94420 + Interview Questions in Chemistry in Class 12 Page 1752 InterviewSolution

87551.	A sample of pure calcium weighing 1.35g was quantitatively converted to 1.88g of pure calcium oxide. Atomic mass of calcium would be:
Answer» 20 40 16 35.5 Answer :B

Discussion

87552.	A sample of potato starch was ground in a ball mill to give a starchlike molecule of lower molecular weight. The product analysed 0086% phosphorus. If each molecule is assumed to contain one atom of phosphorus, what is the molecular weight of the material?
Answer» ANSWER :`3.6 XX 10^(4)` AMU

Discussion

87553.	A sample of potato-starc was ground to give a starch like molecule. The product analysed 0.086% phosphorus, what is the average molecular mass of the material.
Answer» ANSWER :36000 AMU

Discussion

87554.	A sample of pitchblende is found to contain 50% uranium and 2.425% lead 93% of lead was Pb^(206) Isotope. If the disintegration constant is 1.52xx10^(-10)yr^(-1), how old could be the pitchblende deposit?
Answer» ANSWER :`3.3xx10^(8)` YRS.

Discussion

87555.	A sample of pitchblende is found to contain 50% minimum and 2.425% of lead. Of this lead only 93% was Pb^(26) isotope. If the disintegration constant is 1.52 xx 10^(-10) yr^(-1), how old could be the pitchblende deposidts?
Answer» SOLUTION :Mole of `U^(238) = (50)/(100 xx 238) = 2.1 xx 10(-3)` Mole of `Pb^(26) = (2.425)/(100) xx (93)/(100 xx 206) = 0.109 xx 10^(-3)` `N_(0) = (x + y) = 2.1 xx 10^-3) + 0.109 xx 10^(-3) = 2.209 x 10^(-3)` `N = x = 2.1 xx 10^(-3)` `LAMBDA = (2.303)/(t) log ((N_(0))/(N))` `1.52 xx 10^(-10) = (2.303)/(t) log (2.209 xx 10^(-3))/(2.1 x 10^(-3))` `t = 3.3 xx 10^(8)` YEARS

Discussion

87556.	A sample of phosphorus trichloride (PCl_(3)) contains 1.4 moles of the substances. How many atoms are there in the sample?
Answer» `4` `5.6` `8.431xx10^(23)` `3.373xx10^(24)` Solution :1 mole of `PCl_(3)` contains 4 MOLES (1+3) of atoms `:.` 1.4 moles of `PCl_(5)` contains 5.6 moles of atoms `:.` No. of atoms `= 5.6xx6.02xx10^(23)` `= 3.372xx10^(24)`

Discussion

87557.	A sample of phosphorus trichloride (PCl_(3)) contains 1.4 moles of the substance. How many atoms are there in the sample
Answer» 4 5.6 `8.431xx10^(23)` `3.372xx10^(24)` Solution :No. of atoms in ONE MOLECULE = no. of MOLES `xx6.022xx10^(23)=1.4xx6.022xx10^(23)=8.432xx10^(23)xx4=3.372xx10^(24)`

Discussion

87558.	A sample of phosphorus trichloride (PCl_(3))contains 1.4 moles of the substance. How many atoms are there in the sample?
Answer» 4 5.6v `8.431 xx 10^(23)` `3.372 xx 10^(24)` Solution :1 MOLE of `PCl_(3)` contains `4 xx 6.02 xx 10^(23)` atoms. `therefore 1.4` MOLES will CONTAIN `=3.372 xx 10^(24)` atoms.

