94420 + Interview Questions in Chemistry in Class 12

1.	If the concentration of lead iodide in its saturated solution at 25^(@)C be 2 xx 10^(-3) moles per litre, then the solubility product is
Answer» `4 xx 10^(-6)` `8 xx 10^(-12)` `6 xx 10^(-9)` `32 xx 10^(-9)` SOLUTION :`{:(PbI_(2) ,HARR,(Pb^(++)), + ,(2I^(-))),(,,S,,2S):}` `K_(sp) = 4S^(3) = 4 xx [2 xx 10^(-3)]^(3) = 32 xx 10^(-9)`.

Discussion

2.	In 1945 the Allied forces included the US______USSR and ______.
Answer» SOLUTION :1. UK, FRANCE.

Discussion

3.	If the concentration of glucose (C_(6)H_(12)O_(6)) in blood is 0.9 g L^(-1), what will be the molarity of glucose in blood?
Answer» 5 M 50 M 0.005 M 0.5 M Solution :Molar MASS of GLUCOSE `(C_(6)H_(12)O_(6))=72+12+96=180" g mol"^(-1)` `"Molaarity"=("conc. in g L"^(-1))/("Molar mass")=(0.90gL^(-1))/("180 g mol"^(-1))=0.005M.`

Discussion

4.	In 1894, Lord Rayleigh reported that the mass of chemically prepared nitrogen was different from that of nitrogen extracted from the atmosphere, as shown in Tables 1 and 2. Latter, this difference was attributed to the presence of argon in atmospheric nitrogen. The masses of gases were mearsured by using a glass vessel with a known volume under atmospheric pressure (1.013xx10^(5)Pa). \|{:("From nitric oxide",,2.3001 g),("From nitrous oxide",,2.2990 g),("From amonium nitrite purified at a red heat",,2.2987 g),("From urea",,2.2985 g),("From ammonium nitrite purified in the cold",,2.2987 g),("Mean",,2.2990 g):}\| \|{:(O_(2) "was removed by hot copper" (1892),,2.3103 g),(O_(2) "was removed by hot iron" (1893),,2.3100 g),(O_(2) "was removed by ferrous hydrate (1894),2.3102 g,),(Mean,,2.3102 g):}\| Ramsay and cleve discovered helium in cleveite (a mineral consiting of uranium oxide and oxides of lead, thorium, and rare earths, an impure variety of uraninite) independently and virtually simultaneously in 1895. The gas extracted from the rock showed a unique spectroscopic line at around 588 nm (indicated by D3 in Figure 1), which was fist observed in the spectrum of solar prominence during a total eclipse in 1868, near the well-known D_(1) and D_(2) lines of sodium. ul("Which") equation explains the occurrence of argon in rocks among [A] to [D] below? Mark one. One of the strongest evidences for the monoatomicity of argon and helium is the ratio of the heat capacity under constant pressure to that at constant volume, gamma=C_(p)//C_(v), which is exactly 5//3 (1.67 +- 0.01) for a monoatomic gas. The ratio was derived from the measurement of speed of sound v_(s) by using the following equation, where f and the frequency and wavelength of the sound, and R, T, and M denote the molar gas constant, absolute temperature, and molar mass, respectively. v_(s)=flambda=sqrt((gammaRT)/(M)) For an unknown gas sample, the wavelength of the sound was measured to be =0.116m at a frequency of f=3520 Hz (Hz =s^(-1)) and temperature of 15.0^(@)C and under atmospheric pressure (1.013. 10^(5) Pa). The density of the gas for these conditions was measured to be 0.850 +- 0.005 kg m^(-3).
Answer» `ArF_(2)"" Ar+F_(2)` `ArXe rarr Ar+Xe` `^(40)K rarr `^(40)Ar+ epsilon//beta^(+)`` (electron capture//position emission) `^(126)I rarr `^(126)Ar+beta^(-)`` Answer :`[C]` is a well known radioactive decay reaction OCCURRING with a half life of the order of the EARTH age. No STABLE compound of Ar, such as `ArF_(2)` or `ArXe`, can be expected Products of `[D]` should be `^(126)Xe+beta^(-)`. The CORRECT answer is `[C]`

