94420 + Interview Questions in Chemistry in Class 12 Page 1835 InterviewSolution

91701.	A 1.345-g sample of a compound of barium and oxygen was dissolved in hydrochloric acid to give a solution of barium ion, which was then precipitated with an excess of potassium chromate to give 2.012 g of barium chromate, BaCrO_4 . What is the formula of the compound?
Answer» SOLUTION :`BaO_2`

Discussion

91702.	A 1.2 g/1 solution of NACI is isotonic with 7.2 g/1 of solution of glucose. Calculate the van't Hoff's factor of NACI solution
Answer» 2.36 1.5 1.95 1 Solution :VANT HOFF factor of NaCl about 1.95 because it will be ionise into TWO IONS. `NaCl hArr Na^(+)+Cl^(-)`

Discussion

91703.	8g O_(2) has same number of molecules as that in:
Answer» 14G CO 7g CO 11g `CO_(2)` 22g `CO_(2)` ANSWER :B::C

Discussion

91704.	A 110% sample of oleum contains
Answer» 44.4% of `SO_(3)` 55.6% of SULPHURIC ACID 55.6% of `SO_(3)` 44.4% of sulphuric acid ANSWER :AB

Discussion

91705.	8c,c,of gaseous hydrocarbon requires 40c.c. of O_(2) for complete combustion. Identify hydrocarbon.
Answer» Solution :VOLUME of hydrocarbon `=8c.c.`: Volume of`O_(2)=40c.c` Formula No. `1, (8)/(40)=(2)/(3n+)` (For ALKANE) `(1)/(5)=(2)/(3n+1)` or `3n+1=10` or `3n=10-1=9 , n=3` The value of `n` COMES in whole number from `1st` formula it means hydrocarbon is alkane and it is of `3C` atom. `therefore` Hydrocarbon is `C_(3)H_(8)` (Prospane)

Discussion

91706.	A 1:1mixture of pentane and hexane is seperated by fractional distillation in the apparatus shown. At what temperature does the first drop of condensate appear on the thermometer?
Answer» less than `30^(@)`C `36^(@)C` between `36^(@)C` and `69^(@)C` more than `69^(@)C` ANSWER :B

Discussion

91707.	A 10L cylinder of nitrogen at 4.0 atm pressure and 27^(@ C develped a leakage. When the leakage was repaired 2.36 atm of nitrogen remained in the cylinder still at 27^(@0C. How many grams of nitrogen escaped ?
Answer» 18.7 G 0.67g 52.6g 10.0g SOLUTION :Moles of `N_(2)` escaped `= ( 4 XX 10)/( 0.082 xx 300)- ( 2.36 xx 10)/( 0.082 xx 300)= 0.667` MASS of `N_(2) = 0.667 xx 28 = 18.7 g `

Discussion

91708.	._(89)Ac^(231) " given " ._(82)Pb^(207) after emission of some alpha and beta particles. The number of such alpha and beta-particles are respectively
Answer» 5, 6 6, 5 7, 5 5, 7 Solution :Number of `ALPHA-"particles" = (231 -207)/(4) = 6` Number of `beta-"particles" = 2xx 6 - 89 + 82`

Discussion

91709.	A 10L flask at 300K contains a gaseous mixture of CO and CO_(2) at a total pressure of 2.0 atm. If 0.2 mol of CO is present, the partial pressure of CO_(2) is ( Use R = 0.082 atm mol^(-1) K^(-1) )
Answer» <P>0.49 atm 1.49 atm 1.51 atm 2.49 atm Solution :`p = p ( CO ) + p ( CO_(2)) = 2.0 atm` `p(CO) = ( 0.2 XX 0.082 xx 300)/( 10)` `= 0.49` atm `p ( CO_(2)) = 2-0.49 = 1.51 atm`

Discussion

91710.	896 mL. of a mixture ofCO and CO_2 weigh 1.28 g at NTP. Calculate the volume of CO_2 in the mixture at NTP.
Answer» `448 ML ` `672 ml` `224 ml` `500 ml` ANSWER :C

Discussion

91711.	A 10gm mixture of glucose and urea present in 250 ml solutionshows the osmotic pressure of 7.4 atm at 27^(@)C . The precentage compostion of urea in the solution will be
Answer» `17.6%` `82.4%` `20%` `80%` ANSWER :B

Discussion

91712.	._(88)X^(288) - 3alpha + beta rarr Y. The element Y is
Answer» `._(82) Pb^(216)` `._(82)Pb^(217)` `.(83)Bi^(218)` `._(83)Bi^(216)` SOLUTION :ATOMIC No.: `88 - (3 xx 2) + (1xx1) = 83` Mass No.: `228 - (3 xx4) - (1 xx0) = 216`