Discussion

87559.	A sample of PCl_(3) contains 1.4 mole of the substance. How many atoms are there in the sample?
Answer» 4 5.6 `8.431xx10^(23)` `3.372xx10^(24)` ANSWER :D

Discussion

87560.	A sample of phosphorus trichloride (PCl_3)contains 1.4 moles of the substance. How many atoms are there in the sample ?
Answer» 4 5.6 `8.431 XX 10^23` `3.372 xx 10^24 ` SOLUTION :1 mol of `PCl_3` CONTAINS = `4 xx 6.02 xx 10^23` atoms 1.4 mol of `PCl_3` will CONTAIN ` = 4 xx 6.02 xx 10^23 xx 1.4 ` `= 3.372 xx 10^24` atoms

Discussion

87561.	A sample of O_(2) gas initially at NTP is transferred from a 1-litre container to a 2-litre container at a constant temperature. What effect does this change have on (a) the average kinetic energy of O_(2) molecules (b) the average speed of O_(2) molecules (c) the rms speed of O_(2) molecules (d) the total number of collisions of O_(2) molecules with the container waals in a unit time.
Answer» SOLUTION :(a) No change in average kinetic energy as the temperature remains constant. (b) & (c) No change in average speed and RMS speed as the average kinetic energy remains unchanged. (d) Number of collisions per unit time with the CONTAINER walls decrease, as due to increase in volume, the MOLECULES move longer distances between collisions.

Discussion

87562.	A sample of nitrogen occupies a volume of 350 cm^3 at STP. Then, its volume at 550 K and0.5 atm pressure is approximately
Answer» `1280 cm^3` `1409 cm^3` `1050 cm^3` `1428 cm^3` ANSWER :D

Discussion

87563.	A sampleof nitrogen gas is bubbled through liquid water at 25^(@)C and then collected in a volume of 750 cc. The total pressure of the gas which is saturated with water vapour pressure of water at this temperature is 24 mm. How many moles of nitrogen are in the sample ?
Answer» ANSWER :0.028

Discussion

87564.	A sample of NH_(3) gas is 20% dissociated into N_(2) and H_(2) gases. The mass ratio of N_(2) and NH_(3) gases in the final sample is
Answer» `(7)/(38)` `(7)/(17)` `(14)/(17)` `(21)/(17)` ANSWER :A

Discussion

87565.	A sample of natural gas is 85.2% methane, CH_4 , and 14.8% ethane, C_2H_6,by mass. What is the density of this mixture at 18^@Cand 748 mmHg?
Answer» SOLUTION :0.71 g/L

Discussion

87566.	A sample of NaOH weighing 0.40 is dissolved in water and the solution is made to 50.0 cm^(3) in volumetric flask. What is the molarity of the resulting solution?
Answer» ANSWER :0.2 M

Discussion

87567.	A sample of NaNO_(3) wwighing 0.38 g is placed in a 50.0 mL measuring flask. The flask is then filled with water upto the mark on the neck. What is the molarity of the solution?
Answer» SOLUTION :FORMULA mass of `NaNO_(3)=23+14+48=85` `0.38g NaNO_(3)=23+14+48=85` `0.38g NaNO_(3)=(0.38)/(85)mol=0.0045mol` `"MOLARITY of the solution "=(0.0045mol)/(0.050L)=0.090M.`

Discussion

87568.	A sample of NaClO_(3) is converted by heat to Na Cl with a lossof 0 . 16g of oxygen . Theresidue is dissolved in waterand precipitated as AgCl . Themass of AgCl (in g) obtained will be . . . (AgCl = 143 . 5 g mol^(-1))
Answer» `0 . 54` `0 . 4` `0 . 48` `0 . 35` Answer :C

Discussion

87569.	A sample of Na_(2)CO_(3).H_(2)O weighing 0.62 g is added to 100 ml of 0.1 N (NH_(4))_(2)SO_(4) solution. What will be the resulting solution
Answer» ACIDIC Neutral Basic NONE of these Answer :C

Discussion

87570.	A sample of mixture of CaCl_(2) and NaCl weighing 2.22 gm was treated to precipitate all the Ca as CaCO_(3) which was then heated and quantitatively converted to 0.84 gm of CaO. Calculate the percentage (by mass )of CaCl_(2) in the mixture.
Answer» ANSWER :`75%`