Discussion

5.	If the concentration of CrO_(4)^(2-) ions in a saturated solution of silver chromate is 2 xx 10^(-4). Solubility product of silver chromate will be
Answer» `4 xx 10^(-8)` `8 xx 10^(-12)` `12 xx 10^(-12)` `32 xx 10^(-12)` SOLUTION :`K_(sp)` of `Ag_(2)CrO_(4) = [AG^(+)]^(2)[CrO_(4)^(--)]` `CrO_(4)^(--) = 2 xx 10^(-4)` then `Ag^(+_ = 2 xx 2 xx 10^(-4)` `K_(sp) = (4 xx 10^(-4)) (2 xx 10^(-4)) = 32 xx 10^(-12)`.

Discussion

6.	In 1908 Rutherford together with H- Geiger measured the rate of emission of alpha particles (x) by radius (in the nature this element is represented by a single nuclide _(88)^(226)Ra) and found that 1.00 g of radius emits x=3.42xx10^(10) alpha- particle per second. In 1911 Ritherford and American physical chemist B. Boltwood measured the rate of formation of helium from radius. This experiment permits to obtain the most accurate value of Avogadro's number avaliable at the time, given that the value of molar volume of ideal gas was well established. To achieve this goal a sample of radium salt purified from decays products and contaning m=192 mg of Re was put into a device an the volume of the evolved helium was measured. After 83 days (t=83.0 "days") of the experiment 6.58 mm^(3) of He was collected (V_(He)=6.58 mm^(3) "corrected to" 0^(@)C "and" 1 atm). To inderstand the results of this experiment we shall need the kinetic scheme of redioactive decay of Re which is given below (half-lives are the arrows, the type of decay is below the arrows). Ra underset(alpha)overset( gt 1500 "years")(rarr)Rn underset(alpha)overset(3.83 "days")(rarr)RaAunderset(alpha)overset(3.05 "min")(rarr)RaBunderset(beta)overset(26.8)(rarr)RaCunderset(beta)overset(17.7 "min")(rarr) rarrRaC'underset(alpha)overset(1.63xx10^(-4) S)(rarr)RaDunderset(beta)overset(27.1 "years")(rarr)RaEunderset(beta)overset(5 "days")(rarr)Pounderset(alpha)overset(138 "days")(rarr)Pb (stable) (RaA-RaE are intermediate products of radon decay). {:a) How many helium atoms were formed each decayed radius atom after 83 days? {:b) How many helium atom were formed in total during the experiment?
Answer» Answer :{:a) The CORRECT answer: `4` {:B) NUMBER of HELIUM atoms (rough `1.9xx10^(17)` ESTIMATE)

Discussion

7.	If the concentration is expressed in moles per liter, the unit of the rate constant for a first-order reaction is
Answer» MOLE `"litre"^(-1) sec^(-1)` MOL `"litre"^(-1)` `sec^(-1)` `"mole"^(-1) "litre"^(-1) sec^(-1)` Solution :Unit of K for first order reaction = `sec^(-1)` .

