_(4)` are PRESENT in 100 g of solution, i.e., water present = 90 g `"Molality "=(10//98)/(90)xx1000=1.13m`

Discussion

19.	Calculate the molality of a solution that contains 51.2 g of naphthanlene (C_(10)H_(18)) in 500 mL of carbon tetrechloride. Density of C CL_(4) is 1.60 g/ml
Answer» 0.250 m 0.500 m 0.840 m 1.69 m Solution :`"Mass of" C CI_(4)(W_(A))="Volume "xx"density "` `=(500ML)xx(1.60 G mL^(-1))=800G =0.8 kg` `"Molality of "C CI_(4)(m)=("No.ofmoles of napthalene")/("Mass of "C CI_(4)"in kg")` ` =((51.2g)//(128 g mol^(-1)))/((0.8 kg))` 0.5 `mol kg^(-1)=0.5 m`

Discussion

20.	Calculate the molality of a 1-litre solution of 93% H_(2)SO_(4) (wt./vol). The density of the solution is 1.84 g.mL
Answer» Solution :The solution is `1000mL` containing `930g` of `H_(2)SO_(4)` The weight of the solution will be `1840g`. The weight of the SOLVENT `(H_(2)O)` will therefore be `(1840-930)` i.e.910g `:.` molality `=("mole of" H_(2)SO_(4))/("wt.of"H_(2)O(G))xx1000` `=(930//98)/(910)xx1000=10.428m`

Discussion

21.	Calculate the molality of 1 litre solution of 93% H_(2)SO_(4) (weight / volume). The density of the solution is "1.84 g mL"^(-1).
Answer» Solution :`93% H_(2)SO_(4)(w//v)=93 g H_(2)SO_(4)" in 100 cm"^(3)" of the solution = 93 g in 184 g of the solution"` `therefore"Solvent (water) "=184-93 = 91 g = 0.091 kg,"Molality "=("93/98 mol")/("0.091 kg")=10.43 m.`

Discussion

22.	Calculate the molality of 1 litre solution of 93% H_(2)SO_(4) (weight/volume). The density of the solution is "1.84 g mL"^(-1)
Answer» SOLUTION :`93%H_(2)SO_(4)(w//v)=93gH_(2)SO_(4)` in `100CM^(3)` of the solution = 93 g in 184 g of the solution `therefore" SOLVENT (WATER )"=184-93=91g=0.091kg" ,Molality"=(93//98mol)/(0.091kg)=10.43m.`

Discussion

23.	Calculate the molality and mole fraction of 2.5 g of ethanoic acid (CH_(3)COOH) in 75 g of benzene.
Answer» Solution :Mass of solute `(CH_(3)COOH)="2.5 G,Mass of solvent "(C_(6)H_(6))="75 g = 0.075 KG"` `"Molar massof "CH_(3)COOH="60 g mol"^(-1),"Molar mass of "C_(6)H_(6)="78 g mol"^(-1)` Calculation of molality : Moles of the solute `(CH_(3)COOH)=("2.5 g")/("60 g mol"^(-1))=0.0417` `"Molality"=("Moles of the solute")/("Mass of the solvent in kg")=("0.0417 mol")/("0.075 kg"="0.556 mol kg"^(-1)` Calculation of MOLE fraction : Moles of solute `(n_(CH_(3)COOH))="0.0417 calculated above"` `"Moles of solvent "(n_(C_(6)H_(6)))=("75 g")/("78 g mol"^(-1))=0.961` Mole fraction of `CH_(3)COOH` in the solution `=(n_(CH_(3)COOH))/(n_(CH_(3)COOH)+n_(C_(6)H_(6)))=0.0416`

Discussion

24.	Calculate the molality and mole fraction of the solute in aqueous solution containing 3.0 g of urea (molar mass = 60 "g mol"^(-1)) per 250 g of water.
Answer» SOLUTION :`"Molality = 0.2 mol KG"^(-1), "Mole FRACTION "=0.00359`

