94420 + Interview Questions in Chemistry in Class 12 Page 1836 InterviewSolution

91751.	8.4 ml of a gaseous hydrocarbon (A) was burnt with 50 ml of O_(2) in an eudiometer tube. The volume of the products after cooling to room temperature was 37.4 ml, when reacted with NaOH, the volume contracted to 3.8 ml. What is the molecular formula of A?
Answer» Solution :Let `C_(x)H_(y (g))` be the hydrocarbon. `C_(X)H_(Y(g))+ (x+(y)/(4))O_(2(g)) rarr xCo_(2(g))+(y)/(2)H_(2)O_(g)`[Remember this balanced combustion EQUATION for `C_(x)H_(y)`] From Gay Lussac’s Law of combining volume, we GET : 1 VOL. of `C_(x)H_(y) =(x+(y)/(4))` vol. of `O_(2)` = x vol. of `CO_(2) = (y)/(2)` vol. of `H_(2)O` Contraction in volume = `V_(R)- V_(P)` = (8.4 + 50) - (37.4) = 21 ml From equation, we have, the contraction `= 1 + (x + (y)/(4)) - (x + 0)` Contraction = 1 + `(y)/(4)implies`For 8.4 mL of `C_(x)H_(y)`, the contraction = 8.4 (1 + `(y)/(4)`) `8.4 (1 + (y)/(4)) = 21 implies y = 6` After treating with NAOH, there is a contraction of (37.4 - 3.8) = 33.6 ml, which is equal to the volume of `CO_(2)` produced. Volume of `CO_(2)` produced by 8.4 ml of hydrocarbon = `8.4 x implies 8.4 x = 33.6 implies x = 4` So the molecular formula of hydrocarbon is `C_(4)H_(6)`.

Discussion

91752.	A 10% solution (by mass) of sucrose in water has freezing point of 269.15 K. Calculate the freezing point of 10% glucose 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 :Molality of Glucose = `10/180 xx 1000/90` Molality of Sucrose ` = 10/342 xx 1000/90` `DeltaT_f` (Sucrose) = 273.15 - 269.15 = 4 K Applying the relation : `Delta T_f = K_f xx ` molality `Delta T_f` (Glucose) = `K_f`molality (Glucose)..(1) `Delta T_f` (Sucrose) = `K_f xx`molality (Sucrose)...(2) Dividing (1) by (2), we GET `(Delta T_f("GLOCOSE"))/(Delta T_f("sucrose")) = ("Molality (Glucose)")/("Molality (Sucrose) ")` Substituting the values, we have `(Delta T_f("glucose"))/(4) = (10 xx 1000)/(180 xx 90) = (342 xx 90)/(10 xx 1000) = 342/180` `Delta T_f ("glucose") = 342/180 xx 4 = 7.6` Freezing point of glucose solution = 273.15 - 7.6 = 265.55 K

Discussion

91753.	_83Bi^(2H)(t = 130 sec) decays to _81Tl^(207) by alpha-emission. In an experiment starting with 5 moles of _83Bi^(211). how much pressure would be developed in a 350 L closed vessel at 25 C after 760 sec? [Antilog (1.759) = 57 41]
Answer» 0.68 atm 0.22 atm 0.34 atm 0.54 atm Answer :C

Discussion

91754.	._(83)^(214)Bi decays to A by alpha -emission . A then decays to B by beta emission , which further decays to C by another beta emission . Element C decays to D by still another beta emission , and D deacays by alpha-emission to form a stable isotope E. What is element E?
Answer» `_(81)^(207)T1` `_(80)^(206)Hg` `_(79)^(206)Au` `_(82)^(206)Pb` Answer :D

Discussion

91755.	A 10% solution (by mass) of sucrose in water has a freezing point of 269.15 K. Calculatethe freezing point of 10% glucose in water if the 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 : MOLALITY of Glucose = `10/180 xx 1000/90` Molality of Sucrose ` = 10/342 xx 1000/90` ` Delta T_f`(Sucrose) = 273.15 - 269.15 = 4 K Applying the relation : `Delta T_f = K_f xx ` molality `Delta T_f`(Glucose) ` = K_f xx `molality (Glucose)...(1) ` Delta T_f` (Sucrose) ` = K_f xx `molality (Sucrose)..(2) Dividing (1) by (2), we GET ` (Delta T_f ("Glucose"))/(Delta T_f ("Sucrose") ) = ("Molality (Glucose)")/("Molality (Sucrose)")` Substituting the values, we have `(Delta T_f ("Glucose"))/(4) = (10 xx 1000)/(180 xx 90)xx(342 xx 90)/(10 xx 1000)= 342/180` ` Delta T_f ("Glucose") = 342/180 xx 4 = 7.6` Freezing POINT of Glucose solution = 273.15 - 7.6 = 265.55 K.

