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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 31201. |
The current in amp required to liberate Ag from a solution containing 1.7 xx 10^(-3) kg of AgNO_(3) was electrolysed in 1 hr is |
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Answer» `1.342` `170 xx 10^(-3)` KG `AgNO_(3) -= 108 xx 10^(-3) ` kg of Ag `therefore 1.7 xx 10^(-3)` kg `Ag NO_(3) = W_(Ag)` `therefore W _(Ag) = (108 xx 10^(-3) xx 1.7 xx 10^(-3))/(170 xx 10^(-3))` `= 1.08 xx 10^(-3)` kg According to Faraday's `II^(ND)` law `i = (W xx F)/(t xx E) = (1.08 xx 10^(-3) xx 96500)/(1 xx 60 xx 60 xx 108 xx 10^(-3)) = 0.27` amp |
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| 31202. |
The current in a given wire is 1.8 A. The number of colombs that flow in 1.36 minutes will be |
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Answer» 100C |
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| 31203. |
The current flow in electrolytic conductors is due to |
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Answer» MIGRATION of atoms |
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| 31205. |
The CsCl structure is observed in alkali halides only when the radius of the cation is sufficiently large to keep its eight nearest-neighbour anions from touching. What minimum value of r_(+)//r_(-)is needed to prevent this contact? |
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Answer» 0.155 |
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| 31206. |
The cubic unit cell of Al ( molar mass = 27 g mol^(-1)) has an edge lengthof 405 pm. Its density is 2.7 g cm^(-3). The cubic unit cell is : |
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Answer» BODY centred `d = 2.7 g CM^(-3) ,M = 27,a = 405 xx 10^(-10)cm` `2.7 = ( Z xx 27)/( ( 405 xx 10^(-10))^(3) xx ( 6.02 xx 10^(23)))` `:. Z = (( 405 xx 10^(-10))^(3) xx ( 6.02 xx 10^(23))xx 2.7)/( 27)=4` `:.` Lattice is FCC lattice |
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| 31207. |
The cubic unit cell of a metal (molar mass=63.55 g mol^(-1)) has an edge length of 362 pm. Its density is 8.92 g cm^(-3). The type of unit cell is |
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Answer» primitive `z=(rhoN_(A)V)/(M)=(8.92xx6.02xx10^(23)XX(362)^(3)xx10^(-3))/(63.55)` =4 `therefore` it has fcc unit cell. |
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| 31208. |
The crystals of ferrous sulphate on heating give : |
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Answer» `FeO+SO_(2)+H_(2)O` `2FeSO_(4) overset(DELTA)to Fe_(2)O_(3)+SO_(2)+SO_(3)` |
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| 31209. |
The crystals are bounded by plane faces (f), straight edges (e) and interfacial angle (c). The relationship between these is : |
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Answer» f+c = e+2 |
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| 31210. |
The crystalline salt Naw_(2)SO_(4).xH_(2)O on heating loses 55.9% of its weight. The formula of the crystalline salt is |
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Answer» `Na_(2)SO_(4).5H_(2)O` Molar mass of ANHYD. `Na_(2)SO_(4)` `=2xx23+32+4xx16` `=142g` `therefore "142 g of anhyd. "Na_(2)SO_(4)` will be associated with `H_(2)O=(55.9)/(44.1)xx142g=180g="10 MOLES of "H_(2)O` |
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| 31211. |
The crystalline salt Na_2 SO_4.xH_2Oon heating loses 55.9 % of its weight. The formula of crystalline salt is |
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Answer» `Na_2 SO_4. 5H_2O` A crystalline salt on becoming anhydrous loses 55.9% by mass . 44.1 g of anhydrous salt contain `H_2O` = 55.9 142g of anhydrous salt contain `H_2O` ` = (55.9)/(44.1) xx 142 = 180 g ` No. of molecules ` = 180/80 = 10` |
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| 31212. |
The crystal with a metal deficiency defect is…………………………………… |
| Answer» SOLUTION :`FeO` | |
| 31213. |
The crystal very soft in nature is |
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Answer» IONIC CRYSTAL |
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| 31214. |
The crystal system without any element of symmetry is |
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Answer» MONOCLINIC |
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| 31215. |
The crystal system of a compound with unit cell dimensions a = 0.388 , b = 0.388 and c = 0.506 nm and alpha = beta = 90^@ and gamma = 120^@ is |
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Answer» HEXAGONAL |
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| 31216. |
The crystal system of a compound with unit cell dimensions a=0.387,b=0.387 and c=0.504 nm and alpha=beta=90^(@)and gamma=120^(@) is |
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Answer» HEXAGONAL |