Discussion

25.	Calculate the molality and molarity of a solution made by mixing equal volumes of 30% by weight of H_(2)SO_(4) (density =1.218g//mL) and 70% by weight of H_(2)SO_(4) (density =1.610g//mL)
Answer» Solution :Suppose that the solution contains 100mL of each variety of `H_(2)SO_(4)`. Total volume is THEREFORE `200ML` or `0.02` litre WT.of `100mL` of `H_(2)SO_(4)` solution `(30%)=1.218xx100` `=121.8g` and wt.of `100mL` of `H_(2)SO_(4)` solution `(70%)=1.610xx100` `=161g` Wt. of `H_(2)SO_(4)(30%)=121.8xx(30)/(100)=36.54g` Wt. of `H_(2)SO_(4)(70%)=161xx(70)/(100)=112.7g` Total wt of `H_(2)SO_(4)` (solute) =`36.54+112.7=149.24g` `:.` wt of `H_(2)O` (solvent) = wt.of solution `-` wt.of solute `=(121.8+161)-149.24` `=133.56g` Moles of `H_(2)SO_(4)=(149.24)/(8)=1.5228` (mol wt of `H_(2)SO_(4)=98`) MOLALITY `=(1.5228)//(133.56)xx1000=11.4m` Molarity `=(1.5228)/(0.2)=7.6M`

Discussion

26.	Calculate the molality and molarity of a solution made by mixing equal volumes of 30% by weight of H_2SO_4 (density = 1.20 g/mL) and 70% by weight of H_2SO_4 (density = 1.60 g/mL).
Answer» ANSWER :11.44 m, 7.55 M

Discussion

27.	Calculate the molality and molarity of a solution made by mixing equal volumes of 30% by weight of H_(2)SO_(4) (density=1.20g//ml) and 70% by weight of H_(2)SO_(4) (density=1.60g//mL)
Answer» Solution :Suppose that the solution contains `100 ml` of each variety of `H_(2)SO_(4)`. TOTAL solution is, `200 ml` or `0.2` LITRE `WT.` of `100 ml` of `H_(2)SO_(4)` solution `(30%)=1060xx100=160` gram `wt.` of `H_(2)SO_(4)(30%)=120xx(30)/(100)=36` gram `wt.` of `H_(2)SO_(4)(70%)=160xx(70)/(100)=112` gram Total `wt.` of `H_(2)SO_(4)` (SALUTE) `=36+112=148` gram `therefore wt.` of `H_(2)O`(solvent) `=wt.` of solution`-wt.` of solute `=(120+160)-148` `=280-148=132` gram Moles of `H_(2)SO_(4)=(148)/(98)=1.51` (mol `wt.` of `H_(2)SO_(4)=98`) Molality `=(1.51)/(132)xx1000=11.44m` Molarity `=(1.51)/(0.2)=7.55 M`

Discussion

28.	Calculate the molal lowering of vapour pressure for H_(2)O at 100^(@)C.
Answer» <P> Solution :Molal lowering of vapour pressure is the lowering of vapour pressure of water when 1 mole of the SOLUTE is dissolved in `1000g` of the SOLVENT `(H_(2)O)`. Further, vapour pressure of pure water `(p^(0))` at `100^(@)C` will be `760MM` as `100^(@)C` is its boiling point. Now, we have, lowering of v.p. `=p^(0)-p=p^(0)((n)/(n+N))` `=p^(0)((n)/(N))` `=760xx(1)/(1000//18)((n=1),(N=(1000)/(18)))` `=13.68`mm

Discussion

29.	Calculate the molal depression constant of water. Latent heat of fusion of ice at 0^(@) at 80 calories per gram
Answer» ANSWER :`1.84^(@)`

Discussion

30.	Calculate the molal elevation constant of water, it being given that its latent heat of vaporisation is 2.257 kJ//g.
Answer» SOLUTION :`K_(b)=(RT_(0)^(2))/(1000l_(v))=(8.314JK^(-1)"mol"^(-1)xx(373K)^(2))/(1000gkg^(-1)xx2257Jg^(-1))="0.512 K KG mol"^(-1)`