Discussion

91756.	8.2gm of sodium acetate is heated with soda line. What is the volume of methane liberated at S.T.P?
Answer» Solution :The reaction between SODIUM acetate and SODA lime is `CH_3COONa + NaOH underset(Delta)overset(CAO)to CH_4 + Na_2CO_3` 1 MOLE `CH_3 COONa` gives `to` 1 mole `CH_4` at STP 8.2 GM `CH_3COONa` (0.1 mole) give `to ?` `= 0.1 xx 22.4 "lit". = 2.24 "lit"`.

Discussion

91757.	A 1.0 molal aqueous solution of tricloroacetic acid (C C_(3)COOH) is heated to its boiling point. The solution has the boiling point 0f 100^(@)C. Detrmine van't Goff factor for trichoroaccetic acid (K_(b) "for water "= 0.512 "KKg mol"^(-1)).
Answer» SOLUTION :`DeltaT_("observed")=(100.18-100.0)=09.18^(@)C=0.18 K` `DeltaT_("calculated")=K_(b)xxm=(0.512" K KG MOL"^(-1))xx(0.1" mol kg"^(-1))=0.512 K` `(DeltaT_("observed"))/(DeltaT_("calculated"))=((0.18K))/((0.512 K))=0.351.`

Discussion

91758.	8.2 xx 10^(12) litres of water is available in a lake. A power reactor using the electrolysis of water in the lake produces electricity at the rate of 2 xx 10^6 Cs^(-1) at an appropriate voltage. How many years would it like to completely electrolyse the water in the lake. Assume that there is no loss of water except due to electrolysis.
Answer» Solution :Hydrolysis of water: At anode: `2H_2O to 4H^+ + O_2 + 4E^(-) "" …………..(1)` At cathode: `2H_2O + 2e^(-) to H_2 + 2OH^(-)` Overall reaction : `6H_2O to 4H^+ + 4OH^(-) + 2H_2 + O_2` (or) Equation `(1) + (2) xx 2 implies 2H_2O to 2H_2 + O_2` `:.` ACCORDING to Faraday.s Law of electrolysis, to electrolyse two mole of Water `(36g ~= 36 mL " of " H_2O)`, 4F charge is required alternatively, when 36 mL of water is electrolysed,the charge generated = `4 xx 96500 C`. `:.` When the whole water which is avialable on the lake is completely electrolysed the amount of charge generated is equal to `(4 xx 96500 C)/(36 mL) xx 9 xx 10^(12)L = (4 xx 96500 xx 9 xx 10^(12))/(36 xx 10^(-3)) C = 96500 xx 10^(15) C` `:.` Given that in 1 second, `2 xx 10^(6) C` is generated therefore, the TIME required to generate `96500 xx 10^(15) C is = (1S)/(2 xx 10^6C) xx 96500 xx 10^(15) C = 48250 xx 10^9S` `:. ` Number of years = `(48250 xx 10^9)/(365 xx 24 xx 60 xx 60) "1 year= 365 years "` `= 1.5299 xx 10^(6) "years" = 365 xx 24` hours `"" = 365 xx 24 xx 60` min `"" = 365 xx 24 xx 60 xx 60 `sec.