Discussion

91713.	A 10g mixture of iso-butane and iso-butene requires 20g of Br_2(in CCl_4) for complete addition. If 10g of the mixture is catalytically hydrogenated and the entire alkane is monobrominated in the presence of light at 127^(@)C, how much of it would be formed? (Atom weight of bromine=80)
Answer» 24.21g 20.0g 30.0g 12g Solution :Let iso-butane be a g and iso-butene b g. Thus a +b = 10g `CH_3-UNDERSET((56))underset("iso-butene")underset(CH_3)underset(\|)C=CH_2+underset((80xx2))(Br_2)toCH_2-underset(CH_3)underset(\|)OVERSET(Br)overset(\|)C-CH_3Br` `CH_3-underset("iso-butane")underset(CH_3)underset(\|)(CH)-CH_3+Br_2to` No addition REACTION Now 160 g of `Br_2` REACTS with 56 g of iso-butene `:.` 20g of `Br_2`reacts with `(56xx20)/160=7.25` g of iso-butane Total amount of iso-butane available for 10g mixture `=(7.25+3)=10.25 g` `CH_3-underset((58))underset(CH_3)underset(\|) (CH)-CH_3+Br_2tounderset("tert-butyl BROMIDE(137)")(CH_3-underset(CH_3)underset(\|)overset(Br)overset(\|)C-CH_3)` Now 58g of iso-butane gives 137 fo tert-butyl bromide `:.` 10.25g of iso-butane gives `(137xx10.25)/58=24.21g` of tert-butyl bromide.

Discussion

91714.	""_88^@224Rahaving t_(1//2)= 3.64d emits an a particle to form ""_(86)^(220)Rn , which has t_(1//2) = 54.5 s. Given that the molar volume of radon under these conditions is 35.2 dm^3 ,what volume of radon is in secular equilibrium with 1 g of radium?
Answer» SOLUTION :`2.72 XX 10^(-8) m^3`

Discussion

91715.	A 10cm^(3) sample of human urine was found to have 5 milligrams of urea on analysis. Calculate the molarity of given sample with respect to urea (Mol. Mass of urea - 60).
Answer» Solution :CONCENTRATION of urea = 5 mg in 10 `CM^(3)` = `"500 mg L"^(-1)="0.50 g L"^(-1)="0.5/60 MOL L"^(-1)="0.0083 mol L"^(-1)`

Discussion

91716.	""_(88)^(226)Ra emits an o-particle and the daughter element is:
Answer» `""_(90)^(230)Th` `""_(89)^(226)Ra` `""_(86)^(222)Rn` `""_(86)^(226)Rn` ANSWER :C

Discussion

91717.	A 100W and 110V incandescent lamp is connected in series with an electrolytic cell containing CdSO_4 solution. What mass of cadmium will be deposited at the cathode after 4hrs of electricity.
Answer» Solution :Step I: Calculation of the quantity of charge passed: We know, that Watt= Ampere `times` volt `Ampere= (WAT t)/(Vol t)=100/110` Now charg e= Current `times` Time `=(100/110 amp.)times 4 times 60 times 60` `=13091 C` Step II: Calculation of mass of CADMIUM deposited: The cathodic reaction is `Cd^(2+) (aq)+2e^(-) to Cd(s)` `112.2g 2 times 96500 C` `because 2 times 96500 C` of charge deposited Cd=112.2 g `THEREFORE 13091C` charge will DEPOSITE Cd `=(112.2g)/(2 times 96500 C) times 13091C` =7.61g

Discussion

91718.	8:8 coordination of CsCl is found to change into 6:6 coordination :
Answer» HIGH temperature high pressure high temperature and high pressure low temperature and low pressure. Solution :Increase of pressure increases coordination NUMBER. HENCE, by applying high pressure , NACL type crystal can be changed into CsCl type crystal

Discussion

91719.	A 100ml solution of 0.1 N HCl was titrated with 0.2N NaOH solution. The titration was discontinued after adding 30ml of NaOH solution. The remaining titration was completed by adding 0.25NKOH solution. The volume of KOH required for completing the titration is
Answer» 70ml 32ml 35ml 16ml Answer :D

Discussion

91720.	A 100cm^(3) solution of sodium carbonateis prepared by dissolving 8.653 g of the salt in water. The density of solution is 1.0816 g per millilitre. What are the molarity and molality of the solution ? (Atomic mass of Na is 23, of C is 12 and of O is 16).
Answer» SOLUTION :0.816 M, and 0.820 m