Discussion

87571.	A sampleof Mgwas burnt in air to give a mixture of MgO and Mg_(3)N_(2).The ash was disolvedin 60 m.e of HCl and the resulting solution back titrated with NaOH . 12 m.e of Naoh were required to reach the end point . An excess of NaOH was then added and the solution distilled .the ammonia released was then irapped in 10 m.e of second acid solution . Back titration of this solution required 6 m.e of the base . Calculate the percentage of Mg burnt to nitrate .
Answer» Solution :First Method : m.e method m.e of MgO + m.e of `Mg_(3)N_(2)` = m.e of HCL reacted = m.e of totalhCl - m.e of NaOH ` = 60 - 12 = 48` In the dissolution of ash , HCl reacts with total Mg in `Mg_(3)N_(2)` and in MgO and ALSO with N in `Mg_(3)N_(2)` ` :. ` m.e of total Mg + m.e of N = 48 or m.e of total mg + m.e of `NH_(3) = 48` m.e of total Mg = `48 - 4 - 44` Further , Mg converted to `Mg_(3)N_(2)` whose N converts to `NH_(4)CL( or NH_(3))` ` :. ` m.e of Mg converted to `Mg_(3)N_(2) = 3xx m.e of NH_(3)` ` = 3xx (10-6)` = 12 ` :. ` percentage of Mg converted to `Mg_(3)N_(2) = 12/44 xx 100` `= 27.27 % ` Second Method : MOLE method The reactions involved are The reactions involved are`{:(Mg to,MgO,MgO,+ 2HCl = MgCl_(2)+H_(2)O),(" x mmol(say)","x mmol"," x mmol",):}` `{:(Mg to,Mg_(3)N_(2),Mg_(3)N_(2)+ 8HCl ,= 3MgCl_(2),2NH_(4)Cl),(" y mmol (say)",y/3mmol,y/3mmol,,(2Y)/3mmol):}` ` :. 2x ` mmol of `hCl + (8y)/3` mmol of HCl = total mmol of HCl- mmol of NaOH `= 60 - 12 = 48` `2x + (8y)/3 = 48 ""...(1)` Further ,mmol of `NH_(4)Cl` = mmol of `NH_(3) = (10-60)` or `(2y)/3 = 4 "" ...(2)` From eqns . (1) and (2)one cancalculate x = 16 and y = 6 ` :. ` percentage of Mgconvertedto `Mg_(3)N_(2) = y/(x+y) = 27.27 % `

Discussion

87572.	A sample of mixed alkali contains NaOH and Na_2 CO_3 is titrated in the following two schemes. (1) 10 mL of above mixture requires 8 mL of 0.1 N HCl by using phenolphthalein. (ii) 10 mL of above mixture requires 10 mL 0.1 N HCI by using methyl orange. What will be the ratio of the weight of NaOH and Na2CO3 in the sample mixture?
Answer» `1: 132` `2.412` `2.122` `3.142 :1` Solution :Meqif HCIfor10mLofmixtureusingphenolphthleinindicator`= 8 XX 0.1 = 0.8` ` therefore` Meqof `NaOH+1/2 Meq ` of ` Na_2 CO_3 = 0.8` NowMeqof HCIusedfor 10 mLmixtureusingmethylorganeas INDICATOR`=10 xx 0.1 =1` Meq`of NaOH+MeqOf Na_2 CO_3 =1` ` therefore `BYeqn(i ) and (ii ) Meq of ` NaOH=0.6Meq` of`Na_2CO_3= 0.4` ` (w )/( 40) xx 1000 =0.6` ` (w)/( 106 //2) xx 1000 =0.4` WTOF`NaOH= 0.024 `wqof `Na_2 CO_3 = 0.0212` ratioof theweightof `Na_2CO_3= ( 0.024)/( 0.0212) = 1.132:1`

