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19851.	This law was proposed by Dalton in 1803 . According to this law , if two elements combune to form more than one compounds ,the masses of one element that combine with a fixed mass of the other element, are in the ratio of small whole number. For exemple, hydrogen combines with oxygen to form two compounds ,namely, water and hydrogen peroxide. Here, the masses of oxygen (i.e, 16g and 32g) which combine with a fixed mass of hydrogen (2g) bear a simple raio, i.e, 16 :32 or 1 :2 . An element forms two oxides .In one oxide ,one gram of the oxide containd 0.5g of the element. In another oxide, 4g of the oxide contains 0.8g of the element . Show that these data confirm the law of multiple proportion.
Answer» ANSWER :OXIDE = 1 GM `=>` oxygen = `0.5` gm + element = `0.8` gm Ratio `= 1:1` Oxide = 4 gm `=>` oxygen ` = 3.2` gm + element ` = 0.8` gm Ratio `4:1`

19852.	This law was proposed by Dalton in 1803 . According to this law , if two elements combune to form more than one compounds ,the masses of one element that combine with a fixed mass of the other element, are in the ratio of small whole number. For exemple, hydrogen combines with oxygen to form two compounds ,namely, water and hydrogen peroxide. Here, the masses of oxygen (i.e, 16g and 32g) which combine with a fixed mass of hydrogen (2g) bear a simple raio, i.e, 16 :32 or 1 :2 . N_(2) and O_(2) combine to form a number of compounds such as NO,N_(2)O,N_(2)O_(3),N_(2)O_(5). Show that these follow the law of multiple proportion.
Answer» ANSWER :NO (N= 14 gm, 0 = 16 gm or N = 28 gm, O = 32 gm ) `N_(2)O(N=14 gm,O=16gm)` `N_(2)O_(3)(N=28gm,O=48gm)` `N_(2)O_(5)(N=28gm,O=80gm)` Here the masses of oxygen which COMBINE with a fixed mass of introgen (28 gm) bears a simple ratio.

19853.	Which of the following contains vitamin D:
Answer» Calciferol Keratin Tocopherol None ANSWER :A

19854.	Value of K_(p) in the reaction MgCO_(3(s))hArrMgO_((s))+CO_(2(g)) is
Answer» `K_(p)=P_(CO_(2))` `K_(p)=P_(CO_(2))xx(P_(CO_(2))xxP_(MgO))/(P_(MgCO_(3)))` `K_(p)=(P_(CO_(2))xxP_(MgO))/(P_(MgCO_(3)))` `K_(p)=(P_(MgCO_(3)))/(P_(CO_(2))xxP_(MgO))` Solution :In this REACTION gaseous MOLECULE count `MgCO_(3)to MgO_((s))+CO_(2(g))K_(p)=P_(CO_(2))`

19855.	Write the Nernst equation and emf of the following cells at 298 K: (i) Mg(s) \| Mg^(2+) (0.001 M)\|\| Cu^(2+) (0.0001 M) \| Cu (s) (ii) Fe(s) \| Fe^(2+) (0.001 M) \|\| H^(+) ( 1 M) \| H_(2) (g) (1 bar) \| Pt(s) (iii) Sn(s) \| Sn^(2+) (0.050 M) \|\| H^(+) (0.020 M) \| H_(2) (g) (1 bar) \| Pt(s) (iv) Pt(s) \| Br_(2) (l) \| Br^(-) (0.010 M) \|\| H^(+) (0.030 M) \| H_(2) (g) (1 bar) \| Pt(s) Given, E_(Mg^(2+)//Mg)^(@) = -2.37 V, E_(Cu^(2+),Cu)^(@) = +0.34 V, E_(Fe^(2+), Fe)^(@) = -0.44,E_(Sn^(2+)//Sn)^(@) = -0.14 V, E_(1//2 Br^(2),Br^(-))^(@) = +1.08 V
Answer» Solution :(i) Cell reactionP: `Mg + Cu^(2+) to Mg^(2+) + Cu (n=2)` Using Nernst equation and SUBSTITUTING the values, we get `E_("cell") = E_("cell")^(@) - (0.0591)/2 log ([Mg^(2+)])/([Cu^(2+)])` `therefore E_("cell") = 0.34 - (-2.37) - 0.0591/2 log (10^(-3))/(10^(-4)) = 2.71 - 0.02955 = 2.68 V` (ii) Cell reaction: `Fe + 2H^(+) to Fe^(2+) + H_(2) (n=2)` Using nernst equation and substituting the values, we get `E_("cell") = E_("cell")^(@)- (0.0591)/2 log ([Fe^(2+)])/([H^(+)]^(2))` `therefore E_("cell")^(@) = 0-(-0.44) - 0.0591/2 log (10^(-3))/(1)^(2)` `=0.44 - (0.0591)/2 XX (-3) = 0.44 + 0.0887 = 0.523` V (iii) Cell reaction: `Sn + 2H^(+) to Sn^(2+) + H_(2) (n=2)` Using Nernst equation and substituing the values, we get `E_(cell) = E_("cell")^(@) - 0.0591/2 log ([Sn^(2+)])/([[H^(+)])^(2)) = 0- (-0.14) - 0.591/2 log 0.05/(0.02)^(2) = 0.078 V` (iv) Cell reaction : `2Br^(-) + 2H^(+) to Br_(2) + H_(2)` Using Nernst equation and substituing the values, we get `E_("cell") = E_("cell")^(@) - (0.0591)/2 log 1/([Br^(-)]^(2) [H^(+)]^(2))` `therefore =1.08 - 0.0591/2 log (1.111 xx 10^(7)) = -1.08 - 0.0591/2 (7.04567)` `=-1.08 - 0.208 = -1.288` V Thus, oxidation will occur at the hydrogen electrode and reduction on the `Br_(2)` electrode. `E_(cell) = 1.288 V`.