Discussion

91759.	A 10 mL of K_(2)Cr_(2)O_(7) solution , liberated iodine from KI solution .The liberatediodine was titrated by 16 mL of M/25 sodium thiosulphate solution. Calculate the concentration of K_(2)Cr_(2)O_(7) solution per litre .
Answer» Solution :`{:(Cr_(2)O_(7)^(2-) , +14H^(+)+6I^(-), =2Cr^(3+) +7H_(2)O+3I_(2)),(+ 12,,+6):}` `{:(S_(2)O_(3)^(2-), +1/2 I_(2) = 1/2 S_(4)O_(6)^(2-) ,+I^(-1)),(+4,,+5 ):}` EQ. wt of `K_(2)Cr_(2)O_(7) = (" MOL .wt")/( " change in ON per mole") ` ` = (294.18)/6` `= 49.03` m.e of 10 mL of `K_(2)Cr_(2)O_(7)` solution = m.e of iodine = m.e of sodium thiosulphate ` = 1/25xx 16 = 0.64` ` :. ` equivalent of 10 mL of `K_(2)Cr_(2)O_(7) ` solution ` = (0.64)/(1000) = 0.00064` ` :. ` weight per10 mL ` = 0.00064 xx 49.03` ` = 0.0313 `g ` :. ` concentration of `K_(2)Cr_(2)O_(7)` in grams per litre ` = 0.0313 xx 100` ` = 3.13 `g/L

Discussion

91760.	8.2 times 10^(12) litres of water is available in a lake. A power reactor using the electrolysis of water in the lake produces electricity at the rate of 2 times 10^(6)Cs^(-1) at an appropriate voltage. How many years would it like to completely electrolyse the water in the lake. Assume that there is no loss of water except due to electrolysis.
Answer» Solution :Hydrolysis of water At anode: `2H_(2)O rarr 4H^(+)+O_(2)+4e^(-) "...(1)"` At cathode: `2H_(2)O+2e^(-) rarr H_(2)+2OH^(-)` Overall REACTION `6H_(2)O rarr 4H^(+)+4OH^(-)+2H_(2)+O_(2)` `""`(or) Equation (1) + `(2) times 2 rArr 2H_(2)O rarr 2H_(2)+O_(2)` `therefore` According to Faradays Law of electrolysis, to electrolyse two mole of Water `(36gcong36mL " of "H_(2)O)`, 4F charge is required alternatively, when 36 mL of water is electrolysed, the charge generated `=4 times 96500 C`. `therefore` When the whole water which is available on the lake is completely electrolysed the amount of charge generated is equal to `""=(4 times 96500C)/(36mL) times 9 times 10^(12)L` `""=(4 times 96500 times 9 times 10^(12))/(36 times 10^(-3))C` `""=96500 times 10^(15)C` `therefore` Given that in 1 second, `2 times 10^(6)` C is generated therefore, the time required to generate `96500 times 10^(15)C " is " =(1S)/(2 times 10^(6) times C) times 96500 times 10^(15)C` `""=48250 times 10^(9)S` `therefore` NUMBER of years `=(48250 times 10^(9))/(365 times 24 times 60 times 60)` `""=1.5299 times 10^(6) years` 1 year = 365 days `""=365 times 24` hours `""=365 times 24 times 60` min `""=365 times 24 times 60 times 60` sec.

Discussion

91761.	81.4 g sample of ethyl alcohol contains 0.002 g of water. The amount of pure ethyl alcohol (to proper number of significant figures) is :
Answer» 81.398 g 81.40 g 81.4 g 81 g Solution :`81.4 - 0.002 = 81.398 " or " 81.4 `

Discussion

91762.	A 1.0-mg sample of technetium-99 has an activity of 1.7 xx 10^(-5) Cidecaying by beta -emission. What is the decay constant for ""_43^99Tc?
Answer» SOLUTION :`1.0 XX 10^(-13) //s `

Discussion

91763.	80g of oxygen contians as many atoms as in:
Answer» 10 G of hydrogen 5g of hydrogen 80g of hydrogen 1g of hydrogen Answer :B

Discussion

91764.	A 10 mg effervescent tablet containing sodium bicarbonate and oxalic acid releases 0.25 ml of CO_(2) at T = 298.15 K and p = 1 bar. If molar volume of CO_(2) is 25.0 L under such condition, what is the percentage of sodium bicarbonate in each tablet ? [Molar mass of NaHCO_(3) = 84 g mol^(-1)]
Answer» `16.8` `8.4` `0.84` `33.6` ANSWER :B