Discussion

87573.	A sample of magnesium was burnt in air to give a mixture of MgO and Mg_(3)N_(2). The ash was dissolved in 60 meq of HCl and the resulting solution back titrated with NaOH. 12 meq of NaOH were required to reach the end point. An excess of NaOH was then added and the solution distilled. The ammonia released was then trapped in 10 meq of second acid solution. Back titration of this solution required 6 meq of the base. Calculate the percentage of magnesium burnt to the nitride.
Answer» Solution :`MgO+2HCl rarr MgCl_(2)+H_(2)O` `Mg_(3)N_(2)+8HCl rarr 3MgCl_(2)+2NH_(4)Cl` `"12 meq of NaOH "-="12 meq of HCl"` `" i.e,HCl left unreacted = 12 meq"` `THEREFORE"HCl used up by MgO and "Mg_(3)N_(2)=60-12="48 meq = 48 MILLIMOLES"` Suppose in the mixture, there are x millimoles of MgO and y millimoles of `Mg_(3)N_(2)`. Then `2x+8y=48 or x+4y=24.` Further, `NH_(4)Cl+NaOH rarr NaCl+H_(2)O+NH_(3)` Acid used up by `NH_(3)=10-6 = "4 meq."` `therefore""NH_(3)" produced = 4 meq = 4 millimoles"` `"or"NH_(4)Cl" formed in reaction (ii) = 4 millimoles"` This will be formed from `Mg_(3)N_(2)=2`millimoles, i.e., y = 2. Hence,`""x+4xx2=24 or x=16` `""2Mg+O_(2)rarr 2MgO` `""3Mg+N_(2)rarrMg_(3)N_(2)` 16 millimoles of MgO are obtained from Mg = 16 millimoles 2 millimoles of `Mg_(3)N_(2)` are obtained from Mg = 6 millimoles Totl millimoles of Mg `=16+6=22` Hence, Mg CONVERTED to `Mg_(3)N_(2)=(6)/(22)xx100=27.27%`

Discussion

87574.	A sample of liquid in a thermally insulated container is stirred for 1 hr by a mechanical attachment to a motor in the surroundings, which of the following thermodynamic quantity for the system is zero
Answer» Work (W) Change in internal energy `(DeltaE)` Change in enthalpy `(DELTAH)` None of these Solution :For a thermally INSULATED container, `Deltaq=0` But work (W), change in internal energy `(DeltaE)` and change in enthalpy `(DeltaH)` are NON zero.

Discussion

87575.	A sample of lead containing Ag as impurity welghing 0.8g was dissolved in a small quantity of nitric acid to produce an aqueous solution of Pb^(2+) and Ag^*. The volume of the solution was increased to 250 ml by addition of water. A pure silver electrode was immersed in the solution and the potential difference between this electrode and a standard hydrogen Belectrode is 0.50V at 25^@C. Calculate the percentage of silverin the sample.
Answer» ANSWER :0.00175