Discussion

8.	In 1908 Rutherford together with H- Geiger measured the rate of emission of alpha particles (x) by radius (in the nature this element is represented by a single nuclide ._(88)^(226)Ra) and found that 1.00 g of radius emits x=3.42xx10^(10) alpha- particle per second. In 1911 Ritherford and American physical chemist B. Boltwood measured the rate of formation of helium from radius. This experiment permits to obtain the most accurate value of Avogadro's number avaliable at the time, given that the value of molar volume of ideal gas was well established. To achieve this goal a sample of radium salt purified from decays products and contaning m=192 mg of Re was put into a device an the volume of the evolved helium was measured. After 83 days (t=83.0 "days") of the experiment 6.58 mm^(3) of He was collected (V_(He)=6.58 mm^(3) "corrected to" 0^(@)C "and" 1 atm). To inderstand the results of this experiment we shall need the kinetic scheme of redioactive decay of Re which is given below (half-lives are the arrows, the type of decay is below the arrows). Ra underset(alpha)overset( gt 1500 "years")(rarr)Rn underset(alpha)overset(3.83 "days")(rarr)RaAunderset(alpha)overset(3.05 "min")(rarr)RaBunderset(beta)overset(26.8)(rarr)RaCunderset(beta)overset(17.7 "min")(rarr) rarrRaC'underset(alpha)overset(1.63xx10^(-4) S)(rarr)RaDunderset(beta)overset(27.1 "years")(rarr)RaEunderset(beta)overset(5 "days")(rarr)Pounderset(alpha)overset(138 "days")(rarr)Pb (stable) (RaA-RaE are intermediate products of radon decay). Write the first six radioactive dacays using a modern natation showing atomic and mass numbers of all nuclei involved. As a rough first approximation half-lives of all radius decay products, except those of RaD and Po, may be assumed to be negligible compared to the time of measurement t. Using this approximation perform the following calculations.
Answer» Answer :`._(88)^(226)Ra rarr ._(86)^(222)RN rarr ._(2)^(4)He` `._(33)^(222)Rn rarr ._(84)^(218)Po rarr ._(2)^(4)He` `._(88)^(218)Po rarr ._(82)^(214)Pb rarr ._(2)^(4)He` `._(82)^(214)Pb rarr ._(83)^(214)BI rarr e` `._(83)^(214)Bi rarr ._(84)^(214)Po rarr e` `._(84)^(214)Po rarr._(82)^(210)Pb rarr ._(2)^(4)He`

Discussion

9.	IF the concentration of CrO_4^(2-) ion in a saturated solution of silver chromate be 2xx 10^(-4) M solubility of sodium chloride is :
Answer» `4xx10^(-8)` `8xx10^(-12)` `32xx10^(-12)` `6xx10^(-12)` ANSWER :C

Discussion

10.	In 1894, Lord Rayleigh reported that the mass of chemically prepared nitrogen was different from that of nitrogen extracted from the atmosphere, as shown in Tables 1 and 2. Latter, this difference was attributed to the presence of argon in atmospheric nitrogen. The masses of gases were mearsured by using a glass vessel with a known volume under atmospheric pressure (1.013xx10^(5)Pa). \|{:("From nitric oxide",,2.3001 g),("From nitrous oxide",,2.2990 g),("From amonium nitrite purified at a red heat",,2.2987 g),("From urea",,2.2985 g),("From ammonium nitrite purified in the cold",,2.2987 g),("Mean",,2.2990 g):}\| \|{:(O_(2) "was removed by hot copper" (1892),,2.3103 g),(O_(2) "was removed by hot iron" (1893),,2.3100 g),(O_(2) "was removed by ferrous hydrate (1894),2.3102 g,),(Mean,,2.3102 g):}\| Ramsay and cleve discovered helium in cleveite (a mineral consiting of uranium oxide and oxides of lead, thorium, and rare earths, an impure variety of uraninite) independently and virtually simultaneously in 1895. The gas extracted from the rock showed a unique spectroscopic line at around 588 nm (indicated by D3 in Figure 1), which was fist observed in the spectrum of solar prominence during a total eclipse in 1868, near the well-known D_(1) and D_(2) lines of sodium. ul("Identify") the transition relevant to the D_(3) line of helium among the transitions [A] to [E] indicated in Figure 2. Mark one of the following: [A] [B] [C] [D] [E]
Answer» Answer :The CORRECT answer is [E]. The energy `3.380xx10^(-19) J` matches with the energy of the transition between `2p` and `3d` orbitals.