Discussion

31.	Calculate the molal depression constant of a solvent which has freezing point 16.6^(@)C and latent heat of fusion 180.75 Jg^(-1)
Answer» `2.68` `3.86` `4.68` `2.86` SOLUTION :`K_(f)=(RT_(f)^(2))/(1000xx L_(f)), R=8.314 JK^(-1)MOL^(-1)` `T_(f)=273+16.6=289.6K , L_(f)=180.75 JG^(-1)` `K_(f)=(8.314xx289.6xx289.6)/(1000xx180.75)=3.86`

Discussion

32.	Calculate the molal elevation constant for chloroform from the fact its boiling point is 61.2^(@)C and 0.1 molal solution of an organic substance in chloroform boiled at 61.579^(@)C.
Answer» SOLUTION :`"3.79 K KG MOL"^(-1)`

Discussion

33.	Calculate the molal elevation constant of water, it being given that 0.1 molal aqueous solution of a substance boiled at 100.052^(@)C.
Answer» Solution :Here, we are GIVEN that m = 0.1 Boiling point of solution `= 100.052^(@)C "" therefore "" Delta T_(B)=100.52-100=0.052^(@)C` APPLYING the relationship, `Delta T_(b)=K_(b)`. m, we GET `K_(b)=(Delta T_(b))/(m)=(0.052^(@)C)/(0.1 m)=0.52^(@)C//m`

Discussion

34.	Calculate theminimum weight of NaOH required to be added in R.H.S. to consume all the H^(+) present in R.H.S of cell of e.m.f. +0.701 V at 25^(@)C before its use. Also report the e.m.f. of cell after addition of NaOH. {:Zn\|{:(Zn^(2+)),(0.1 M):}\|\|{:(HCl),(1 "litre"):}\|{:(Pt_(H_(2(g)))),(1 "atm"):}, E_(Zn//Zn^(2+))^(@) = +0.760 V
Answer» ANSWER :`1.264g, 0.3765 V ;`

Discussion

35.	Calculate the minimum uncertainty in velocity of a particle of mass 1.1 xx 10^(-27) kg if uncertainty in its position is 3 xx 10^(-10)cm. (h= 6.62 xx 10^(-34)kg.m^(2) s^(-1))
Answer» Solution :We have, `Deltax.Deltap= (h)/(4pi)` `DELTA x. (m Delta V)= (h)/(4pi)` or `Delta v= (h)/(4pi) .(1)/(m Delta x)` `=(6.62 xx 10^(-34) (kg.m^(2)s^(-1)))/(4 xx 3.14 xx (1.1 xx 10^(-27)kg) xx (3 xx 10^(-12) m))` `=1.6 xx 10^(4) ms^(-1)`

Discussion

36.	Calculate the minimum potential (eV) which must be applied to a fr ee electron so that it has enough energy to excite, upon impact, the electron in a hydrogen atom from its ground state to a state of n = 5.
Answer» SOLUTION :For hydrogen ATOM, `DeltaE = E_(5)-E_(1)` `=(13.6)/(5)^(2) -(-13.6)/(1)^(2)` `=(-13.6)/25 + 13.6` `=-0.544 + 13.6` `=13.05` EV

Discussion

37.	Calculate the minimum number of kWh of electricity required to produce 1.0 kg of Mg from electrolysis of molten MgCl_(2) if the applied emf is 5.0 V. (1kWh = 3.6 xx 10^(6)J)
Answer» ANSWER :11.0 KWH

Discussion

38.	Calculate the minimum number of kilowatt-hours of electricity required to produce 100 kg of Al by electrolysis of Al^(3+) if the required emf is 4.50 V.
Answer» SOLUTION :The no. of FARADAY required for electrolysis = number of eq. of AL deposited `= (10^(6))/(27//3)F = (10^(6))/(9)F`. Charge `= (10^(6))/(9) xx 96500` coulombs. `= 1.07 xx 10^(10)C`. `therefore` electric energy `= 1.07 xx 10^(10) xx 4.50 J = 4.815 xx 10^(10)J ""(because J = C xx V)` kilowatt-hours `= (4.815 xx 10^(10))/(3.6 xx 10^(6))kWh ""(because 1 kWh = 3.6 xx 10^(6)J)` `= 1.34 xx 10^(4) kWh`.