Discussion

91721.	8:8 coordination of CsCI is found to change into 6:6 coordination on:
Answer» APPLYING pressure Increasing temperature Both (a) and (B) None Answer :B

Discussion

91722.	A 100-watt, 110-volt incandescent lamp is connected in series with an electrolytic cell containing cadmium sulphate solution. What weight of cadmium will be deposited by the current flowing for 10 hours ?
Answer» Solution :We know current in amperes `= ("wattage")/("voltage") = (100)/(110)`……….(Eqn. 5) Further, CHARGE in coulombs = current in amperes `xx` TIME in seconds `= (100)/(110) xx 10xx 60 xx 60 = 32727.27` `therefore` charge in faraday `= (32727.27)/(96500)F = 0.34 F` `therefore` mole of electricity = 0.34 F. Amount of Cd deposited = 0.34 eq. `= (0.34 xx eq. wt.)g` `= 0.34 xx (112.4)/(2) = 19.11 g` `("eq. wt. of CADMIUM " = (112.4)/(2))`

Discussion

91723.	87.5% decomposition of a radioactive substance completes in 3 hours. What is the half-life of that substance
Answer» 2 hours 3 hours 90 minutes 1 hours Solution :`(0.693)/(t_(1//2)) = (2.303)/(180) XX "log" (100)/(12.5)` `t_(1//2) = (0.693 xx 180)/(2.303 xx 3 xx 0.3010) = 60` mini = 1 hr.

Discussion

91724.	A 100 W, 220 V incandescent lamp is connected in series with an electrolytic cell containing copper sulphate solution. What weight of copper will be deposited by 1 A current flowing for 5 hours? (at. Wt. of Cu=63.54).
Answer» SOLUTION :CURRENT=`("POWER")/("Voltage")=(100)/(220)"ampere"=(5)/(11)`ampere.

Discussion

91725.	8.7 g of pure MnO_(2) is heated with an excess of HCl and the gas. evolved is passed into a solution of KI. Calculate the amount of .he iodine liberated (Mn=55,Cl=35.5, I=127):
Answer» ANSWER :a,B

Discussion

91726.	""_(86)^(226)Rato_(84)^(222)Rn+(4)/(2)He Order of reaction is..
Answer» 0 `1^(st)` `2^(ND)` `3^(rd)` ANSWER :B

Discussion

91727.	A 100% pure sample of a divalent metal carbonate weighing 2 g on complete thermal decomposition releases 448 cc of carbon dioxide at STP. The equivalent mass of the metal is
Answer» 40 20 28 12 Solution :`underset(1 "mol"^3 (2g) ) (MCO_3) overset(DELTA)(to) underset(22400 "cc")(MO + CO_2)` 448 cc of `CO_2` is evolved from `MCO_3 = 2g` 22400 cc of `CO_2` will be evolved from `MCO_3 = (2)/(448) xx 22400 = 100 G` or molecular mass of `MCO_3 = 100 g "mol"^(-1)` If M is the atomic mass of METAL `M + 12 +3 xx 16 = 100` ` therefore M = 100 - 12 - 48= 40 ` Equivalent wt.= atomic mass / valency= 40/2 = 20

Discussion

91728.	""_(86)^(223) Ra is a member of actinium series . Which of the following nuclide belong to the same series ?
Answer» `""_(91)^(235) PA` `""_(90)^(232) Th` `""_(92)^(238) U` `""_(88)^(237) Ra` Solution :`(4 n + 3)` SERIES `implies Pa""_(91)^(235) , Ra""_(88)^(227)`

Discussion

91729.	A 1.00 molar .aqueous solution of trichloroacetic acid ("CCl"_(3)"COOH") is heated to its boiling point. The solution has the boiling point of 100.18 ""^(@)C. Determine the Van't Hoff factor for trichloroacetic acid. (K_(b) "for water" = 0.512 "kg mol"^(-1)).
Answer» SOLUTION :`DELTA T_(b) = iK_(b)`m `(100.18 - 100)^(@)C = I XX 0.512 k" mol"^(-1) xx 1 ` m 0.18 K = i`xx 0.512` K g `mol^(-1) xx 1` m i= 0.35 OR (i) Mole fraction is the ratio of number of moles of one component to the total number of moles in a mixture.`"" `[ Or mathematical expression ] `x_(A) = (n_(A))/(n_(A) + n_(B))` Where `n_(A)`= number of moles of component A. `n_(B)` = number of moles of component B. (ii) Solution which obey Raoult.s law over the ENTIRE range of concentration are called ideal solutions.