Discussion

11.	In 1894, Lord Rayleigh reported that the mass of chemically prepared nitrogen was different from that of nitrogen extracted from the atmosphere, as shown in Tables 1 and 2. Latter, this difference was attributed to the presence of argon in atmospheric nitrogen. The masses of gases were mearsured by using a glass vessel with a known volume under atmospheric pressure (1.013xx10^(5)Pa). \|{:("From nitric oxide",,2.3001 g),("From nitrous oxide",,2.2990 g),("From amonium nitrite purified at a red heat",,2.2987 g),("From urea",,2.2985 g),("From ammonium nitrite purified in the cold",,2.2987 g),("Mean",,2.2990 g):}\| \|{:(O_(2) "was removed by hot copper" (1892),,2.3103 g),(O_(2) "was removed by hot iron" (1893),,2.3100 g),(O_(2) "was removed by ferrous hydrate (1894),2.3102 g,),(Mean,,2.3102 g):}\| Ramsay and cleve discovered helium in cleveite (a mineral consiting of uranium oxide and oxides of lead, thorium, and rare earths, an impure variety of uraninite) independently and virtually simultaneously in 1895. The gas extracted from the rock showed a unique spectroscopic line at around 588 nm (indicated by D3 in Figure 1), which was fist observed in the spectrum of solar prominence during a total eclipse in 1868, near the well-known D_(1) and D_(2) lines of sodium. ul("Which") equation explains the occurrence of helium in cleveite among [A] to [D] below? Mark one.
Answer» `.^(238)U RARR .^(234)Th+alpha` `Uhe_(2) rarr U+2 He` `.^(240)U rarr .^(240)Np+beta^(-)` `.^(235)U+n rarr .^(95)Y+^(139)l+2 n` Answer :Considering that the a particle is the NUCLEUS of helium a DECAY `[A]` is the relevent source of helium in such ROCKS. No compounds of He such as `UHe2` in `[B]` is known to be stable at ambient temperature. `[C]` is a radioactive decay of `240 U` in the thorium series. `[D]` is a nuclear fission reaction of `235U` occurring in nuclear reactors. Thus, the correct answer is `[A]`

Discussion

12.	If the concentration is expressed in mol dm^(-3) and time in seconds, the unit of the rate constant for a first-order reaction will be
Answer» MOL `DM^(-3)s^(-1)` `s^(-1)` `dm^(3)mol^(-1)s^(-1)` `dm^(-3)mol^(-1)s^(-1)` SOLUTION :For a first order REACTION, `dx/dt=k[A]` `k=("mol "dm^(-3) s^(-1))/("mol "dm^(-3))=s^(-1)`

Discussion

13.	In 1894, Lord Rayleigh reported that the mass of chemically prepared nitrogen was different from that of nitrogen extracted from the atmosphere, as shown in Tables 1 and 2. Latter, this difference was attributed to the presence of argon in atmospheric nitrogen. The masses of gases were mearsured by using a glass vessel with a known volume under atmospheric pressure (1.013xx10^(5)Pa). \|{:("From nitric oxide",,2.3001 g),("From nitrous oxide",,2.2990 g),("From amonium nitrite purified at a red heat",,2.2987 g),("From urea",,2.2985 g),("From ammonium nitrite purified in the cold",,2.2987 g),("Mean",,2.2990 g):}\| \|{:(O_(2) "was removed by hot copper" (1892),,2.3103 g),(O_(2) "was removed by hot iron" (1893),,2.3100 g),(O_(2) "was removed by ferrous hydrate (1894),2.3102 g,),(Mean,,2.3102 g):}\| Ramsay and cleve discovered helium in cleveite (a mineral consiting of uranium oxide and oxides of lead, thorium, and rare earths, an impure variety of uraninite) independently and virtually simultaneously in 1895. The gas extracted from the rock showed a unique spectroscopic line at around 588 nm (indicated by D3 in Figure 1), which was fist observed in the spectrum of solar prominence during a total eclipse in 1868, near the well-known D_(1) and D_(2) lines of sodium. ul(Calculate) the energy E[J] of a photon with the wavelength of the D_(3) line of helium shown in Figure 1 Figure 2. Energy diagram of atomic orbitals of helium when an electron resides in the 1s orbital. Figure 2 shows an energy diagram of the atomic orbitals of helium. The arrows indicate the ''allowed'' transitions according to the spectroscopic principle.
Answer» Answer :According to figure `1`, the wavelength of the `D3` LINE is approximately `587.7 nm` The CORRESPONDING PHOTON energy is `E=(HC)/lambda=(6.62610^(-34)xx2.99810^(8))/(587.7*10^(9))=3.380xx10^(-19) J`