Discussion

91765.	8.0575 xx 10^(-2) of Glauber's salt is dissolved in water to obtain 1 dm^3of a solutionof density 1077.2 kg m^(-3). Calculate the molarity, molality and mole fraction of Na_2SO_4in the solution.
Answer» Solution :`1dm^3 = 1L, Kg m^(-3)= GL^(-1)` 0.5674 M, 0.5693 m, 0.01

Discussion

91766.	A 1.0 M solution of Cd^(+2) is added to excess iron and the system is allowed to reach equilibrium . What is the concentration of Cd^(+2) ? Cd^(2+)(aq) + Fe(s) to Cd(s) + Fe^(2+) (aq) , E^(0) = 0.037
Answer» `0.195` `0.097` `0.053` `0.145` Solution :`CD^(+) + Fe to Fe^(+2) + Cd , E=E^(0) -(0.0591)/(N) LOG Q` ` 0=0.037 - (0.0591)/(2) log""([Fe^(+2)])/([Cd^(+2)]) ,log""([Fe^(+2)])/([Cd^(+2)]) = (0.037 xx 2)/( 0.0591) = 1.25` `log""((y)/(1-y)) = 1.252 , ((y)/(1-y)) = 10 xx 1.79 , (y)/(1-y) =17.9` `y = 17.9 - 17.9 y , y+ 17.9 y = 17.9 , y=(17.9)/(18.9) = 0.947 , (1-y) = 1- 0.947 = 0.0529 =0.053`

Discussion

91767.	800mL 0.l(M) HCl solution is mixed with 200mL 0.5(M) CH_3NH_2 solution. In the resulting solution , concentration of H_3O^(+) ions is [K_b(CH_3NH_2)=5xx10^(-4)] is
Answer» `3xx10^(-5)(M)` `1.25xx10^(-4)(M)` `8xx10^(-11)(M)` `7.2xx10^(-10)(M)` ANSWER :C

Discussion

91768.	800 ml of 0.1 MH_(2) SO_(4) is mixed with 200ml of 0.8 MH_(2) SO_(4) The molarity of the mixture is
Answer» 0.32 0.24 0.12 0.48 Solution :`M_("MIX")=(M_(1) V_(1) + M_(2) V_(2))/(V_(1) + V_(2))` `((0.1 XX 0.8)+(0.2 xx 0.8))/(0.8 + 0.2)= 0.24 M `

Discussion

91769.	A 10 L flask contains 0.2 mole of CH_4 and 0.3 mole of hydrogen at 25^@ C ansd which makes non- reacting gaseous mixture. Then, the total pressure inside the flask is
Answer» `1.22 ATM` 0.5 atm 0.61 atm 2.20 atm Answer :A

Discussion

91770.	A 1.0 g sample of substance A at 100^(@)C is added to 100 ml of H_(2)O at 25^(@)C. Using separate 100 ml portions of H_(2)O the procedure is repeated with substance B and then with substance C. How will the final temperature of the water compare{:("Substance","Specific Heat"),(A,0.60 Jg^(-1^(@))C^(-1)),(B,0.40 Jg^(-1^(@))C^(-1)),(C,0.20 Jg^(-1^(@))C^(-1)):}
Answer» `T_(C)gtT_(B)gtT_(A)` `T_(B)gtT_(A)gtT_(C)` `T_(A)gtT_(B)gtT_(C)` `T_(A)=T_(B)=T_(C)` ANSWER :C

Discussion

91771.	80 gm of SO_(X) gas occupies 14 litre at 2atm & 273K. The value of X is :
Answer» ANSWER :2

Discussion

91772.	A 1.0 g sample of Fe_(2)o_(3) solid of 55.2 % purity is dissolved in acid and reduced by heating the solution with zinc dust. The resultant solution is cooled and made upto 100.0 mL. An aliquot of 25.0 mL of this solution requires 17.0 mL of 0.0167 M solution of an oxidant for titration. Calculate the number of moles of electrons taken up by the oxidant in the reaction of the above titration.
Answer» ANSWER :`6.07~~6`

Discussion

91773.	8.0 moles of SO_2and 4.0 moles of O_2are mixed in a closed vessel. The reactionproceeds at constant temperature. By the moment equilibrium sets in, 80% of the initial amount of SO_2enters the reaction. Determine the pressure of the gas mixture in equilibrium if the initial pressure was 2.96 atm.
Answer» SOLUTION :2.17 ATM