Discussion

87576.	A sample of KCIO_(3)on decomposition yield 448 ml of oxygen gas at NTP calculate (i) weight of oxygen produced, (ii) weight of KCIO_(3)originally taken, and (iii) weight of KCI produced (K = 39, CI = 35.5 and O =16)
Answer» Solution : (i) Mole of oxygen = `(448)/( 22400) = 0.02"" ` (RULE3 ) Wt.of oxygen ` = 0 . 0 2 xx 32= 0 . 64 g ""` (Rule 1 ) (ii)`KClO_(3) toKCl+ O_(2)` ApplyingPOAC for O atoms, MOLES of O atomsin `KClO_(3) = ` moles of O atomsin `O_(2)` `3 xx` moles of `KCiO_(3) = 2 xx` moles of `O_(2)` (1 moles of `KCiO_(3)`contains 3 moles of O and 1 mole of `O_(2)` contains2 MOLESOF O) `3 xx (wt. of KClO_(3))/(underset(("Rule 1 "))(mol. wt . of KClO_(3))) = 2 xx ("vol . at NTP (litres )")/(underset(("Rule 3" ))(22.4))` `3 xx (w.t of KClO_(3))/( 122. 5) = 2 xx (0. 448)/( 22 . 4)` W.tof `KCiO_(3) = 1 . 634 g.` (iii)Again applyingPOAC for Katoms, moles of K atomsi `KClO_(3)` = moles of K atoms in KCl or`1 xx ` moles of `KClO_(3) = 1 xx` moles of KCl (1 mole of `KCiO_(3)` contains 1 moles of Kand 1 mole of KCl contains 1 moleof K) ` 1 xx (wt. of KClO_(3))/(mol. wt. of KClO_(3)) = 1 xx (wt . of KCl)/( mol. wt. of KCl)` ` (1 . 634)/( 122.5) = (wt . of KCl)/( 74 . 5 )` Wt . of KCl = 0 . 00037g

Discussion

87577.	A sample of KCIO_3 on decomposition yielded 448 mL of oxygen gas at STP, then the weight of KCIO_3 originally taken was
Answer» `0.815 G` `1.63 g` `3.27 g` `2.45 g` ANSWER :`B`

Discussion

87578.	A sample of iron ore, weighing 0.700 g, is dissolved in nitric acid. The solution is then, diluted with water, following with sufficient concentrated aqueous ammonia, to quantitative precipitation the iron as Fe(OH)_(3). The precipitate is filtered, ignited and weighed as Fe_(2)O_(3). If the mass of the ignited and dried preciipitate is 0.541 g, what is the mass percent of iron in the original iron ore sample (Fe=56)
Answer» `27.0 %` `48.1 %` `54.1%` `81.1 %` ANSWER :C

Discussion

87579.	A sample of impure cuprite, Cu_2Ocontains 66.6% copper. Calculate the percentage of pure Cu_2Oin the sample.
Answer» ANSWER :0.75

Discussion

87580.	A sample of hyrdazine sulphate (N_(2)H_(6)SO_(4)) was dissovled in 100ml 10ml of this solution was reacted with excess of FeCl_(3) solution and warmed to complete the reaction Ferrous ions formed required 20ml of (M)/(50) KMnO_(4) solution Given: 4Fe^(+3)+N_(2)H_(4)rarrN_(2)+4Fe^(+2)+4H^(+) MnO_(4)^(-)+5Fe^(+2)+8H^(+)rarrMn^(2+)+5Fe^(+3)+4H_(2)O The amount in gm of hydrazinc sulphate in one litre is:
Answer» 1.30gm 6.5gm 3.25gm 8.66gm Solution :`N_(kmNo_(4))=(20)/(1000)xx(1)/(50)` `n_(Fe^(+2))xxn_(KMnO_(4))` `=5xx(20)/(1000)xx(1)/(50)=0.002` `n_(N)_(2)H_(4))` is 10 ml `=(1)/(4) n_(Fe^(+2))=(0.002)/(4)` `n_(N_(2)H_(4))` is 1L`=(0.002)/(4)xx100=(0.2)/(4)` mass of `N_(2)H_(6)SO_(4)` in `1L=(0.2)/(4)xx130=6.5gm`