19856.	Which one of the following statements is not correct
Answer» The value of equlibrium constant is changed in the PRESENCE of a CATALYST in the reacton at equlibrium Enzymes catalyes MAINLY bio-chemical reactions Coenzymes increases the catalytic activity Catalyst does not INITIATE any REACTION Solution :Equlibrium constant is not affected by presence of catalyst.

19857.	Which does not obey EAN rule:
Answer» `[Cu(NH_3)_4]^(2+)` `[Zn(OH)_4]^(2-)` `[HgI_4]^(2-)` `[FE(CO)_5]` Answer :A

19858.	What happens when acetone reacts with : HCN
Answer» SOLUTION :Propan-2-ol or ISOPROPYL ALCOHOL is FORMED.

19859.	The use of chemicals for treatment of diseases is called as
Answer» Homoeotherapy Isothermotherapy Angiotherapy Chemotherapy Answer :D

19860.	Write the name of the reaction.
Answer» SOLUTION :ALDOL CONDENSATION

19861.	The relation between equilibrium constant K_(p) and K_(c) is
Answer» <P>`K_(C)=K_(p)(RT)^(DELTAN)` `K_(p)=K_(c)(RT)^(Deltan)` `K_(p)=((K_(c))/(RT))^(Deltan)` `K_(p)=K_(c)=(RT)^(Deltan)` Answer :B

19862.	What would be the standard free energy change forthe formation of methane at 298 K? The value of Delta H^@ for CH_4(g) is -74.81 kJ "mol"^(1) and S^@ values for C_("(grap)") H_2(g) and CH_4(g) are 5.70, 130.7 and 186.3 J K^(-1) "mol"^(-1)respectively.
Answer» `-50.71 ` `47.56` `60.45` `-74.8` Solution :For the reaction , ` C + 2H_2 to CH_4` GIVEN `Delta H^@ = - 74.81 kJ "mol"^(-1) , Delta S_m^@` can be CALCULATED as `Delta S_m^@ CH_4(g)- [S_m^@ C_("graphite") + 2S_m^@ H_2 (g) ]` ` =[ 186.3 - (5.70 + 2 xx 130.7)] JK^(-1) "mol"^(-1)` `= -80.8 JK^(-1) "mol"^(-1)` SINCE `Delta G^@ = Delta H^@ - T Delta S^@` ` = - 74.81 -[(298) (-80.8xx 10^(-3) ) ]` `= -74.81 kJ "mol"^(-1) + 24.1 kJ"mol"^(-1) = -50.71 kJ "mol"^(-1)`