Discussion

91774.	A 1.0 g sample of Fe_(2)O_(3) solid of 55.2% purity is dissolved in acid and reduced by heating the solution with zinc dust. The resultant solution is cooled and made up to 100.0 mL. An aliquot of 25.0 mL of this solution requires 17.0 mL of 0.0167 M solution of an oxidant for titration. Calculate the number of electrons taken up by the oxidant in the reaction of the above titration.
Answer» Solution :The redox changes are : For reduction of `Fe_(2)O_(3)` by zinc dust `Fe_(2)^(6+) + 2e^(-) rarr 2 FE^(2+)` `Fe^(2+) rarr Fe^(3+) + e^(-)` oxidant + ne rarr reductant Meq. of `Fe_(2)O_(3)` in25 mL = Meq. of `Fe^(2+)` formed = Meq. of oxidant used to oxidize `Fe^(2+)` again Meq. of `Fe_(2)O_(3)` in 25 mL = Meq. of oxidant = `17 XX 0.0167 xx n` Where n is the number of ELECTRONS gained by 1 molecule of oxidant Meq. of `Fe_(2)O_(3)`in 100 mL = `17 xx 0.0167 xx n xx 100/25` `:. 1 xx 55.2 xx 1000/(100 xx M/2) = 17 xx 0.0167 xx n xx 4` Molecule wt. of `Fe_(2)O_(3)` = 160 `n = 1 xx 55.2 xx 2 xx 1000/(100 xx 160 xx 17 xx 0.0167 xx 4) = 6` Hence, number of electrons gained by one molecule of oxidant = 6

Discussion

91775.	A 1 g sample of Fe_(2)O_(3) solid of 55.2% purity is dissolved in acid and reduced by heating the solution with zinc dust. The resultant solution is cooled and made upto 100 mL. An aliquot of 25 mL of this solution requires 17 mL of 0.0167M solution of an oxidant for titration. Calculate no.of electrons taken up by oxidant in the above titration.
Answer» Solution :Weight of `Fe_(2)O_(3) = 0.552 g ` Number of eq. OD `Fe_(2)O_(3) = (0.552)/80 {:(,Fe_(2)O_(3)to,2FeO),("+6",," +4"),(eq. wt . of ,Fe_(2)O_(3)= 160/2,=80):}` Let the number of electrons taken upby the oxidant in the reaction be n ( i..ethe CHANGE in oxidation number ) ` :. ` NORMALITYOF the oxidant = 0.0167 n N ` :. ` m.e of 25 mL of `Fe^(2+)`solution= 0.0167 n xx 17 ` :. ` m.eof 100 mLof `Fe^(2+)`solution= `(68 xx 0.0167 n)/(1000)` eq . of `Fe_(2)O_(3) ` = eq of FEO `(0.552)/80 = (68 xx 0.167n)/1000 n = 6`

Discussion

91776.	80 g of oxygen contains as many atoms as in
Answer» 10 g of hydrogen 5 g of hydrogen 80 g of hydrogen 1 g of hydrogen Solution :No . Ofmole of oxygen `=(80)/(16)=5`mol 1 mol contains avogadro no . Atom i.e., `6.02xx10^(23)` atoms. `therefore` 5 mol contains `6.02xx10^(23)XX5`atom `=3.1xx10^(23)` SIMILARLY in option (b) No . Of mol of hydrogen `=(5)/(1)=5` mol 5 mol of hydrogen contains `3.1xx10^(23)` atoms.

Discussion

91777.	A 10 g sample of a mixture of calcium chloride and sodium chloride is treated with Na_(2)CO_(3) to precipitate calcium as calcium carbonate. This CaCO_(3) is heated to convert all the calcium to CaO and the final mass of CaO IS 1.12 gm. Calculate % by massof NaCl in the original mixture.
Answer» ANSWER :`%NACL = 77.8%`

Discussion

91778.	8 mole of a gas AB_3 are introduced into a 1.0 dm^3 vessel. It dissociates as, 2AB_3 (g) ⇌ A_2 (g) + 3B_2 (g) At equilibrium, 2 mole of A_2 are found ot be present. What is the equilibrium constant of rection ?
Answer» 2 3 27 36 Answer :C

