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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.
| 28351. |
The mass of precious stones is expressed in terms of 'carat'. Given that 1 carat = 3.168 grains and 1 gram = 15.4 grains, calculate the total mass of a ring in grams and kilograms which contains 0.500 carat diamond and 7.00 gram gold. |
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Answer» SOLUTION :The unit CONVERSION factors will be : `1=("1 carat")/(3.168" grains")=("3.168 grains")/("1 carat")` `1=("1 gram")/("15.4 grains")=("15.4 grains")/("1 gram")` `"0.500 carat"="0.500 carat"xx("3.168 grains")/("1 carat")xx("1 gram")/("15.4 grains")="0.10 gram"` `THEREFORE"Total MASS of the ring"=7.00+0.10g=7.10gxx(1kg)/(1000G)=0.0071kg.` |
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| 28352. |
The molecule/ion having pyramidal shape is : |
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Answer» `PCl_(3)` |
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| 28353. |
The mass of P_(4)O_(10) produced if 440 gm P_(4)S_(3) is mixed with 384 gm of O_(2) is |
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Answer» 568gm |
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| 28354. |
The molecule in which one of the nitrogen bases is bonded with a sugar molecule is called ………………….. |
| Answer» SOLUTION :NUCLEOSIDE | |
| 28355. |
The mass of potassium dichromate crystals required to oxidise 750 cm^3 of 0.6 M Mohr's salt solution is: (Given molar mass : potassium dichromate = 294, Mohr's salt = 392) |
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Answer» 0.45 g Gram equiv. of mohr.s salt in `750 cm^3` of 0.6 M solution ` = 0.6 xx 1 xx 750/1000` = 0.45 Let the mass of `K_2Cr_2O_7 = X g ` ` therefore(x)/(294//6)= 0.45` `(becauseEq. wt of K_2Cr_2O_7 = ("mol. wt)/(6) )` `x= (0.45 xx 294)/(6) = 22.05 g ` |
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| 28357. |
The mass of oxygen gas which occupies 5.6 litres at STP would be |
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Answer» The gram atomic mass of OXYGEN 5.6 L of `O_2` gas at STP has mass ` = (32 XX 5.6)/(22.4) = 8 g ` 8 g of corresponds to half of the atomic mass oxygen . |
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| 28358. |
The molecule having smallest bondangle is : |
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Answer» `NCl_(3)` `{:(NCl_(3),PCl_(3),AsCl_(3),SbCl_(3)),(107^(6),109.5^(@),98^(@),98.2^(@)):}` |
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| 28359. |
The mass of one molecule of oxygen is: |
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Answer» 32 G ` = (32)/(6.022 xx 10^23) g ` |
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| 28360. |
The molecule having smallest bond angle is : |
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Answer» `PCl_(3)` |
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| 28361. |
The mass of one litre sample of ozonised oxygen at N.T.P. was found to be 1.5 g. When 100 mL of this mixture at N.T.P. were treated with turpentine oil, the volume was reduced to 90 mL. Calculate the molecular mass of ozone. |
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Answer» Solution :As ozone is absorbed by turpentine oil, therefore, volume of ozone in 100ML of the mixture `=100-90=10mL` `therefore""O_(2)" in the mixture"=100-10=90mL` As 1 L of the mixture WEIGH = 1.5 g, therefore, average MOLAR mass of the mixture (mass of 22.4 L at S.T.P.)`=1.5xx22.4=33.6"g mol"^(-1)` `"Ratio of ozone " : "oxygen in the mixture "=10:90` It m is the molecular mass of ozone, then `"Average mol. mass of mixture "=(10xxm+90xx32)/(100)="33.6 (CALCULATED above)"` `"or"m+288=336 "or"m=48` |
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| 28363. |
The mass of Mg_(3)N_(2) produced if 48 gm metal is reacted with 34 gm NH_(3) gas isMg + NH_(3) to Mg_(3)N_(2) + H_(2) |
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Answer» 200/3 gm |
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| 28364. |
The mass of molecule A is twice the mass of molecule B. The rms speed of A is twice the rms speed of B. If two samples of A and B contain same no. of molecules, what will be the ratio of P of two samples in separate containers of equal volume. |
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Answer» <P> Solution :Given, `M_(A) =2M_(B)``therefore` Mol wt. of `A=2xx` mol wt. of B `"…(1)"` `U_("rma")` of `A=2xxU_("rms")` of B `"…(2)"` Also, the no. of molecules of A =NO. of Molecules of B `"….(3)"` For GAS `A P_(A) V_(A) =(1)/(3) M_(A) U_("rms")^(2) A` `(P_(A)V_(A))/(P_(B)V_(B))=(M_(A))/(M_(B))xx(U_(A))/(U_(B))".....(4)"` Given `V_(A) =V_(B)".....(5)"` `therefore` By equation (1) (2) (4) & (5) `(P_(A))/(P_(B)) =2xx(2)^(2)=8` `P_(A) =8P_(B)` |
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| 28365. |
The molecule having least dipole moment (Assume benzene molecule to be a regular hexagon) : |
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Answer»
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| 28366. |