19863.	Write the IUPAC names of the following compounds: (i) (C_(2)H_(5))_(2)CH-CH=CH_(2) (ii) CH_(2)=CH-overset(C_(2)H_(5))overset(\|)(C)H-CH_(2)-C-=CH (iii) CH_(2)=underset(C(CH_(3))_(2))underset(\|)(C)-CH_(2)-CH_(3) (iv) CH_(3)-CH=CH-C-=CH (v) CH_(2)=CH-overset(CH_(3))overset(\|)(C)H-CH_(2)-CH=CH_(2) (vi) CH_(3)CH_(2)C-=underset(H_(2)C=CH)underset(\|)(C CH)-CH_(2)CH(CH_(3))_(2) (vii) CH_(3)CH_(2)CH_(2)overset(CH=CH_(2))overset(\|)(C)HCH_(2)CH_(2)CH_(2)CH_(3) (viii) CH_(2)=CH-overset(CH_(3))overset(\|)(C)H-overset(CH_(2))overset(\|\|)(C)-CH_(2)-C-=CH (ix) HC-=C-C=CH-CH=CH_(2) (x) CH_(2)=C=CH-CH_(2)-underset(CH=CH_(2))underset(\|)(C)H-CH_(3)
Answer» Solution :(i) 3 - ethtylpent - 1 - ene, (ii) 3 - ethylhex - 1 - 5 - yne (iii) 2 - ETHYL - 3,3 - dimethylbut - 1 - ene (iii) 2 - ethyl - 3,3 dimethylbut - 1 - ene (iv) pent - 3 - en - 1yne (V) 3 - methylhexa - 1,5 diene (vi) 3 - isobutyhept - 1 -en - 4 - yne (vii) 3 - propylhept - 1 - ene (viii) 3 - methyl - 4 - methylidenehept - 1 - en - 6 - yne (ix) Hexa - 1,3 - DIEN - 5 - yne (X) 5 - methylhepta - 1,2,6 - TRIENE

19864.	The species having tetrahedral shape is :
Answer» `[PdCl_(4)]^(2-)` `[NI(CN)_(4)]^(2-)` `[Pd(CN)_(4)]^(2-)` `[NiCl_(4)]^(2-)` Answer :D

19865.	The resistance of (N)/(10)solutionis foundto be2.5xx10^(3) omegatheequivalent conductance of the solutionis (cellconstant=1.25 cm^(-1))
Answer» 2.5 `omega^(-1) cm^(-2) equiv^(-1)` 5.0 `omega^(-1) cm^(2) equiv^(-01)` 2.5 `omega^(-1) cm^(-2 equiv^(-1)` 5.0 `omega^(-1) cm^(-2) equiv^(-1)` Solution :Resistance of `(N)/(10)` solution`=2.5 xx10^(3) omega` `K=(1)/("resistance ")xx` cell constant `=(1)/(2.5 xx10^(-3))xx1.25` `=(1.25 xx10^(-3))/(2.5)=5x10^(-4) omega^(-1) cm^(-1)` EQUIVALENT conductance `=(Kxx1000)/(M)=(5XX10^(-4)xx1000)/(1//10)` `= omega^(-1) cm^(2) equiv^(-1)`

19866.	Which Mutarotation is not observed in……………..
Answer» SUCROSE D-Glucose L-Glucose None of these SOLUTION :Sucrose

19867.	The salt which responds to dilute and concentrated H_2SO_4 is
Answer» `Na_3PO_4` `Na_2SO_4` `BA(NO_3)_2` `CaF_2` SOLUTION :`Ba(NO_3)_2`

19868.	Which of the following aqueous solutions remains neutral after electrolysis ?
Answer» `K_(2)SO_(4)` `NaCl` `AgNO_(3)` `CuSO_(4)` SOLUTION :Since reduction potential of `H_(2)Ogt` reduction potential of `K^(o+)` IONS, so the reduction of `H_(2)O` takes place, to give `H_(2)(g)` Similarly, oxidation potential of `H_(2)Ogt` oxidation potential of `SO_(4)^(2-)` , so again the oxidation of `H_(2)O` TAKE place to give `O_(2)(g)` Reduction of `H_(2)O` produces same equivalent of `overset(c-)(O)H` as the oxidation of `H_(2)O` produces same equivalent of `H^(o+)` ions to give neutral solution. In option `(b), Na^(o+)` is not reducted, but `H_(2)O` is REDUCED to give `overset(c-)(O)H` ions. So solution is basic. `Cl^(c-)` ions are oxidized at anode to give `Cl_(2)(g)`. In option `(c)`, `Ag^(o+)` ions are reducted at cathode to give `Ag` , while `NO_(3)^(c-)` ions are not oxidized but `H_(2)O` is oxidized to give `H^(o+)` ions, so solution is acidic. In option `(d), Cu^(2+)` ions are reduced at cathode to give `Cu`, while `SO_(4)^(2-)` ions are not oxidized, but `H_(2)O` is oxidized to give `H^(o+)` ions, so solution is acidic.