Discussion

87581.	A sample of hydrazine sulphate [N_(2)H_(6)SO_(4)] was dissolved in 100 mL water. 10 mL of this solution was treated with excess of FeCl_(3) Sol. Ferrous ions formed were estimated and it required 20 mL of M/50 KMnO_(4) solution in acidic medium. Fe^(+)+N_(2)H_(4)rarrN_(2)+Fe^(2+)+H^(+) MnO_(4^(-)) + Fe^(2+)+H^(+)rarrMN^(2+)+Fe^(3+)+H_(2)O (a) Write the balanced redox reactions. (b) Estimate the amount of hydrazine sulphate in one litre of Sol.
Answer» Solution :(a) Given `4Fe^(3+)+N_(2)H_(4)rarrN_(2)+4Fe^(2+)+4H^(+)` `MnO_(4^(-))+5Fe^(2+)+8H^(+)rarr Mn^(2+)+5Fe^(3+)+4H_(2)O` (b) In 10 mL solution , EQ. of `N_(2)H_(6)SO_(4)`=Eq. of `Fe^(2+)` = Eq. of `KMnO_(4)` `=20xx(1)/(50)xx5xx10^(-3)=2xx10^(-3)` v.f. of `N_(2)H_(6)SO_(4)=4` so, WEIGHT of `N_(2)H_(6)SO_(4)` in 1 L solution `=(2xx10^(-3)xx1000)/(4xx10)xx130=6.5 g.`

Discussion

87582.	A sample of hydrazine sulphate (N_(2)H_(6)SO_(4)) was dissolved in 100 ml of water, 10 ml of this solution was reached with excess of FeCl_(3) solution and warmed to complete the reaction. Fe^(2+) ions formed were estimated and it required 20 ml of M/50 KmNO_(4) solution. Estimate the amount of hydrazine sulphate in the sample. Given: 4Fe^(2+) + N_(2)H_(4) to N_(2) + 4Fe^(2+) + 4H^(-) MnO_(4)^(-) + 5Fe^(2+) + 8H^(+) to Mn^(2+) + 5Fe^(3+) + 4H_(2)O
Answer» ANSWER :`0.0007`

Discussion

87583.	A sample of hydrogen (in the form of atoms), is made to absorb white light. 52% of the hydrogen atoms got ionised. In order to calculate the ionisation energy of hydrogen from its absorption spectrum ("assuming the electrons that got ejected have" KE=0), it is possible by measuring the frequency of the:
Answer» line of SHORTEST wavelength line of LONGEST wavelength line of greatest intensity line of SMALLEST intensity Answer :1

Discussion

87584.	A sample of HI (9.30 x× 10^(-3) mol) was placed in an empty 2.00 L container at 1000 K. After equilibrium was reached, the concentration of I_(2) was 6.29 x× 10^(-4) M. Calculate the value of K_(c) at 1000 K fo the reaction H_(2)(g)+I_(2)(g)hArr2HI (g).
Answer» ANSWER :29

Discussion

87585.	A sample of He have as much atoms as there are letters in “ALLEN”. Then find the mass of He sample :-
Answer» 16 a.m.u 16 g.m 20 g.m 20 a.m.u Answer :A

Discussion

87586.	A sample of hard water with hardness 150 ppm is found to containCa(HCO_(3))_(2) and CaCl_(2). If one litre of the smaple on boiling produce 0.108 g CaCO_3), then its permanent hardness is
Answer» 42 ppm 32 ppm 15 ppm 10.8 ppm SOLUTION :TOTAL hardness of water sample = 150 ppm. 1 litre of water on BOILING produces 0.108 g `CaCO_(3)`. `rArr"1000 PARTS by weight of water contains 0.108 parts by weight of "CaCO_(3)`. `=108"parts by weight of "CaCO_(3)`. `rArr"The temporary hardness of water = 108 ppm."` `therefore"Permanent hardness of the water sample"` `=150-108="42 ppm."`

Discussion

87587.	A sampleof hardwaterwas foundto contain278 ppm of calciumion . Into 1.00 L of thiswater , 1.06g of Na_(2)CO_(3) was added . What is the newconcentrationof Ca^(2+)ionin ppm ? Densityof all solutionis 1.0g//cc and K_(sp)of CaCO_(3) is 4.5 xx10^(-9)
Answer» SOLUTION :`0.059 ` PPM of `CA^(2+)`