Discussion

91730.	A 1.00 molal aqueous solution of trichloroacetic acid ("CCl"_3COOH)is heated to its boilingpoint. The solution has a boiling point of 100.18^@C . Determine the van't Hoff factor for trichloro acetic acid (K_b " for water" = 0.512 K kg "mol"^(-1) ) .
Answer» Solution :Apply the relation ` Delta T_b = K_f xx m` ` Delta T_b = 0.512 xx 1 = 0.512` van.t HOFF FACTO =Observed ELEVATION in boiling POINT / CALCULATED elevation in boiling point `= (0.18)/(0.512) = 0.3516`

Discussion

91731.	860 fine silver means:
Answer» 14% Cu + 86%AG 8.6% Ag +91.4% Cu 86%Ag+14%Cu 86%Ag + 26%Cu Answer :C

Discussion

91732.	A 100 mL solution of KOH contains 10 milliequivalents of KOH .Calculateits strength in normality and grams /litre .
Answer» SOLUTION :Normality `=("no.of m..E")/("VOLUME in mL")…(Eqn.1) ` `=10/100 = 0.1 ` ` :. ` strength of the solution = N/10 Again, strength in grams/LITRE = normality `xx` eq.wt = `1/10 = 56 = 5.6` grams /litre . `(" eq.wt of KOH" ("molecular wt")/("acidity") =56/1 =56)`

Discussion

91733.	""_(84)Po^(218)(t_(1//2)=3.05mins) decays to ""_(82)Pb^(214)(t_(1//2)=2.68min) by alpha-emission while Pb^(214) is a beta-emitter. In an experiment starting with 1g atom of pure Po_(218), how much time would be required for the number of nuclei of ""_(82)Pb^(214) to reach maximum?
Answer» ANSWER :`4.12mins`

Discussion

91734.	A 100 ml solution of 0.1N-HCl was titrated with 0.2 N-NaOH solution. The titration was discontinued after adding 30 ml of NaOH solution. The remaining titration was completed by adding 0.25N-KOH solution. The volume of KOH required for completing the titration is
Answer» 16 ml 32 ml 35 ml 70 ml Solution :In the neutralization of acid and base `NxxV` of both must be EQUIVALENT `NxxV` of HCl=0.1`xx100=10` `NxxV` of NaOH=0.2`xx30=6` as to OBTAIN `10NxxV` of base `4NxxV` of base is required `NxxV` of KOH=0.25`xx16=4` `N_(1)V_(1)=underset(NaOH)(NxxV)+underset(KOH)(NxxV)` `0.1xx100=0.2xx30+0.25xxV` `10=6+0.25V` `V=(400)/(0.25)impliesV=16ml`

Discussion

91735.	A 100 mL solution of 0.1 N HCI was titrated with 0.2 N NaOH solution. The titration was discontinued after adding NaOH solution.The remaining titration was completed by adding 0.25 N KOH solution. The volume of KOH required for completing the titration is
Answer» 70 mL 32 mL 35 mL 16 mL Solution :`N_(1)V_(1) = N_(2)V_(2) + N_(3)V_(3)` i.e. `0.1 xx 100 = 0.2 xx 30 + 0.25 V_(3)` `0.25 V_(3) =4` or `V_(3) = 16 mL`

Discussion

91736.	A 100 ml solution of 0.1 N HCl was titrated with 0.2 N NaOH solution. The titration was discontinued after adding 30 ml of NaOH solution. The remaining titration was completed by adding 0.25 N KOH solution. The volume of KOH required for completing the titration is
Answer» 70 ml 32 ml 35 ml 16 ml Solution :`N_(1)V_(1)=N_(2)V_(2)+N_(3)V_(3)` i.e., `0.1xx100=0.2xx30+0.25V_(3)` `"or"0.25V_(3)=4 or V_(3)=16ml`

Discussion

91737.	A 100 mL sample of water was treated to convert any iron present to Fe^(2+). Addition of 25 mL of 0.002 M K_(2)Cr_(2)O_(7) resulted in thereaction : 6Fe^(2+)+Cr_(2)O_(7)^(2-)+14H^(+)rarr6Fe^(3+)+2Cr^(3+) + 7H_(2)O The excess K_(2)Cr_(2)O_(7) was back-titrated with 7.5 mL of 0.01 M Fe^(2+) solution. Calcution the parts per million (ppm) of iron in the water sample.
Answer» ANSWER :126 PPM