19869.	Which reagent cannot be used to prepare haloakane from an alcohol?
Answer» HCl<BR>HBr HI `Br_(2)` Solution :With `HX(X=Cl,Br,I)` the reaction occurs via protonation. The protonated ALCOHOL then react with `X^(-)` to fromhaloalkane. An alcohol does not react with `Br_(2)`because there is a no protonating agne and secondly, the extermely WEAK base `Br^(-)` cannot displace very STRONG base `OH^(-)`

19870.	Which of the following species contains two lone pairs ant two bond pairs around the central atom ?
Answer» `CO_2` `H_2O` `NH_3` `BF_2` ANSWER :B

19871.	Which of the following compounds will undergo racemisation when solution of alcoholic KOH ?
Answer» `H_(3)C-OVERSET(CH_(3))overset("\|")"CH"-CH_(2)Cl` Answer :C

19872.	Which of the following "is"//"are" correct?
Answer» 1 FERMI `= 10^(3) DPS` 1 curie `= 3.7 XX 10^(10) dps` 1 RUTHERFORD `= 10^(6) dps` 1 becquerel `= 1 dps` Answer :C::D

19873.	The tallest trees known are the redwoods in California. Assuming the height of a redwood to be 105 m, estimate the osmotic pressure required to push water up from the roots to the treetop
Answer» SOLUTION :`1.029 XX 10^6 PA`

19874.	Whatare food preservatives ? Give an example .
Answer» Solution :CHEMICAL ADDED in food to preserve are called PRESERVATIVES. e.g. SODIUM benzoate.

19875.	Which one of the following is a covalent crystal?
Answer» ROCK salt Ice Quartz Dry ice Solution :Quartz is a covalent CRYSTAL.

19876.	The sharp melting point of crystalline solids is due to :
Answer» a regular arrangement of constituent particles OBSERVED over a short distance in the CRYSTAL LATTICE. a regular arrangement of constituent particles observed over a LONG distance in the crystal lattice. same arrangement of constituent particles in DIFFERENT directions. different arrangement of constituent particles in different directions. Solution :a regular arrangement of constituent particles observed over a long distance in the crystal lattice.

19877.	When does a rain is considered to be acid rain ?
Answer» Solution :When the pH of RAIN falls below 5-6 , it BECOMES ACID rain.

19878.	What is electrochemical series ? Write different applications of the series.
Answer» ANSWER :

19879.	Which among the following is correct regarding nucleic acids?
Answer» Hydrolysis of DNA and RNA gives two components, a pentose SUGAR and PHOSPHATE group Both DNA and RNA have thymine as one of the pyrimidines The two major purine bases of both RNA and DNA are adenine and GUANINE. The PYRIMIDINE base present only in DNA is uracil. Solution :The two major purine bases of both RNA and DNA are denine and guanine.

19880.	Whichof thefollowing seriescontainselectrophiles only ?
Answer» `H_(2) O , SO_(3) ,H_(3) O^(+)` `NH_(3) ,H_(2) O , AICI_(3)` `AICI_(3) ,SO_(3) ,NO_(2)^(+)` `H_(2)O, CI^(+), NO_(3)^(-)` ANSWER :C

19881.	Write IUPAC name of (##DBT_SM_CHE_XII_U_12_E02_004_Q01.png" width="80%">
Answer» SOLUTION :3- Methyl-N, N-dimethyl pentanamine

19882.	Which of the following is the odd one out
Answer» POTASSIUM ferrocyanide Ferrous AMMONIUM sulphate Potassium ferricyanide Tetrammine COPPER (II) sulphate Answer :B

19883.	Which one of the following thermo setting plastic is used in paints?
Answer» melamine hard bakelite NAVOLAC SOFT bakelite Solution :navolac

19884.	There is large difference between the melting and boiling points of oxygen and sulphur.
Answer» Solution :DUE to small atomic size and high electronegativity, oxygen atom prefers to form `p pi-p pi` double BOND with other oxygen atom forming diatomic oxygen `(O_(2))` molecule. In contrast due to bigger size and lower electronegativity, sulphur atom does not form `p pi-p pi`double bond with other sulphur atoms and hence does not EXIST as diatomic `(S_(2))` molecule. Instead, it prefers to form single bonds with other sulphur atoms. In other words, sulphur atom shows greater tendency for catenation than oxygen atom and hence exists as octa-atomic sulphur `(S_(8))` molecule. Due to much larger molecular size of `S_(8)` molecule as compared to `O_(2)` molecule, the van der WAALS forces of attraction are much stronger in `S_(8)` MOLECULES than in `O_(2)` molecules. As a result, there is a larger difference between the melting and boiling points of oxygen and sulphur.

19885.	Which of the following compounds are diprotic
Answer» `H_(2)PO_(5)` `H_(2)S` `HClO_(3)` `H_(3)PO_(3)` Solution :Diprotic SOLVENTS give `2H^(+)` ions or `OH^(-)` ions.