The mass of metal, with equivalent mass 31.75 which would combine with 8 g of oxygen is |
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Answer» 1 Mass of metal = X Combine with 8 g of oxygen Equivalent mass of metal= `("mass of metal")/("mass of oxygen") XX 8` i.e. `31.75 = (x)/(8) xx(8) Rightarrow x = 31.75` |
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| 28368. |
The mass of helium atom is 4.0026 amu, while that of the neutron and proton are 1.0087 and 1.0078 amu respectively on the same scale. Hence, the nuclear binding energy per nucleon in the helium atom is about |
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Answer» Solution :Mass defect of `._(2)He^(4), Delta m = [2(m_(p) + m_(n)) - m_(He)]` `Deltam = [2(1.0087 + 1.0078) - 4.0026]` `= 4.033 - 4.0026 rArr Delta m = 0.0304` `:.` BINDING energy PER nucleon `= (Deltam xx 931.5)/(4) = (0.0304 xx 931.5)/(4) = 7.01 MeV` |
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| 28369. |
The molecule having largest bond angle is :- |
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Answer» `H_(2)O` |
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| 28370. |
The mass of gold in an 18 carat gold ring of 4 g is : |
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Answer» Solution :24 carat gold is pure gold 4g ring of 18 carat gold ` = ( 4XX 18)/(24) = 3g` |
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| 28371. |
The mass of glucose that should be dissolved in 50 g of water in order to produce the same lowering of vapour pressure as produced by dissolving 1 g of urea in the same quantity of water is |
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Answer» <P>1 g As (`p^(@)-p_(s))//p^(@)` is same in the two CASES `((w_(2)M_(1))/(w_(1)M_(2)))_("glucose")=((w_(2)M_(1))/(w_(1)M_(2)))_("urea")` `(w_(2)xx18)/(50xx180)=(1xx18)/(50xx60) or w_(2)=3 g.` |
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| 28372. |
The molecule having highest bond dissociation energy is : |
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Answer» `O_(2)` |
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| 28373. |
The mass of dinitrogen in grams produced by the thermal decomposition of 2 moles of ammonium dichromate. |
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Answer» `{:("1 mole","1mole"),("2 mole","2mole"):}` Mass of `N_(2)=` mole of `N_(2)xx` molar mass of `N_(2)` `=2xx28=56 g` |
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| 28374. |
The mass of gases evolved in Question is |
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Answer» `0.06g` `:. 2xx96500Cimplies120g` `0.1xx965C=(120xx0.1xx965)/(2xx96500)implies0.06g` |
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| 28375. |
The molecule has: |
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Answer» ONE asymmetric carbon and one MESO FORM and two optically active isomers.  Two asterisk SHOW asymmetric carbon. It has two optically active isomers and one meso form. |
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| 28376. |
The mass of glucose that should be dissolved in 50 g of water in order to produce the same lowering of vapour pressure as is produced by dissolving 1 g of urea in the same quantity of water is |
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Answer» 1 G To produce same lowering of vapour PRESSURE `((p_(0)-p_(s))/(p_(s)))` TERM will be same for both the cases (solvent is same). `THEREFORE (w_(g)xx18)/(50xx180)=(w_(u)xx18)/(50xx60)` (`w_(g)=`weight of glucose, `w_(u)=`weight of urea) or,`(w_(g)xx18)/(50xx180)=(1xx18)/(50xx60)`or,`w_(g)=3`. |
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| 28377. |
The mass of CO_(2) produced from 620 gm mixture of C_(2)H_(4)O_(2) & O_(2) prepared to produce maximum energy is ( combustion reaction is exothermic ) |
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Answer» 413.33 gm |
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| 28378. |
Themolecule ClCH_(2)CH_(2)SCH_(2)CH_(2)Cl is commonlycalled as |
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Answer» TEAR GAS |
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| 28379. |
The mass of carbon anode consumed (giving only carbon dioxide) in the production of 270kg of Al metal from bauxite by Hall process is: |
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Answer» 270kg `because` 108g Al is produced by consuming =36G carbon `therefore 270xx10^(3)g` Al will be produced by consuming `=(36)/(108)xx270xx10^(3)g` of carbon `=90xx10^(3)g`=90 kg carbon. |
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| 28380. |
The mass of carbon present in 0.5 mole of k_(4)[Fe(CN)_(6)] is: |
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Answer» 1.8g |
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| 28381. |
The molecule exhibiting maximum number of non-bonding electron pairs (l.p.) around the central atom is : |
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Answer» `XeOF_(4)` |
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| 28382. |
The mass of carbon anode consumed (giving only carbondioxide) in the production of 270 kg of aluminium metal from bauxite by the hall process is |