Discussion

91738.	._(84)Po^(210) rarr ._(82)Pb^(206) + ._(2)He^(4). From the above equation, deduce the position of polonium in the periodic table (lead belongs to group IV A)
Answer» IIA IV B VI B VI A Solution :An ELEMENT formed by losing one `alpha-`PARTICLE occupies who position left to parent element, Pb in IVA, thus PO should be in VIA

Discussion

91739.	""_(84)Po^(210) decays with alpha-particle to ""_(82)Pb^(206) with a half -life of 138.4 days . If 1.0g of ""_(84)Po^(210) is placed in a sealed tube, how much helium will accumulate in 69.2 days . Express the answer in cm^(3) at STP
Answer» ANSWER :`31.25cm^(3)`

Discussion

91740.	A 10.0 ml of (NH_(4))_(2) SO_(4) solution was treated with excess of NaOH. The ammonia evolved was absorbed in 50 ml of 0.1 N HCl. The excess HCl required 20 ml of 0.1 N. NaOH. Calculate the strength of (NH_(4))_(2) SO_(4) in the solution.
Answer» Solution :`(NH_(4))_(2)SO_(4)+2NaOHrarr Na_(2)SO_(4)+2Nh_(4)OH` `NH_(4)OH rarr NH_(3)+H_(2)O` `"EXCESS "HCl+NaOH rarr NaCl +H_(2)O` `NaOH` used by excess `HCl="20 ml of 0.1 N"` `"Excess HCl "="20 ml of 0.1 N"` `HCl" used by "NH_(3)="30 ml of 0.1 N"` `"Ammonia produced "= "30 ml of 0.1 N"` `=" 30 ml of 0.1 M"` `(NH_(4))_(2)SO_(4)" used "="30 ml of 0.05 M"` `"MASS of "(NH_(4))_(2)SO_(4)=(30xx0.05)/(1000)` `=0.198g` `"Thus "(NH_(4))_(2)SO_(4)="0.198 g/litre"`

Discussion

91741.	._(84)Pb^(219) is a member of actinium series. The other member of this series is
Answer» `._(89)Ac^(225)` `._(90)Th^(232)` `._(15)P^(35)` `._(92)U^(235)` Solution :The seires `4n + 3` is KNOWN as actinium series. It shows that if mass NUMBER is divided by 4 then the remainder MUST be 3. Hence, the another member of this series is `._(92)U^(235)` `:' (235)/(4) = 58 +` Remainder 3

Discussion

91742.	A 100 mL 0.1 M solution of ammonium acetate is diluted by adding 100 mL of water. The pH of the resulting solution will be (PK_a of acetic acid is nearly equal to pK_a of NH_4 OH)
Answer» `4.9` `5.0` `7.0` `10.0` SOLUTION :HYDROLYSISOF a saltof weakacidandweakandbase : `PH =7 +1/2(pK_a -pK_b) =7` `as pk_a ~~ pJK_b`

Discussion

91743.	._(84)^(218)Po(t_(1//2)=183 sec) decay to ._(82)Pb(t_(1//2)=161 sec) by alpha-emission, while Pb^(214) is a beta-emitter. In an experiment starting with 1 mole of pure Po^(218), how many time would be required for the number of nuclei of ._(82)^(214)Pb to reach maximum ?
Answer» 147.5 247.5 182 304 Answer :B

Discussion

91744.	._(84)^(218)Po (t_(1//2) = 183 sec) decay to ._(82)^(214)Pb (t_(1//2) = 161) sec by alpha emission, while ._(82)^(214)Pb decay by beta-emission. In how much time the number of nuclei of ._(82)^(214)Pb will reach to the maximum?
Answer» 182 SEC 247.5 sec 308 sec 194.8 sec Solution :`._(84)^(218)Po overset(lambda_(1) = (0.693)/(183) = 3.786 xx 10^(-3) sec^(-1))(to)` `._(82)^(214)Pb overset(lambda_(2)= (0.693)/(161) = 4304 xx 10^(-3) sec^(-1))(to)` `t_("MAX") = (2.303)/(lambda_(1) - lambda_(2)) log (lambda_(1))/(lambda_(2))` `(2.303)/(3.786 xx 10^(-3) - 4.304xx 10^(-3)) log (3.786 xx 10^(-3))/(4.304 xx 10^(-3))` `= - (2.303) xx 5.183 xx 10^(-4)) (-0.05569)` `= 247.5 sec`