Discussion

14.	If the compound contains C,H and halogen. When C and H are to be estimated the combustion tube at the exit should contain a :
Answer» COPPER spiral Silver spiral Lead spiral Iron Spiral Answer :B

Discussion

15.	In 1811, Avogadro calculated the formula of camphor by means of elemental chemical analysis and by measuring the density of its vapour. Avogadro found the density to be 3.84 g/L when he made the measurement at 210^(@)C at 1 atm pressure. Which of the following is the correct formula of camphor?
Answer» `C_(10)H_(14)O` `C_(10)H_(16)O` `C_(10)H_(16)O_(2)` `C_(10)H_(18)O` SOLUTION :`Pm=DRT` `m=(dRT)/(P)=(3.84xx0.0821xx483)/(1)=152.27` `therefore C_(10)H_(10)O` will be the CORRECT formula.

Discussion

16.	If the complex [Mo(CO)_(4) Br(CPh_(3))] follow 18 electron rule or EAN rule, find the bond order of Mo-C bond .
Answer» <BR> Solution :`[OVERSET(+2)(Mo)(overset(0)CO)_(4)overset(-1)(Br)(CP overset(-1"")(h_(3)))]` `impliesd^(4),2+4+4x=18` `IMPLIES"" x=3`

Discussion

17.	In 16th group element, which element does not show negative oxidation state ?
Answer» O S Po Te Answer :C

Discussion

18.	If the circumference of Bohr orbit is 3.3 xx 10^(-10) m and represents one wavelength, find the velocity of electron revolving in that orbit.
Answer» SOLUTION :`2.19xx10^(6)MS^(-1)`

Discussion

19.	In 1.2 year half of 160 mg of a radioactive isotope decays . The amount present after 6 years is :
Answer» a) 0 mg b) 4 mg c) 5 mg d) 16 mg ANSWER :C

Discussion

20.	In 10^3 Litre sample of hard water CaSO_4 and MgSO_4 is present.If elevation in Boiling point is 0.000052^@C.Calculate the degree of Hardness of hard water .(K_b for H_2O=0.52)
Answer» SOLUTION :`DeltaT_b=DeltaT_(b_1)=DeltaT_(b_2)` `DeltaT_b=0.0000104=(m_1i_1+m_2i_2)xxk_b` `{:(CaSO_4,MgSO_4),(i=i_1,i'=i_2),(n=n_1,n'=n_2):}` `0.000052=((n_1+n_2)xx0.52xx2)/10^6xx1000` `n_1+n_2=0.05` for DEGREE of Hardness `n_(CaSO_4)+n_(MgSO_4)=n_(CaCO_3) " " n_1+n_2=n_(CaCO_3)` `n_(CaCO_3)=0.05` `m_(CaCO_3)=0.05xx100=5` gm degree of hardness=`5/10^6xx10^6=5` ppm

Discussion

21.	If the central atom is of third row or below this in the periodic table, then lone pair will occupy a stereochemically inactive s-orbital and bonding will be through almost pure p-orbitals and bond angles are nearly 90^(@), if the substituent's electronegativity value is le 2.5. The hybridisation of atomic orbitals of central atom "Xe" in XeO_(4), XeO_(2)F_(2) and XeOF_(4) respectively.
Answer» `SP^(3),sp^(3)d^(2),sp^(3)d^(2)` `sp^(3)d,sp^(3)d,sp^(3)d^(2)` `sp^(3),sp^(3)d^(2),sp^(3)d` `sp^(3),sp^(3)d,sp^(3)d^(2)` ANSWER :D