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Answer» 180kg |
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| 28383. |
The molecule ClF_(3) has same number of lone pairs as are present in : |
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Answer» `SF_(4)` |
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| 28384. |
The mass of carbon anode consumed (giving only carbon dioxide) in the production of 270 kg of aluminium metal from bauxite by Hall process is: |
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Answer» Solution :`2Al_2O_3 + 3C to 3CO_2 + 4Al` `4 xx 27` g of Al are OBTAINED when carbon ANODE consumed= `12 xx 3g` 270 g of the Al are obtained when carbon anodeconsumed ` = (12 xx 3)/(4 xx 27)xx 270 = 90 kg ` |
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| 28385. |
The molecularity of the reaction,6FeSO_4+3H_2SO_4+KClO_3rarrKCl+3Fe_2(SO_4)_3+3H_2O is |
| Answer» Solution : The total number of reactant MOLECULES participating in a chemical reaction is KNOWN as its MOLECULARITY, hence the molecularity = 6 +3+ 1 = 10 | |
| 28386. |
The mass of carbon anode consumed (giving only carbon dioxide) in the production of 270 kg of aluminium metal from bauxite by Hall process is |
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Answer» `180 KG` `4xx27g` of Al are OBTAINED when CARBON anode consumed `= 12xx3 g` |
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| 28387. |
The mass of carbon anode consumed (giving only carbon dioxide) in the production of 270 kg of aluminium metal from bauxite by the Hall process is : |
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Answer» 90 KG PRODUCTION of `4xx27Kg=36kg` `therefore"Production of 270 kg of Al CONSUMERS C of anode"` `=(36)/(4xx27)xx270=90kg.` |
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| 28388. |
'The molecularity of a reaction can be 0,1,3 etc.'' The statement is |
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Answer» 1Both the above |
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| 28389. |
The molecularity of reaction 2NO+O_2to2NO_2is |
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Answer» 5 |
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| 28390. |
The mass of BaCO_(3) produced when excess CO_(2) is bubbled through a solution of 0.205 mol Ba(OH)_(2) is |
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Answer» 81 g Atomic wt. of `BaCO_(3)=137+12+16xx3=197` No. of mole `=("wt. of substance")/("mol wt.")` `:'` 1 mole of `Ba(OH)_(2)` GIVES 1 mole of `BaCO_(3)` `:.` 0.205 mole of `Ba(OH)_(2)` will give 0.205 mole of `BaCO_(3)` `:.` wt. of 0.205 mole of `BaCO_(3)` will be `0.205xx197=40.385 GM ~~ 40.5 gm`. |
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| 28391. |
The molecular weight of O_(2) and SO_(2) are 32 and 64 respectively. At 15°C and 150 mm Hg pressure, one litre of O_(2)contains N molecules. The number of molecules in two litres of SO_(2) , under the same conditions of temperature and pressure will be |
| Answer» Solution :If 1 L of `O_(2)` gas contains N molecules, 2L of any gas under the same conditions will contain 2 N molecules. | |
| 28392. |
The mass of an electron is m, its charge e and it is accelerated from rest through a potential difference V. The Kinetic energy of the electron in joules will be: |
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Answer» V |
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| 28393. |
The molecular weight of protein is |
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Answer» lt10000 |
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| 28394. |
The mass of an electron is 9.1xx10^(-31)kg . Ifits K.E . Is 3.0xx10^(-25)J , calculate its wavelength . |
| Answer» SOLUTION :896.7 NM | |
| 28395. |
The mass of an atom of carbon is: |
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Answer» 1G |
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| 28396. |
The molecular weight of NaCl determined from colligative properties is always |
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Answer» `GT 58.5` |
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| 28397. |
The mass of an atom of carbon is : |
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Answer» 1g ` = (12)/(6.02 xx 10^23) =1.99 xx 10^(-23)g ` |
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| 28398. |
The molecular weight of NaCl determined by studying freezing point depression of its 0.5% aqueous solution is 30. The apparent degree of dissociation of NaCl is: |
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Answer» 0.95 |
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| 28399. |
The mass of an Al block (in grams) whosedimensions are 2.0 inch x 3.0 inch x 4.0 inch having density 2.78 g cm^(-3)is |
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Answer» 64.8 G |
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| 28400. |
The molecular weight of hydrogen peroxide is 34.What is the unit of the molecular weight ? |
| Answer» Answer :C | |