Discussion

91745.	A 1.00- g sample of H_(2)O_(2)solution containing x percent H_(2)O_(2)by weightrequiresx mLof a KMnO_(4)solution for complete oxidation under acidic conditions .Calculate the normality of the KMnO_(4) solution .
Answer» Solution :`{:(2MnO_(4),+ 5H_(2)O_(2),6H^(+),to 5O_(2),+ 2Mn^(2+),+8H_(2)O),((+7),,,,(+2),):}` FROMTHE equation , we see that change in oxidation number of Mn = `7-2 = 5` ` :. ` 1 MOLE of `KMnO_(4)` = 5 EQ. og `KMnO_(4)` ` :. ` 10 eq. of ` KMnO_(4)`= 5 eq. of `KMnO_(4)` ` :. ` 10 eqof `KMnO_(4)`combines with 5 moles of `H_(2)O_(2)` . ` :. ` 1 eq . of `KMnO_(4)` combines with `1/2 ` mole of `H_(2)O_(2)` ` :. ` eq.wt of `H_(2)O_(2)= 34/2 = 17` Now , ` :.100 ` g of `H_(2)O_(2)`solution contains X g of `H_(2)O_(2)` ` :. 100` g of `H_(2)O_(2)`contains `x/17`equivalent of `H_(2)O_(2)` ` :. 1 ` g of `H_(2)O_(2)` contains `x/(17 xx100) ` eq. of `H_(2)O_(2)` ` :. ` number of m.e of `H_(2)O_(2)` in 1 g solution = `x/(17 xx100) xx 1000` ` = (10 x)/17` m.e of `H_(2)O_(2)` = m.eof `KMnO_(4)` `(10 x)/17 = x N ` ( N = normality of `KMnO_(4)`) `:. ` normality of `KMnO_(4)` solution = `10/17eq. " lit"^(-1)`

Discussion

91746.	""_(84)^(210)Po to ""_(82)^(206)Pb + ""_(2)^(4)He From the above nuclear reaction, predict the position of polonium in the periodic table (leads belongs to group IV A).
Answer» IIA IV B VI B VI A SOLUTION :Group IV A MEANS Group 14. Po will be in Group 16 i.e., VI A.

Discussion

91747.	""_(84)^(210)Po decayswith emission of alpha-particle to ""_(82)^(206)Pb with a half-life period of 138.4 days. If 1g of ""_(84)^(210)Po is placed in a sealed tube, how much helium will be accumulated in 69.2 days ? Express the answer in cm^(3) at STP ?
Answer» ANSWER :`31.23cm^(3)`

Discussion

91748.	A 10.0 g sample of a mixture of calcium chloride and sodium chloride is treated with Na_(2)CO_(3) to precipitate the calcium as calcium carbonate . This CaCO_(3) is heated to convert all the calcium to CaO and the final mass of CaO is 1.62 gms. The % by mass of CaCl_(2) in the original mixture is
Answer» `15.2 %` `32.1 %` `21.8 %` `11.07 %` SOLUTION :N/A

Discussion

91749.	""_(84)^(210) Po to ""_(82)^(206) Pb + ""_(2)^(4) He . In the reaction , the position of Po in the periodic table is ___ A group when lead belongs to IV A
Answer» Solution :`underset(VIA) (""_(84) PO^(210)) to underset(""_(82) Pb^(206)) + ""_(2)He^(4)`

Discussion

91750.	A 10% solution (by mass) of sucrose in water has freezing point of 269.15 K. Calculate the freezing point of 10% glucose solution in water, if freezing point of pure water is 273.15 K. Given : (Molar mass of sucrose = "342 g mol"^(-1), molar mass of glucose ="180 g mol"^(-1)).
Answer» Solution :`10%` solution in water by mass MENAS 10 g of sucrose is present in 100 g of the solution, i.e., Solute (sucrose) = 10 g, solvent (water) = 90 g `M_(2)=(1000K_(f)w_(2))/(w_(1)xxDeltaT_(f))` `"or"K_(f)=(M_(2)xxw_(1)xxDeltaT_(f))/(1000xxw_(2))=(342xx90xx(273.15-269.15))/(1000xx10)="12.312 K kg mol"^(-1)` `10%` of glucose solution in water means `w_(2)=10, w_(1)=90g` `DeltaT_(f)=(1000K_(f)w_(2))/(w_(1)M_(2))=(1000xx12.312xx10)/(90xx180)=7.6^(@)` `THEREFORE"FREEZING point of solution "= 273.15 - 7.6=265.55 K`

Discussion

Explore topic-wise InterviewSolutions in .

A 1.345-g sample of a compound of barium and oxygen was dissolved in hydrochloric acid to give a solution of barium ion, which was then precipitated with an excess of potassium chromate to give 2.012 g of barium chromate, BaCrO_4 . What is the formula of the compound?