Discussion

91779.	A 1.0 g of substance of molecular formula AB_(2)when dissoved in 25 g of benzene reduced the freezing point by 1.25^(@)C, Determine the atomic mass of A and B (K_(f) for benzene= 5.1 kg mol^(-1))
Answer» Solution :`M_((AB_(2)))=(K_(F)xxW)/(DeltaT_(f)xxW)=((5.1"kg MOL"^(-1))xx(1g))/((1.6K)xx(0.025kg))=127.5" g mol"^(-1)` `M_((AB_(3)))=(K_(f)xxW)/(DeltaT_(f)xxW)=((5.1"kg mol"^(-1))xx(1g))/((1.25K)xx(0.025kg))=163.2" g mol"^(-1)` LET the atomic mass of ELEMENT A = a Let the atimic mass of element B = b `therefore""a+2b=127.5` a+3b=163.2 On subracting eqn. (i) FORM eqn. (ii) `3b-2b=163.2-127.5=35.7 or b = 35.7` `a+bxx(35.7)=127.5 or a=127.5-71.14=56.36` `DeltaT_(f)=0.201^(@)C=0.201 K, K_(f)=1.86" K kg mol"^(-1), m=0.10"mol"^(-1)kg`

Discussion

91780.	8 gm of CH_(4) is completely burnt in air. The number of moles of water produced are
Answer» 0.5 1 2 18 Solution :16GM `CH_(4)` PRODUCED 2 mol water therefore 8GM will PRODUCE 1 mol of water.

Discussion

91781.	A 10 g samle of only CuS and Cu_(2)S was treated with 100 ml of 1.25 M K_(2) Cr_(2) O_(7).The Products obtained were Cr^(3+) , Cu^(2+) " and " SO_(2) . The excess oxidant was reacted with 50 ml of Fe^(2+) Solution : 25 ml of the same Fe^(2+) solution required 0.875 M KMnO_(4) under acidic condition , the volume of KMnO_(4) used was 20 ml . Find the % of CuS and Cu_(2)S in the sample .
Answer» Solution :EQUIVALENTS of DICHROMATE initially` = (1.25 xx 6 xx 100)/1000 = 0.75` Equivalents of `Fe^(2+) ` in 25 ml` = (0.875 xx 5 xx 5 xx 20)/1000 = 0.0875` Equivalents ofexcess dichromate = `0.175` ` :. ` Equivalents of dichromate consumed by `(CuS" and " Cu_(2)S)` ` = 0.75 - 0.175 = 0.575` Ifx G is the mass of CuS, the mass of `Cu_(2)S " is " (10 - x) ` g `x/(95.5) xx 6 + ((10 -x))/(159) xx 8 = 0.575` ` :. x = 5.74 ` gm ` %CuS = (5.74)/10 xx 100 = 57.4 %` ` % Cu_(2)S = 42.6 %`

Discussion

91782.	8 g of the radioactive isotope, cesium-137 were collected on Feb. 1 and kept in a sealed tube. On July 1, it was found that only 0.25 g of it remained. So the half life period of the isotope is:
Answer» 37.5 days 30 days 25 days 50 days. Answer :B

Discussion

91783.	8 g of HBr is added in 100 g of H_(2)The freezing point will be (K_(f) = 1.86, H=1, br = 80 )
Answer» `-0.75^(@)C` `0^(@) ` `-3.67 ^(@) C` `-7.6 ^(@) C ` Solution :`Delta T_(F) = iK_(f) xx (W_(2) xx 1000)/(M_(2) xx W_(1)) = 2 xx 1.86 xx (8 xx 100)/(81 xx 100)=3.67^(@)` `T_(f) = T^(@) - Delta T_(f) = -3.67^(@) C`

Discussion

91784.	(a) 10 g of non-electrolyte solute dissolved in 50 g of benzene lowered the freexing point of benzene by 0.4 K. Find the molar mass of the solute.[Given : Freezing point depression constant of benzene = 5.12 K. kg mor^(-1)] (b) How solubility of a gas in liquid varies with (i) Temperature and (ii) pressure?
Answer» Solution :(a) `M_2 = (k_f xx w_2 xx 1000)/(DELTA T_f xx w_1)` `M_2 = (5.12 k KG mol^(-1) xx 1.0 g xx 1000 g kg^(-1))/(0.4 xx 50 g)` `M_2 = 256 g mol^(-1)`. (B)(i) Solubility decreases with increases in temperature. OR Solubility increases with DECREASE in temperature. OR Solubility varies INVERSELY with temperature. (ii) Solubility increases with increases in pressure. OR Solubility decreases with decrease in pressure. OR Solubility varies directly in pressure OR Solubility varies directly with pressure.