Discussion

22.	In 100 ml sample of hard water, 100 ml of N/50 Na_2 CO_3 was added and the mixture was boiled and filtered. The filtrate was then neutralized with 60 ml of N/50 HCI. The permanent hardness of water in ppm will be:
Answer» 800 ppm 400 ppm 1000 ppm 200ppm Answer :B

Discussion

23.	If the central atom is of third row or below this in the periodic table, then lone pair will occupy a stereochemically inactive s-orbital and bonding will be through almost pure p-orbitals and bond angles are nearly 90^(@), if the substituent's electronegativity value is le 2.5. Select incorrect statement regarding P_(4) molecule.
Answer» Each P atom is ioined with THREE P-atoms `P_(4)` molecule contains total 12 bond ANGLES LONE PAIR of each P atom is present in ALMOST pure s-orbital Lone pair of each P atom present in hybrid orbital Answer :D

Discussion

24.	In 100 ml of 5 M H_(2)SO_(4) (aq) solution 'x' ml of H_(2)O (l) is added. If 20 ml of this diluted H_(2)SO_(4) solution is completely neutralised by 100 ml of 0.4 M NaOH solution. What is the value of (x)/(80)?
Answer» SOLUTION :`chi_("PAULING")=(chi_("MULLIKEN"))/(2.8)=(l.E+E.A)/(2xx2.8)=(13.0+3.8)/(2xx2.8)=3.0`

Discussion

25.	If the central atom is of third row or below this in the periodic table, then lone pair will occupy a stereochemically inactive s-orbital and bonding will be through almost pure p-orbitals and bond angles are nearly 90^(@), if the substituent's electronegativity value is le 2.5. In which of the following option, covalent bond is having maximum s% character ?
Answer» S-H bond in `H_(2)S` P-H bond in `PH_(3)` N-H bond in `NH_(3)` All have equal s% character Answer :C

Discussion

26.	In 100 mL of an aqueous HCl of pH 1.0, 900 mL of distilled water is added, the pH of the resultant becomes:
Answer» 1 2 4 7 Answer :B

Discussion

27.	If the cell reaction is spontaneous then:
Answer» `DELTAG^(o) `is +ve `E^(o) "" _(RED)` is - ve `E^(o) "" _(red)` is +ve `DeltaG^(o) `is - ve Answer :D

Discussion

28.	In 100 g of naphthalene, 2.423 g of S was dissolved. Melting point of naphthalene =80.1^(@)C,DeltaT_(f)=0.661^(@)C,L_(f)=35.7"cal/g" of naphtahalene. Molecular formula of sulphur added is
Answer» `S_(2)` `S_(4)` `S_(6)` `S_(8)` SOLUTION :`M_(2)=(1000K_(f)w_(2))/(w_(1)DeltaT_(f))=(1000w_(2))/(w_(1)DeltaT_(f))xx(RT_(0)^(2))/(1000L_(f))` `=(1000xx2.423)/(100xx0.661)xx(2XX(353.1)^(2))/(1000xx35.7)="256 g mol"^(-1)` If `S_(x)` is the formula, then MOLECULAR mass `x xx 32 = 56 or x = 8`.

Discussion

29.	If the cell reaction is spontaneous
Answer» `E^(@)` is -ve `DELTAG` ispositive `E^(@)` is +ve `(DeltaG+E^(@))` is POSITIVE

Discussion

30.	In 1 litre volume reaction vessel, the equilibrium constant K_(c) of the reaction PCl_(5)hArrPCl_(3)+Cl_(2) is 2xx10^(-4)" lit"^(-1). What will be the degree of dissociation assuming only a small of 1 mole of PCl_(5) has dissociated ?
Answer» ANSWER :`x=1.414xx10^(-2)`