Discussion

87588.	A sample of hard water contains96 ppm of SO_(4)^(2-) and 183ppm of HCO_(3)^(-) from1000 kgof this water ? If 1000 kg of this water is treated with the amount of CaO calculated above ,what will be the concentration( in ppm)of residual Ca^(2+)ions? (AssumeCaCO_(3) to be completelyexchanged with hydrogen ions , in one litreof the treated water are completely exchanged with hydrogen ions, what will be its pH? (One ppm means one part of the substance in one million parts of water , weight/weight )
Answer» Solution :`Ca(HCO_(3))_(2)+CaO = 2CaCO_(3) +H_(2)O` Mole of CaO = mole of `Ca(HCO_(3))_(2) " in " 10^(6) ` g of solution ` = 1/2 xx " mole of " HCO_(3)^(-)` ` = 1/2 xx 183/61 = 1.5` As `CaCO_(3)` is assumedto be completely insoluble in WATER `Ca^(2+)`IONS left are , therefore , those associated only with `SO_(4)^(2-)` ion ( 96 ppm) For `CaSO_(4)` , we have mole of `Ca^(2+) //10^(6) g = " mole of " SO_(4)^(2-) //10^(6) g = 96/96 " mole/"10^(6) g ` ` = 1 " mole /"10^(6) g`40 ppm. Now ,assuming densityof solution to be 1 g/mL mole of `Ca^(2+)` is replaced by `H^(+)` `[H^(+)] = 2xx 10^(-3) M ` PH = `-log (2 xx 10^(-3)) = 2.7 `

Discussion

87589.	A sample of hard water contains 20 mg of Ca^(2+) ions per litre. How many milliequivalents of Na_(2)CO_(3) would be required to soften 1 litre of the sample ?
Answer» Solution :`underset(40G)(Ca^(2+))+underset(2xx23+12xx3xx16=106g)(Na_(2)CO_(3))rarrCaCO_(3)+2Na^(+)` `"40 g "Ca^(2+)" ions react with "Na_(2)CO_(3)=106g` `THEREFORE"20 mg, i.e., "20xx10^(-3)gCa^(2+)` ions will react with `Na_(2)CO_(3)=(106)/(40)xx20xx10^(-3)g=5.3xx10^(-2)g` `"Equivalent wt. of "Na_(2)CO_(3)=("Mol. wt.")/(2)=(2xx23+12+3xx16)/(2)=53` `therefore""5.3xx10^(-2)g Na_(2)CO_(3)=(5.3xx10^(-2))/(53)"g EQ."=10^(-3)"g eq. = 1 millieq."`

Discussion

87590.	A sample of H_(2)SO_(4) (density 1.787g/mL) is labelled as 86 %by weight . What is the molarity ofacid ? Whatvolume of acid has to be used to make 1 litre of 0.2M H_(2)SO_(4)
Answer» Solution :Volume of 100 GM `H_(2)SO_(4) " solution " = 100/(1.787) ` mL ` = 100/(1.787 xx 1000) ` Moles of `H_(2)SO_(4) " in " 100 gm(86)/(98) ` `M_(H_(2)SO_(4)) = 86/(98 xx 100/(1.787 xx 1000)) ` LET V volume of `H_(2)SO_(4)` used to prepare 1 litre of `0.2 MH_(2)SO_(4)` ` :.V xx 15.68 = 1000 xx 0.2rArrV = 12.75 ` mL

Discussion

87591.	A sample of H_(2)SO_(4) is labelled as 9.8% by weight . Its specific gravity is 1.8g//c c. Volume of this solution required to prepare 1000 ml of 0.18molar H_(2)SO_(4) solution is
Answer» 100 ml 40 ml 10 ml 50 ml Answer :A

Discussion

87592.	A sample of hard water contains 0.005 mole of CaCl_2per litre. What is the minimum concentration of Na_2SO_4which must be exceeded for removing the calcium ions from the water sample? Ksp (CaSO_4) = 2.4 xx 10^(-5)at 25^@C .
Answer» SOLUTION :0.0048 mole/L