Discussion

91785.	8 g of a radioactive susbtance is reduced to 0.5 g after 1 hours. The t_(1//2) of the radioactive substance is
Answer» 15 min 30 min 45 min 10 min Solution :`N_(0) = 8g, N = 0.5 g and t = 1 HR. = 60` min. Find `t_(1//2)` by `t = (2.303 XX t_(1//2))/(0.693) "log" (N_(0))/(N)`

Discussion

91786.	A 10 g mixture of Cu_(2)S and CuS was treated with 200 ml of 0.75 M KMnO_(4) in acid solution producing SO_(2),Cu^(2+) and Mn^(2+). The SO_(2) was boiled off and the excess of KMnO_(3) was treated with 175 ml of 1 M Fe^(2+) solution. Calculate % of CuS in the mixture.
Answer» ANSWER :`0.0058`

Discussion

91787.	8 g O_(2) has the same number of molecules as
Answer» 7 g CO `14gN_(2)` `11gCO_(2)` `16gSO_(2)` SOLUTION :`8gO_(2)=(8)/(32)mol=0.25mol.` `7gCO=(7)/(28)=0.25mol,` `14gN_(2)=(14)/(28)=0.5mol.` `11gCO_(2)=(11)/(44)=0.25mol.` `16gSO_(2)=(16)/(64)=0.25mol.` Equal moles contain equal number of MOLECULES.

Discussion

91788.	A 10% aqueous solution of cane sugar (molecular mass 342) is isotonic with 1.754% aqueous solution of urea. Find the molecular mass of urea.
Answer» Solution :`pi_("glucose")=(W_(B)xxRxxT)/(M_(B)xxV)=((10g)xxRxxT)/((342" g mol"^(-1))xxRxxT)` `pi_("urea")=(W_(B)xxRxxT)/(M_(B)xxV)=((1.754g)xxRxxT)/(M_(B)xxRxxT)` `"SINCE solutions area ISOTONIC," pi_("glucose")=P_("urea")` `((10g))/((342" g mol"^(-1)))=((1.754 g))/(MB)or M_(B)=((1.754g)XX(342"g mol"^(-1)))/((10g))` `=59.98~~60" g mol"^(-1)`.

Discussion

91789.	8 g O_2has same number of atoms as that in :-
Answer» 14G CO 7g CO 11g `CO_2` 22 G `CO_2` Solution :`8gO_2 =8/32xx2` MOL O =0.5 mol O atoms 7 g CO = `7/28xx2` mol =0.5 mol atoms

Discussion

91790.	A 1% (wt/ wt) solution of KCl (I) , NaCl(II) , BaCl_(2) (III) and urea (IV) have their osmotic pressures at the same temperature in the ascending order (molar masses of NaCl , KCl , BaCl_(2) and urea are 58.5 , 74.5 , 208.4 and 60 g mol^(-1) ) assuming 100 % ionisation of the electrolytes at this temperature :
Answer» `I LT III lt II lt IV ` `III lt I lt II lt IV ` `I lt II lt III lt IV` `III lt IV lt I lt II` ANSWER :D

Discussion

91791.	8 : 8 coordination is noticed in:
Answer» MGO `AI_2O_3` CsCI All Answer :C

Discussion

91792.	A 1% solution of phenol in water is used as an antiseptic
Answer» ANSWER :F

Discussion

91793.	7g of MgCO_3 after reaction with 14 g solution of H_2SO_4 left 0.7 g of MgCO_3 unreacted calculate the percentage strength of H_2SO_4.
Answer» ANSWER :`52.5%` (w/w)