A 1.2 g/1 solution of NACI is isotonic with 7.2 g/1 of solution of glucose. Calculate the van't Hoff's factor of NACI solution

8g O_(2) has same number of molecules as that in:

A 110% sample of oleum contains

8c,c,of gaseous hydrocarbon requires 40c.c. of O_(2) for complete combustion. Identify hydrocarbon.

A 1:1mixture of pentane and hexane is seperated by fractional distillation in the apparatus shown. At what temperature does the first drop of condensate appear on the thermometer?

A 10L cylinder of nitrogen at 4.0 atm pressure and 27^(@ C develped a leakage. When the leakage was repaired 2.36 atm of nitrogen remained in the cylinder still at 27^(@0C. How many grams of nitrogen escaped ?

._(89)Ac^(231) " given " ._(82)Pb^(207) after emission of some alpha and beta particles. The number of such alpha and beta-particles are respectively

A 10L flask at 300K contains a gaseous mixture of CO and CO_(2) at a total pressure of 2.0 atm. If 0.2 mol of CO is present, the partial pressure of CO_(2) is ( Use R = 0.082 atm mol^(-1) K^(-1) )

896 mL. of a mixture ofCO and CO_2 weigh 1.28 g at NTP. Calculate the volume of CO_2 in the mixture at NTP.

A 10gm mixture of glucose and urea present in 250 ml solutionshows the osmotic pressure of 7.4 atm at 27^(@)C . The precentage compostion of urea in the solution will be

._(88)X^(288) - 3alpha + beta rarr Y. The element Y is

A 10g mixture of iso-butane and iso-butene requires 20g of Br_2(in CCl_4) for complete addition. If 10g of the mixture is catalytically hydrogenated and the entire alkane is monobrominated in the presence of light at 127^(@)C, how much of it would be formed? (Atom weight of bromine=80)

""_88^@224Rahaving t_(1//2)= 3.64d emits an a particle to form ""_(86)^(220)Rn , which has t_(1//2) = 54.5 s. Given that the molar volume of radon under these conditions is 35.2 dm^3 ,what volume of radon is in secular equilibrium with 1 g of radium?

A 10cm^(3) sample of human urine was found to have 5 milligrams of urea on analysis. Calculate the molarity of given sample with respect to urea (Mol. Mass of urea - 60).

""_(88)^(226)Ra emits an o-particle and the daughter element is:

A 100W and 110V incandescent lamp is connected in series with an electrolytic cell containing CdSO_4 solution. What mass of cadmium will be deposited at the cathode after 4hrs of electricity.

8:8 coordination of CsCl is found to change into 6:6 coordination :

A 100ml solution of 0.1 N HCl was titrated with 0.2N NaOH solution. The titration was discontinued after adding 30ml of NaOH solution. The remaining titration was completed by adding 0.25NKOH solution. The volume of KOH required for completing the titration is

A 100cm^(3) solution of sodium carbonateis prepared by dissolving 8.653 g of the salt in water. The density of solution is 1.0816 g per millilitre. What are the molarity and molality of the solution ? (Atomic mass of Na is 23, of C is 12 and of O is 16).

8:8 coordination of CsCI is found to change into 6:6 coordination on:

A 100-watt, 110-volt incandescent lamp is connected in series with an electrolytic cell containing cadmium sulphate solution. What weight of cadmium will be deposited by the current flowing for 10 hours ?

87.5% decomposition of a radioactive substance completes in 3 hours. What is the half-life of that substance

A 100 W, 220 V incandescent lamp is connected in series with an electrolytic cell containing copper sulphate solution. What weight of copper will be deposited by 1 A current flowing for 5 hours? (at. Wt. of Cu=63.54).

8.7 g of pure MnO_(2) is heated with an excess of HCl and the gas. evolved is passed into a solution of KI. Calculate the amount of .he iodine liberated (Mn=55,Cl=35.5, I=127):

""_(86)^(226)Rato_(84)^(222)Rn+(4)/(2)He Order of reaction is..

A 100% pure sample of a divalent metal carbonate weighing 2 g on complete thermal decomposition releases 448 cc of carbon dioxide at STP. The equivalent mass of the metal is

""_(86)^(223) Ra is a member of actinium series . Which of the following nuclide belong to the same series ?

A 1.00 molar .aqueous solution of trichloroacetic acid ("CCl"_(3)"COOH") is heated to its boiling point. The solution has the boiling point of 100.18 ""^(@)C. Determine the Van't Hoff factor for trichloroacetic acid. (K_(b) "for water" = 0.512 "kg mol"^(-1)).