Discussion

31.	In 1 L saturated solution of AgCl[K_(sp) (AgCl) = 1.6 xx 10^(-10)] 0.1 mol of CuCl [K_(sp)(CuCl) = 1.0 xx 10^(-6)] is added. The resultant concentration of Ag^(+) in the solution is 1.6 xx 10^(-x). The value of "x" is
Answer» Solution :`{:(AgCl(s)hArr,"AG"^(+),+"Cl"^(-),K_(sp)(AgCl) = 1.6 xx 10^(-10)),(,Z,Z+Y,),(CuCl(s) hArr,Cu^(+),+Cl^(-),K_(sp)(CuCl) = 10^(-6)),(,Y,Z+Y,):}` `Z(Z+Y) = 1.6 xx 10^(-10)` `Y(Z+Y) = 10^(-6)` `rArr (Z + Y)^(2) = 1.6 xx 10^(-10) + 10^(-6) rArr (Z + Y)^(2) ~~ 10^(-6)` `rArr Z + Y = 10^(-3) rArr Z(Z + Y) = 1.6 xx 10^(-10)` `rArr Z xx 10^(-3) = 1.6 xx 10^(-10) rArr Z = 1.6 xx 10^(-7)` `rArr 1.6 xx 10^(-X) = 1.6 xx 10^(-7) rArr x = 7`.

Discussion

32.	In 1 L saturated solution of AgCl[K_(sp)(AgCl)=1.6xx10^(-19)], 0.1" mol of "CuCl[K_(sp)(CuCl)=1.0xx10^(-6)] is added. The resultant concentration of Ag^(+) in the solution is 1.6xx10^(-x). The value of''x'' is
Answer» ANSWER :7

Discussion

33.	If the cell reactin is spontaneous then :
Answer» `DELTAG^@is + ve` `E_(RED)^@ is -ve` `E_(red)^@ is + ve` `DeltaG is -ve` ANSWER :D

Discussion

34.	In 1 gram of ametal oxide, metal precipitated is 0.68 gm. What is the equivalent weight of metal
Answer» 17 34 68 52 Solution :`1-0.68=0.32g` oxygen `(0.68)/(eq." "WT.)=(0.32)/(8)IMPLIES"eq. wt."=(0.68xx8)/(0.32)=17`

Discussion

35.	If the boundary of system moves by an infinitiesimal amount, the work involved is given by dw= -P_("ext") dV, for irreversible process W= -P_("ext") Delta V (where DeltaV= V_(f)- V_(i)). For reversible process P_("ext") = P_("int") +- dP ~~ P_("int"), so for reversible isothermal process W= -nRTln (V_(f))/(V_(i)). 2 mole of an ideal gas undergoes isothermal compression along three different paths: (i) reversible compression from P_(i)= 2 bar and V_(i)= 8 to P_(f)=20 bar (ii) a single stage compression against a constant external pressure of 20 bar (iii) a two stage compression consisting initially of compression against a constant external pressure of 10 bar until P_("gas") = P_("ext"), followed by compression against a constant pressure of 20 bar until P_("gas") = P_("ext"). Total work done on the gas in two stage compression is
Answer» 40 80 16 None of these Answer :B

Discussion

36.	In 0.020 M carbonic acid solution
Answer» `H_2 CO_3` isstronger acidthan ` HCO_(3)^(-)` `H_2 CO_3 hArr2H^(+) +CO_(3)^(3- ) ,k_(EQ) = K_(a_1) .K_(a_2)` `[HCO_(3)^(- )] ~~[CO_(3)^(2-)]` It canbe SAID`k_(a_i)gt gtK_(w` Solution :As` K_(a_1 )gt gtK_(w)` `HCO_(3)^(-)` isa verypooracid

Discussion

37.	Impurity along SO_(2) is ____and it is removed by
Answer» `As_2O_3 , Fe_(2)O_(3), xH_2O` `H_2S, AL(OH)_(3)` `H_2S, Fe_(2)O_(3), xH_(2)O` `AS_(2)O_(3), Al_(2)O, xH_2O` Answer :A

Discussion

38.	Impurity along SO_(2) is __ and it is removed by
Answer» `As_(2)O_(3),Fe_(2)O_(3),xH_(2)O` `H_(2)S,Al(OH)_(3)` `H_(2)S,Fe_(2)O_(3),xH_(2)O` `AS_(2)O_(3),Al_(2)O_(3),xH_(2)O` ANSWER :A