Discussion

91794.	(A) 1-propanol & 2-propanol can be distinguished by haloform test(R) The trihalomethyl group (-CX_3)ofCH_3- oversetunderset(\|\|)(O)(C) - CX_3is a good leaving group for nucleophilic substitution reaction
Answer» Both (A) and (R) are TRUE and (R) is the CORRECT EXPLANATION of (A) Both (A) and (R) are true and (R) is not the correct explanation of (A) (A) is true but (R) is FALSE Both (A) and (R) are false Answer :B

Discussion

91795.	._(7)^(7)Li + _(1)^(1)P rarr X, Identify X if reaction is (p,alpha) type.
Answer» `._(4)^(8)Be` `._(2)^(4)He` `._(0)^(0)GAMMA` none of these Answer :B

Discussion

91796.	(A) 1 mole Ra^226 on alpha disintegration will give 44.8 lit of gaseous product at S.T.P (R) 1 mole Ra^226on alpha disintegration will give 1 mole radon and one mole helium
Answer» Both (A) and (R) are TRUE and (R) is the CORRECT EXPLANATION of (A) Both (A) and (R) are true and (R) is not the correct explanation of (A) (A) is true but (R) is false Both (A) and (R) are false Answer :A

Discussion

91797.	75% of the first order reaction was completed in 32 minutes. When was 50% of the reaction completed
Answer» 24 minutes 16 minutes 8minutes 4 minutes Solution :`N = (25)/(100) N_(0)` (at t = 32 minutes) THUS, `t = (2.303)/(0.693) XX t_(1//2) "LOG"(N_(0))/(N)`

Discussion

91798.	(A) 1 mole oxygen and N_(2) have same volume at same temperature pressure. (R) 1 mole gas at NTP occupies 22.4 litre volume at STP.
Answer» If both (A) and (R) are CORRECT and (R) is the correct EXPLANATION of (A). If both (A) and (R) are correct but (R) is not the correct explanation of (A). If (A) is correct but (R) is WRONG. If (A) is wrong but (R) is correct. Answer :B

Discussion

91799.	75% of the first order reaction was completed in 32 min. 50% of the reaction was completed in
Answer» 24 MIN. 8 min. 16 min. 4 min. Solution :( c ) : `75%` of REACTION is completed in two half-lives, i.e., `2xxt_(1//2)=32" min"` or `t_(1//2)=16" min"`

Discussion

91800.	A 1 M solution of glucose reaches dissociation equilibrium according to equation given below 6HCHO hArrC_(6)H_(12)O_(6). What is the concentration of HCHO at equilibrium of equilibrium constant is 6xx10^(22)
Answer» `1.6xx10^(-8)M` `3.2xx10^(-6)M` `3.2xx10^(-4)M` `1.6xx10^(-4)M` Solution :`6HCHO hArrH_(6)H_(12)O_(6)` FORWARD reaction `C_(6)H_(12)O_(6)overset(K_(1))hArr6HCHO` backward reaction `K_(2)=[(1)/(K_(1))]^(1//6),K_(2)=[(1)/(6xx10^(22))]^(1//6)` `K_(2)=1.6xx10^(-4)M`

Discussion

Explore topic-wise InterviewSolutions in .

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A 10% solution (by mass) of sucrose in water has freezing point of 269.15 K. Calculate the freezing point of 10% glucose 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) ]

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A 10% solution (by mass) of sucrose in water has a freezing point of 269.15 K. Calculatethe freezing point of 10% glucose in water if the 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) ]

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80 g of oxygen contains as many atoms as in

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7g of MgCO_3 after reaction with 14 g solution of H_2SO_4 left 0.7 g of MgCO_3 unreacted calculate the percentage strength of H_2SO_4.

(A) 1-propanol & 2-propanol can be distinguished by haloform test(R) The trihalomethyl group (-CX_3)ofCH_3- oversetunderset(||)(O)(C) - CX_3is a good leaving group for nucleophilic substitution reaction

._(7)^(7)Li + _(1)^(1)P rarr X, Identify X if reaction is (p,alpha) type.

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75% of the first order reaction was completed in 32 minutes. When was 50% of the reaction completed

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75% of the first order reaction was completed in 32 min. 50% of the reaction was completed in

A 1 M solution of glucose reaches dissociation equilibrium according to equation given below 6HCHO hArrC_(6)H_(12)O_(6). What is the concentration of HCHO at equilibrium of equilibrium constant is 6xx10^(22)