Discussion

39.	Calculate the minimum mass of AB_(2) (s) which must be added to 100 mL water (in mg) to form a saturated solution. K_(sp) (AB_(2)) = 3.2 xx 10^(-11) M_(w.t) [AB_(2)(s)] = 100 g//"mole"
Answer» ANSWER :`0002`

Discussion

40.	Calculate the Miller indices of crystal planes which cut through the crystal axes at (a, b, c), (2a, b, c) and (2s, - 3b, - 3c).
Answer» ANSWER :[(1,1,1),(1,2,2), (3,2,2)]`

Discussion

41.	Calculatethe maximum workwhen24 g of oxygenare expandedisothermallyand reversiblyfrom apressureof1.6 xx 10^(5)Pa to 100Kpaat 298 K.
Answer» Solution :Given :mass ofoxygen= W=24 g Initialpressure`= P_(1)= 1.6 xx 10^(5)pa` Finalpressure`= P_(2)=100kP_(a) = 100 xx 10^(3)Pa= 1 xx 10^(5) Pa` Temperature=T =298 K MOLARMASS ofoxygen`(O_(2)) = M_(O_(2)) = 32g mol^(-1)` `W_(max)= ?` Numberof molesof `O_(2)= n_(O_(2)) = (W)/(M_(O_(2)))= (24)/(32) = 0.75mol` `W_(max) =- 2.303nRT log_(10) .(P_(1))/(P_(2))` `=- 2.303xx 0.75 xx 8.314xx 298log_(10).(1.6 xx 10^(5))/(1xx 10^(5))` `=- 2.303xx 0.75xx 8. 314xx 298xx log_(10) 1.6` `=- 2.303 xx 0.75xx 8.314xx 298xx 0. 2041` `=- 873.5 J`

Discussion

42.	Calculate the maximum work done when pressure on 10 g of hydrogen is reduced from 20 to 1 atm at a constant temperature of 273 K. The gas behaves ideally. Will there be any change in internal energy ? Also, calculate 'q'.
Answer» Solution :We have, `W=-2.303"nRT"log.(p_(1))/(p_(2))` …(Eqn. 4a) n=number of moles of hydrogen`=("wt.in grams")/("mol.wt.")=(10)/(2)=5 "moles" ` Thus `W=-2.303xx5xx2xx273xxlog.(20)/(1)` `=-8180 "CALORIES"` Further, the CHANGE in state of the SYSTEM is from a GAS to a gas and THEREFORE, at constant temperature, internal energy will not change, i.e., `DeltaU=0` Again, `q=DeltaU-W` `0-(-8180)=8180 "calories"`.

Discussion

43.	Calculate the maximum work and log K_(c)for the given reaction at 298 K : Ni(s)+2Ag^(+)(aq) iff Ni^(2+)(aq)+2Ag (s) ["Given :"E^(@)_(Ni^(2+)//Ni)=-0.25V,E^(@)_(Ag^(+)//Ag)=+0.80V,1F=96500C mol^(-1)]
Answer» Solution :`Ni(s)+2Ag^(+)(aq)iff Ni^(2+)(aq)+2Ag(s)` `E^(@)_("cell")=0.80V-(-0.25V)=1.05V` `E^(@)_("cell")=(0.059)/(2)LOG K_(C)` or `""1.05V=(0.059)/(2)log K_(c)` or `""log K_(c)=(1.05xx2)/(0.059)=35.59` `Delta_(R)G^(@)=-NF E^(@)_("cell")` `=-2xx96500C mol^(-1)xx1.05V` `=-202650" J mol"^(-1)` `=-202.65" kJ mol"^(-1)`

Discussion

44.	Calculate the maximum possible electrical work that can be obtined from the cell under the standard conditions at 298 K Zn\|Zn^(2+)(aq)\|\|Ni^(2+)(aq)\|Ni(s) Given E_(Zn^(2+))^(@)(aq)\|Zn(s)=-0.76" V ", E_(Ni^(2+))^(@)(aq)\|Ni(s)=-0.25" V "
Answer» SOLUTION :`E_(cell)^(@)=E_((CATHODE))^(@)-E_((anode))^(@)=[(-0.25)-(-0.76)]=0.51" V"` n=2,F=96500" C"` DeltaG^(@)=-nFE_(cell)^(@)=(-2)xx(96500C)xx(0.51" V")=-98430" CV"=-98430" J"=-98.430" kJ"`.