19886.	Which among the following compounds contains quaternary carbon?
Answer» ANSWER :B

19887.	Which one of the following is wrongly matched ?
Answer» `[Cu(NH_(3))_(4)]^(2+)` - square planar `[NI(CO)_(4)]` - neutral ligand `[Fe(CN)_(6)]^(3-) - sp^(3)d^(2)` `[Cr(en)_(3)]^(3+)`-follows EAN RULE Solution :`[Fe(CN)_(6)]^(3-)` is an inner orbital octahedral complex with the hybridisation `d^(2)sp^(3)`.

19888.	Which is the correct statement
Answer» Isotopes are ALWAYS radioactive `beta-`RAYS are always negatively charged particles `alpha-`rays are always negatively charged particles `GAMMA`-rays can be deflected in MAGNETIC FIELD Solution :`alpha`-rays positively charged, `beta-`rays are negatively charged, `gamma`-rays carry no charge and thus are not deflected in field

19889.	Which of the following chemical systems is/are nonaromatic?
Answer» ANSWER :D

19890.	Write mathematical expression for Vant HoffBoyle's law ?
Answer» Solution :`pi prop ` C at CONSTANT T. OR `pi prop (1)/(V)` at Constant T. OR `pi` V = ACONSTANT at constant T.

19891.	Write the IUPAC name of CH_(3)CH(CH_(3))CH_(2)C(CH_(3))_(2)COCH_(3)
Answer» SOLUTION :3, 3, 5 - Trimethylhexan -2- OEN.

19892.	Which of the following fuels will have the highest calorific value (kJ//kg)
Answer» Charcoal KEROSENE Wood Dung Solution :Out of GIVEN SUBSTANCES, kerosene oil has maximum calorific value.

19893.	Which set of elements is called chalcogens ?
Answer» <P>CL, Br, I O, S, Se N, P, As C, Si, Ge Answer :B

19894.	Velocity of the S_(N^(2)) reaction depends on the concentration of the substrate as well as te nucleophile. The reaction is favoured by strong Nu^(overset(..)(ө)), and in the presence of polar aprotic solvent, optically active halides give walden inversion by S_(N^(2)) mechanism. the presence of hetro group (atom as beta-C atom, unsaturation at beta-C, and (-overset(O)overset(\|\|)(C)-) Allyl halides also give S_(N^(2)) mechanism. EDG at ortho-and para-positions in benzyl halides favours S_(N^(1)) mechanism, whereas EWG favours S_(N^(2)) mechanism. Q. Which of the following will give Walden inversion
Answer» `C_(4)H_(9)-CHD-Br` All of these Solution :(a) is only OPTICALLY active and will have WALDEN inversion.

19895.	Velocity of the S_(N^(2)) reaction depends on the concentration of the substrate as well as te nucleophile. The reaction is favoured by strong Nu^(overset(..)(ө)), and in the presence of polar aprotic solvent, optically active halides give walden inversion by S_(N^(2)) mechanism. the presence of hetro group (atom as beta-C atom, unsaturation at beta-C, and (-overset(O)overset(\|\|)(C)-) Allyl halides also give S_(N^(2)) mechanism. EDG at ortho-and para-positions in benzyl halides favours S_(N^(1)) mechanism, whereas EWG favours S_(N^(2)) mechanism. Q. Which of the following will give S_(N^(2)) mechanism
Answer» MeBr All of these Solution :All, (a) `implies1^(@)1^(@)RX,(b) IMPLIES"ALLYL HALIDE, "(C)implies" Benzyl halide"`

19896.	Water is important to living being because
Answer» It is a compound of hydrogen and OXYGEN It can be obtained in pureform Itis a good solvent and its boiling point is MODERATELY HIGH It is COLOURLESS liquid Answer :C

19897.	Which of the following can cause depletion of ozone ?
Answer» `H_2S` NO Smoke Aerosols Answer :D

19898.	What are coinage metals ? Are they transition elements ?
Answer» Solution :IB group elements copper, silver and GOLD are called coinage elements. In OLDEN days these metals are used to prepare coins. They do not have incompletely filled subshell in elemental FORM but they have PARTIALLY filled d-shell in their CHEMICALLY significant oxidation states. Thus they are transition elements.

19899.	Which of the following is a non biodegradable polymer ?
Answer» PHBV PGA PMMA PCL Solution :PGA

19900.	Which one of the following is used for detection of pressure of charge on sol particles?
Answer» Cataphoresis ELECTRO dialysis Dialysis Ultrafilteration Solution :Cataphoresis

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