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| 31217. |
The crystal system having one 6 fold axis is |
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Answer» hexagonal |
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| 31218. |
The crystal system having rectangular prisms is |
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Answer» Triclinic |
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| 31219. |
The crystal of CsBr has edge length of 437 pm. If the density of the crystal is 4.24 "g cm"^(-3) , determine the type of crystal structure of CsBr(At. mass of Cs = 133, Br = 80) |
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Answer» |
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| 31220. |
The crystal structure of solid Mn(II) oxide is |
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Answer» NaCl STRUCTURE |
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| 31221. |
The crystal having hcp is A_2B_3. Which atom hashcp structure and by other molecules of tetrahedral voids how much space is occupied ? |
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Answer» hcp crystal - A, `2//3` TETRAHEDRAL VOIDS - B |
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| 31222. |
The crystal having highest melting point is |
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Answer» IONIC CRYSTAL |
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| 31223. |
The crystal field theory assumes interaction between metal ion and the ligands as a purely electrostatic and ligands are supposed to be point charges. Q. Amongst the following complexes which has square planar geometry? |
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Answer» `[RhCl(CO)(P Ph_(3))_(2)]` |
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| 31224. |
The crystal field theory assumes interaction between metal ion and the ligands as a purely electrostatic and ligands are supposed to be point charges. Q. Which of the following match are incorrect? |
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Answer» |
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| 31225. |
The crystal field theory assumes interaction between metal ion and the ligands as a purely electrostatic and ligands are supposed to be point charges. Q. Which of the following order of CFSE in incorrect? |
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Answer» `[Cr(NO_(2))_(6)]^(3-) GT [Cr(NH_(3))_(6)]^(3+) gt [Cr(H_(2)O)_(6)]^(3+)` |
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| 31226. |
The crystal field stabilisation energy of [Co(NH_(3))_(6)]Cl_(3) is: |
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Answer» `-7.2Delta_(0)` |
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| 31227. |
The crystal field stabilization energy (CFSE) is the highest for |
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Answer» `[CoF_4]^(2-)` |
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| 31228. |
The crystal field-splitting for Cr^(3+) ion in octahedral field changes for ligands I^(-), H_(2)O, NH_(3), CN^(-) in increasing order is : |
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Answer» `I^(-) LT H_(2)O lt NH_(3) lt CN^(-)` |
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| 31229. |
The crystal field-splitting for Cr^(3+)ion in octahedral field changes for I^("ϴ"), H_(2)O, NH_(2), CN^("ϴ")and the increasing order is :- |
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Answer» `I^("ϴ") lt H_(2)O lt NH_(3) lt CN^("ϴ")` |
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| 31230. |
The crystal field splitting energy of Ti^(3+) ion complexes such as [TiBr_(6)]^(3-), [TiF_(6)]^(3-), [Ti(H_(2)O)_(6)]^(3+) the ligands are in the order..................... |
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Answer» |
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| 31231. |
The crystal field splitting energy for octahedral (Delta_(0)) andtetrahedral (Delta_(1)) complexes is related as |
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Answer» `Delta_(t)=-(1)/(2)Delta_(0)` `Delta_(t)=-(4)/(9)Delta_(0)`. |
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| 31232. |
The crystal field splitting energy for octahedral (Delta_(o)) and tetrahedral (Delta_(t)) complexes is related as: |
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Answer» `Delta_(t)=(1)/(2)Delta_(o)` |
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| 31233. |
The crystal field splitting energy for octahedral complex (Delta_(0)) and that for tetrahedral complex (Delta_(t)) are related as: |
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Answer» `Delta_(t)=4/9 Delta_(0)` |
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| 31234. |