A 1.00 molal aqueous solution of trichloroacetic acid ("CCl"_3COOH)is heated to its boilingpoint. The solution has a boiling point of 100.18^@C . Determine the van't Hoff factor for trichloro acetic acid (K_b " for water" = 0.512 K kg "mol"^(-1) ) .

860 fine silver means:

A 100 mL solution of KOH contains 10 milliequivalents of KOH .Calculateits strength in normality and grams /litre .

""_(84)Po^(218)(t_(1//2)=3.05mins) decays to ""_(82)Pb^(214)(t_(1//2)=2.68min) by alpha-emission while Pb^(214) is a beta-emitter. In an experiment starting with 1g atom of pure Po_(218), how much time would be required for the number of nuclei of ""_(82)Pb^(214) to reach maximum?

A 100 ml solution of 0.1N-HCl was titrated with 0.2 N-NaOH solution. The titration was discontinued after adding 30 ml of NaOH solution. The remaining titration was completed by adding 0.25N-KOH solution. The volume of KOH required for completing the titration is

A 100 mL solution of 0.1 N HCI was titrated with 0.2 N NaOH solution. The titration was discontinued after adding NaOH solution.The remaining titration was completed by adding 0.25 N KOH solution. The volume of KOH required for completing the titration is

A 100 ml solution of 0.1 N HCl was titrated with 0.2 N NaOH solution. The titration was discontinued after adding 30 ml of NaOH solution. The remaining titration was completed by adding 0.25 N KOH solution. The volume of KOH required for completing the titration is

._(84)Po^(210) rarr ._(82)Pb^(206) + ._(2)He^(4). From the above equation, deduce the position of polonium in the periodic table (lead belongs to group IV A)

""_(84)Po^(210) decays with alpha-particle to ""_(82)Pb^(206) with a half -life of 138.4 days . If 1.0g of ""_(84)Po^(210) is placed in a sealed tube, how much helium will accumulate in 69.2 days . Express the answer in cm^(3) at STP

A 10.0 ml of (NH_(4))_(2) SO_(4) solution was treated with excess of NaOH. The ammonia evolved was absorbed in 50 ml of 0.1 N HCl. The excess HCl required 20 ml of 0.1 N. NaOH. Calculate the strength of (NH_(4))_(2) SO_(4) in the solution.

._(84)Pb^(219) is a member of actinium series. The other member of this series is

A 100 mL 0.1 M solution of ammonium acetate is diluted by adding 100 mL of water. The pH of the resulting solution will be (PK_a of acetic acid is nearly equal to pK_a of NH_4 OH)

._(84)^(218)Po(t_(1//2)=183 sec) decay to ._(82)Pb(t_(1//2)=161 sec) by alpha-emission, while Pb^(214) is a beta-emitter. In an experiment starting with 1 mole of pure Po^(218), how many time would be required for the number of nuclei of ._(82)^(214)Pb to reach maximum ?

._(84)^(218)Po (t_(1//2) = 183 sec) decay to ._(82)^(214)Pb (t_(1//2) = 161) sec by alpha emission, while ._(82)^(214)Pb decay by beta-emission. In how much time the number of nuclei of ._(82)^(214)Pb will reach to the maximum?

A 1.00- g sample of H_(2)O_(2)solution containing x percent H_(2)O_(2)by weightrequiresx mLof a KMnO_(4)solution for complete oxidation under acidic conditions .Calculate the normality of the KMnO_(4) solution .

""_(84)^(210)Po to ""_(82)^(206)Pb + ""_(2)^(4)He From the above nuclear reaction, predict the position of polonium in the periodic table (leads belongs to group IV A).

""_(84)^(210)Po decayswith emission of alpha-particle to ""_(82)^(206)Pb with a half-life period of 138.4 days. If 1g of ""_(84)^(210)Po is placed in a sealed tube, how much helium will be accumulated in 69.2 days ? Express the answer in cm^(3) at STP ?

A 10.0 g sample of a mixture of calcium chloride and sodium chloride is treated with Na_(2)CO_(3) to precipitate the calcium as calcium carbonate . This CaCO_(3) is heated to convert all the calcium to CaO and the final mass of CaO is 1.62 gms. The % by mass of CaCl_(2) in the original mixture is

""_(84)^(210) Po to ""_(82)^(206) Pb + ""_(2)^(4) He . In the reaction , the position of Po in the periodic table is ___ A group when lead belongs to IV A

A 10% solution (by mass) of sucrose in water has freezing point of 269.15 K. Calculate the freezing point of 10% glucose solution in water, if freezing point of pure water is 273.15 K. Given : (Molar mass of sucrose = "342 g mol"^(-1), molar mass of glucose ="180 g mol"^(-1)).