Discussion

87593.	A sample of H_(2)O_(2) solution labelled as "28 volume" has density of 26.5 g/L. Mark the correct option(s) representing concentration of same solution in other units
Answer» `M_(H_(2)O_(2))=2.5` `% W/V=17` Mole fraction of `H_(2)O_(2)=0.2` m`m_(H_(2)O_(2))=13.88` Solution :`V_("strength"=56`' `:. M=28/11.2 =2.5` `:.` 1 L contain 2.5 moles of `H_(2)O_(2)` or `2.5xx34=85 g H_(2)O_(2)` wt. of 1 litre solution `=265 g""( :' d=265 g//L)` `:. w_(H_(2)O)=180 g` or number of moles of `H_(2)O=180/18=10` `x_(H_(2)O_(2))=2.5/(2.5+10)=0.2` `% w/v=(2.5xx34)/1000xx100=8.5` `m=2.5/180xx1000=13.88`.

Discussion

87594.	A sample of H-atoms in ground state absorb the electromagnetic radiation of x eV/quanta. Subsequently, the sample emit radiation of 15 different photon energies. The energy of emitted photon of longest wavelength is-
Answer» `(11x)/875 eV` `(35X)/36 eV` `x/36 eV` `(9X)/(384)` eV Answer :A

Discussion

87595.	A sample of gaseous hydrocarbon occupying 1.12 "litre" at NTP, when completely burnt in air produced 2.2 g CO_(2) and 1.8 g H_(2)O. Calculate the weight of hydrocarbon taken and the volume of O_(2) at NTP required for its combustion.
Answer» SOLUTION :(0.8g, 2.24 LITRES, `CH_(4)`)

Discussion

87596.	A sample of gaseous hydro carbon occupying 1 lit at NTP. When completely burnt in air produced 1.964g CO_(2) and 1.607 g H_(2)O. Calculate the volume of O_(2) (lit) at NTP required for its combustion.
Answer» Solution :Wt PF oxygen in `CO_(2)`, `W_(O)=(32)/(44) xx W_(CO_(2))=(32)/(44)xx1.964=1.4284` Wt. of oxygen in `H_(2)O`, `W_(O)=(16)/(18) xx W_(H_(2)O)=(16)/(18) xx 1.607=1.4284` Total Wt of oxygen = 2.857 G, `C_(STP)=(2.857g)/(32g) xx 22.4 l=2` LITRE

Discussion

87597.	A sample of gascous hydrocarbon occupying 11.2 lit at NTP. When completely burnt in air produced 22 g CO_(2) and 18 g H_(2)O. Calculate the weight of hydrocarbon taken (in grams)?
Answer» Solution :`W_(C)=(12)/(44) XX W_(CO_(2))=(12)/(44) xx 22=6 gr`, `W_(H)=(2)/(18) xx W_(H_(2)O)=(2)/(18) xx 18=2g` TOTAL WT 6+2=8g (11.2 ml), MW=16 (22.4 ml at STP)

Discussion

87598.	A sample of gas is to be identified by means of its behaviour in the presence of a glowing splint. Which of the following gases will neither itself burn nor cause the spling to burn ?
Answer» oxygen nitrogen hydrogen methane Solution :Nitrogen NEITHER itself burns nor cause the SPLINT to BURN because nitrogen is an inert gas and an inert gas does not undergo CHEMICAL reactions under a set of given conditions.

Discussion

87599.	A sample of gas is at 0^@C The temperature at which its rms speed of molecules will be doubled is :
Answer» `1103^@C` `273^@C` `723^@C` `819^@C` ANSWER :D

Discussion

87600.	A sample of gas has a molar volume od 10.1 L at a pressure of 745 mm Hg and a temperature of -138^(@)C. Is the gas behaving ideally ?
Answer» ANSWER :No

Discussion

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