The cryoscopic constant and freezing point of benzene is "5.12 K kg mol"^(-1) and 278.6 K respectively. At what temperature will one molal solution of benzene containing a nonelectrolyte (i=1) freeze? |
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Answer» SOLUTION :Depression in breezing point `(DeltaT_(f))` for dilute solution is directly proportional to molality `DeltaT_(f)alpham` `DeltaT_(f)=K_(f)m` `"GIVEN : "K_(f)="5.12 K kg MOL"^(-1)` `T_(f)^(@)=278.6K` `m="1 MOLAL"DeltaT_(f)=5.12 xx1` `""DeltaT_(f)=5.12K` `W.K.T""Delta_(f)=T_(f)^(@)-T_(f)` `5.12 =278.6-T_(f)` `T_(f)=278.6-5.12` `T_(f)=273.48K` |
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| 31235. |
The cryoscopic constant value depends upon |
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Answer» The number of PARTICLES of the solute in the SOLUTION `K_(f)` depends upon `Delta H_(f)` and `T_(f)`. Hence, choices (c ) and (d) are correct. It follows that (a) and (b) are not correct. |
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| 31236. |
The crossed aldol condensation product of the reaction between Formaldehyde and Acetaldehyde is .......... |
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Answer» 3 – HYDROXY PROPANOL |
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| 31237. |
The cryolite is : |
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Answer» `Na_3AlF_6` |
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| 31238. |
The cross aldol product formed when propanal acts as the electrophile and butanal as nucleophile is |
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Answer» 3-hydroxy-2-methylpentanal 
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| 31239. |
The critical temperature of a substance is :- |
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Answer» The temperature above which the substance UNDERGOES decomposition |
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| 31240. |
The critical temperature of agas is that temperature : |
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Answer» Above which it can no longer REMAIN in the GASEOUS state |
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| 31241. |
The critical temperature of a gas is the temperature |
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Answer» below which it cannot be LIQUIFIED |
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| 31242. |
The critical micelle concentration (CMC) is: |
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Answer» the concentration at which micellization STARTS |
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| 31243. |
The criteria for spontaneity of a process is/are |
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Answer» `(dG)_(T.P) LT 0` |
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| 31244. |
The creative species in chlorine bleach is |
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Answer» `CI_(2)O` |
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| 31245. |
The Cr metal (atomic weight 52) crystallises BBC structure. The unit cell edge length is 2.88 Å. The density of Cr is 7.2 g ml^(-1). The number of atom is 52 g of Cr is |
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Answer» `3 xx 10^(23)` |
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| 31246. |
The couplings between base units of DNA is through |
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Answer» HYDROGEN bonding |
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| 31247. |
The coupling of alkyl halides to form an alkane is called: |
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Answer» WURTZ synthesis |
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| 31248. |
The cost of table salt (NaCl) and table sugar (C_(2)H_(22)O_(11)) is Rs. 2 per kg and Rs. 6 per kg respectively. Calculate their costs per mole. |
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Answer» `THEREFORE"Cost of NaCl per mole"=(2)/(1000)xx58.5"Rs. = 0.117 Re = 11.7 p = 12p."` 1 mole of sugar `(C_(12)H_(22)O_(11))=342g` `therefore"Cost of sugar per mole"=(6)/(1000)xx342"Rs. = Rs. 2.05 p."` |
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| 31249. |
The counting rate observed from a radioactive source at t = 0 seconds was 1600 counts / sec and at t = 8 sec it was 100 counts /sec . The counting rate obserbed as count per sec at t = 6 sec will be |
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Answer» 400 `t_(1//2) = (0.693)/(lambda) = (0.693)/(0.693//2) = 2` sec , `1600 overset(2) (to) 800 overset(2)(to) 400 overset(2) (to) 200 , T = 6` |
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| 31250. |
The coupling between base units of DNA is through : |
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Answer» hydrogen bonding |
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