Discussion

39.	Impurities present in the ore react with a substance to from a fusible product known as....
Answer» FLUX GANGUE NUGGET MINERAL ANSWER :A

Discussion

40.	If the bond length of CO bond in carbon monoxide is 1.128 A, then what is the value of CO bond length in Fe(CO)_(s)
Answer» `1.15Aoverset(@)` `1.128Aoverset(@)` `1.172Aoverset(@)` `1.118Aoverset(@)` ANSWER :D

Discussion

41.	Impurity present in red bauxite is
Answer» ZNO `Fe_(2)O_(3)` `SiO_(2)` `Al_(2)O_(3)` Answer :B

Discussion

42.	If the bond energies of H - H, Br - Br and HBr are 433, 192 and 364 kJ mol^(-1) respectively, the DeltaH^(@) for the reaction, H_(2)(g)+Br_(2)(g)rarr2HBr(g) is
Answer» `+ 261` kJ `- 103` kJ `- 261` kJ `+ 103` kJ Solution :`H-H+Br-Brrarr2H-Br` `{:(433+192,2xx364),(625,728):}` Energy ABSORBED = Energy RELEASED Not energy released = 728-625=103 kJ i.e., = `DeltaH=-103 kJ`.

Discussion

43.	Impurities present in ore are called
Answer» FLUX slag matrix gangue Answer :C::D

Discussion

44.	If the bond energies of H-H, Br-Br and H-Br are 433, 192 and 364 kJ "mol"^(-1) respectively, then DeltaH^@ for the reaction : H_(2(g)) + Br_(2(g)) to 2HBr_((g)) is
Answer» `-261KJ` `+103KJ` `+261KJ` `-103KJ` Solution : `Delta_(R)H=sumB.E._("REACTANTS")-sumB.E._("PRODUCTS")` `=[B.E_(H_(2))+B.E._(Br_(2))]-[2B.E._(HBr)]` `=[433+192-2xx364]KJ=--103KJ`

Discussion

45.	Impurities physically associated with minerals are:
Answer» Slag Flux Alloy Matrix Answer :D

Discussion

46.	If the bond dissociation energy of XY, x_ (2) ANDy_(2) (all diatomic molecules) are in the ratio of 1 : 1 : 0.5 and Delta_(f)H for the formation of XY is - 200 kJ mol^(-1). The bond dissociation energy of X_(2) will be :
Answer» `100 kJ MOL^(-1)` `200 kJ mol^(-1)` `800 kJ mol^(-1)` `400 kJ mol^(-1)` Solution :Let the BOND dissociation energies of XY, `X_(2) and Y_(2)` be x, x and `0.5 x kJ mol^(-1)` respectively. `(1)/(2) X_(2) + (1)/(2) Y_(2) RARR XY Delta H_(f) = - 200 kJ mol^(-1)` `Delta H = (1)/(2) Delta H (X_(2)) + (1)/(2) Delta H (Y_(2)) - Delta H (XY) - 200 = (1)/(2) x + (0.5 x)/(2) - x` or `- 0.25 x = - 200` `x = (200)/(0.25) = 800 kJ mol^(-1)`

Discussion

47.	Impurities in the zinc spelter are
Answer» AG and Au CD and PD Cd and Pb Cd As and Au Answer :C

Discussion

48.	If the bond dissociation energies of XY,X_2 and Y_2(all diatomic molecules)are in the ratio of 1:1:0.5 and Delta_fH for the formation of XY is -200kJ mol^(-1).The bond dissociation energy of X_2 will be
Answer» 100kJ `mol ^(-1)` 800KJ `mol ^(-1)` `300kJ` `mol ^(-1)` `400KJ` `mol ^(-1)` ANSWER :B

Discussion

49.	Impure glycerine is purified by :
Answer» STEAM distillation simple distillation vacuum distillation none of these Answer :C

Discussion

50.	Impure glycerine can be purified by :
Answer» STEAM distillation Vacuum distillation Simple distillation Extraction with a solvent Answer :B

Discussion

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