Discussion

45.	Calculate the maximum possible electric work that can be obtained from the following cell under the standard conditions at 25^(@)C : Given Zn\|Zn^(2+)(aq)\|\|Sn^(2+)(aq)\|Sn(s) At 25^(@)C,E_(Zn^(2+)(aq)\|Zn(s))^(@)=-0.76V E_(Sn^(+)(aq)\|Sn(s)=-0.14V.)^(@) F=96500C" mol"^(-1)
Answer» Solution :`E_("CELL")^(@)=E_("cathode")^(@)-E_("anode")^(@)` `E_("cell")^(@)=-0.14-(-0.76)` `=-0.14+0.76` `=+0.62` `DELTAG^(@)=-nFE_("cell")^(@)` `=-2xx96500xx0.62` `=-11966CV(J)` `DeltaG^(@)=-11.966kJ`.

Discussion

46.	Calculate the max. work done by system in an irreversible (single step) adiabatic expansion of 1 mole of a polyatmic gas (gamma=1.33) from 300 K and pressure 10 atm to 1 atm.
Answer» SOLUTION :`-1.683kJ`

Discussion

47.	Calculate the massof oxygen obtained by complete decomposition of 10kg of pure potassium chlorate (Atomic mass K=39,O=16 and Cl=35.5).
Answer» 39.2kg 392kg 3.92kg 3kg Solution :`2KCIO_(3)(s)rarr2KCI(s)+3O_(2)(g)` 2 mol 3 mol `2xx122.5g 3xx32g` =245g =96g Mass of `O_(2)` obtained from 10kg `KCIO_(3)=(96)/(245)xx10=3.92` kg

Discussion

48.	Calculate the mass present of calcium, phosphorus and oxygen in calcium phosphate Ca_(3)(PO_(4))_(2).
Answer» Solution :Molar MASS of `Ca_(3)(PO_(4))_(2)=3xx40+(31+64)xx2=120+190=310" g MOL"^(-1)` Ca PRESENT = 120 g. P present `=2xx31=62g, "O present "=2xx64=128g` `%" of Ca"=(120)/(310)xx100=38.71%` `%" of P"=(62)/(310)xx100=20.0%` `%" of (O)=(128)/(310)xx100=41.29%.`

Discussion

49.	Calculate the mass percentage of benzene (C_6H_6)and carbon tetrachloride (CCl_4) if 22g of benzene is dissolved in 122g of CCl_4
Answer» SOLUTION :MASS PERCENTAGE of `C_6H_6`=(Mass of `C_6H_6`)/(Mass of solution)`XX100 =(22xx100)/(22+122)=15.28%` Mass percentage of `CCl_4= (122xx100)/(22+122)=84.72%`

Discussion

50.	Calculate the mass percentage of benzene (C_(6)H_(6)) and carbon tetrachloride (C Cl_(4)) if 22 of benzene is dissolved in 122 g of carbon tetrachloride.
Answer» SOLUTION :Mass PERCENTAGE of `C_(5)H_(6)` `= ("Mass of " C_(6)H_(6))/("Total mass of the solution")XX100%` `= ("Mass of "C_(6)H_(6))/("Mass of "C_(6)H_(6)+"Mass of " C Cl_(4))xx100%` `=(22)/(22+122)xx100%=15.28%` Mass percentage of `C Cl_(4)` `=("Mass of "C Cl_(4))/("Total mass of the solution")xx100 %` `= ("Mass of "C Cl_(4))/("Mass of "C_(6)H_(6)+" Mass of "C Cl_(4))xx 100%` `=(122)/(22+122)xx100%=84.72 %` Alternatively, Mass percentage of `C Cl_(4)=(100-15.28)%` = 84.72 %.

Discussion

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