52671 + Interview Questions in Chemistry in Class 11 Page 1017 InterviewSolution

50801.	(A) .^(20)Ne and .^(22)Ne are isotones. (R) Noble gases do not exist as isotopes as they are not reactive.
Answer» Both (R) and (A) are true and reason is the CORRECT EXPLANATION of assertion. Both (R) and (A) are true but reason is not correct explanation of assertion Assertion (A) is true but reason (R) is false Assertion (A) and reason (R) both are false Answer :D

Discussion

50802.	A 20g chunk of dry ice is placed in an empty 0.75 litre wire bottle tightly closed what would be the final pressure in the bottle after all CO_2 has been evaporated and temperature reaches to 25^(@)C?
Answer» <P> SOLUTION :Pressure INSIDE the BOTTLE `= P+` atm pressure `=14.828+1=15.828atm`

Discussion

50803.	A 2.0g mixture of Na_(2)CO_(3) and NaHCO_(3) suffered a loss of 0.12g on heating. Percentage of Na_(2)CO_(3) in the mixture-
Answer» 83.8 16.2 38.8 61.2 Answer :A

Discussion

50804.	A 200 mL solution of I_(2) divided into two unequal parts. Part I reacts with pypo solution in acidic medium and required 8 " mL of " 2 M hypo solution for complete neutralisation. Part II was added with 300 " mL of " 0.1 M NaOH solution residual base required 30 " mL of " 0.1 M H_(2)SO_(4) solution for complete neutralisation. Calculate the value of 20 times the initiall concentration I_(2)?
Answer» 1 2 3 4 Solution :First part: mmoles of `Na_(2)S_(2)O_(3)` used `=8xx2xx1 ("n-factor")=16` `I_(2)+2Na_(2)S_(2)O_(3)to2NaI+Na_(2)S_(4)O_(6)` m" mol of "`I_(2)` used `=(1)/(2)` m" mol of "`Na_(2)S_(2)O_(3)=(16)/(2)=8`..(i) Second Part: `3I_(2)+6NaOHto5NaI+NaIO_(3)+3H_(2)O` mmoles of `H_(2)SO_(4)=` excess `NaOH=30xx0.1xx2` `xx("n-factor")=6` m" mol of "total `NaOH=300xx0.1xx1("n-factor")=30` m" mol of "NaOH `used=30-6=24` m" mol of "`I_(2)` used `=(1)/(2)` m" mol of "NaOH used `=(24)/(2)=12 m" mol of "I_(2) used` Total m" mol of "`I_(2)` used `=` part I`+` part II `=8+12=20mmol` `M of I_(2)=(mmol)/(V_(mL))=(20)/(200)=0.1M` 20 TIMES the initial `M_(I_(2))=0.1xx20=2`

Discussion

50805.	A 200 mL sample of a gaseous hydrocarbons has a density of 2.52gL^(-1) at 55^(@)C and 720mm Hg. What Is its formula?
Answer» `C_(2)H_(6)` `C_(4)H_(10)` `C_(5)H_(12)` `C_(6)H_(6)` Answer :C

Discussion

50806.	A 200 mL sample of a citrus fruit drinks containing ascorbic acid(vitamin C, mol. We 176.13) was acidified with H_(2)SO_(4) and 10mLof 0.250M I_(2) was added. Some of the iodine was reduced by theascorbicacid to I^(-). The excess of I_(2)required 4.6mL of 0.01M Na_(2)S_(2)O_(3) for reduction. What was the vitaminC contentof the drink in mg vitamin per mL drink? The reactions are: C_(6)H_(8) O_(6) + I_(2) rarrC_(6) H_(6) O_(6) + 2HI 5H_(2)O + S_(2)O_(3)^(2-) + 4I_(2) rarr 2SO_(4)^(2-) + 8I^(-) + 10H^(-)
Answer» ANSWER :`0.058mg//mL`

Discussion

50807.	A 2.00 litre evacuated container has a mass of 1050.0g. When the container is filled with an unknown gas at 800mm Hg pressure and 25.0^(@)C the mass is 1052.4g. What is the molar mass of the gas (in g"mol"^(-1))?
Answer» 28 31 54 56 Answer :a

Discussion

50808.	A 20 mL sample of a Ba(OH)_(2) solution is titrated with 0.245 M HCl. If 27.15 mL of HCl is required, what is the molarity of the Ba(OH)_(2) solution ?
Answer» 0.166 M 0.180 M 0.333 M 0.666 M Solution :N//A

Discussion

50809.	A 20 mL mixture of ethane, ethylene, and CO_2 is heated with O_2. After the explosion, there was a contraction of 28 mL and after treatment with KOH, there was a further contraction of 30 mL. What is the composition of the mixture?
Answer» <P> Solution :Let the volume of ethane, ethylene, and `CO_(2)`, respectively, are a,b and `(20-a-b)mL` Now contraction after cooling`=28mL` `thereforeV_(R)-V_(P)=25` `V_(R)=` Volume of `C_(2)H_(6)+` volume of `C_(2)H_(4)+` volume of `CO_(2)+` volume of `O_(2)` used for combustion `V_(P)=` volume of `CO_(2)` produced (Volume of `H_(2)O` is not taken SINCE it condenses) Considering the combustion GASES. (a). `C_(2)H_(6)+(7)/(2)O_(2)to2CO_(2)+3H_(2)O` (b). `C_(2)H_(4)+3O_(2)to2CO_(2)+2H_(2)O` (c). `CO_(2)to` no REACTION `V_(R)=(a+(7)/(2)a)+(b+3b)+(20-a-b)` `V_(P)=(2a+2b)+(20-a-b)` `V_(P)-V_(R)=(5)/(2)a+2b=28` `therefore5a+4b=56`..(i) There is a further contraction of 32 mL on treatment with KOH. Volume of `CO_(2)` produced `+` Volume of `CO_(2)` original `=32` `(2a+2b)+(20-a-b)=32` `therefore a+b=12`...(ii) On solving (i) and (ii) we get `a=8mL` (volume of `C_(2)H_(6))` `b=4mL` (Volume of `C_(2)H_(4))` Volume of `CO_(2)=8mL`

Discussion

50810.	A 20 ml mixture of C_(2)H_(4)"and"C_(2)H_(2) undergoes sparking in gas eudimeter with just sufficient
Answer» ANSWER :5

Discussion

50811.	A 20 litre container at 400 K contains CO_2(g) at pressure 0.4 atm and an excess of SrO (neglect the volume of solid SrO). The volume of the container is now decreased by moving the mavable piston fitted in the container. The maximumvolume of the conatiner, when pressure of CO_2 attains its maximum value will be : Give that : SrCO_3 (s)hArr SrO(s) + CO_2(g) K_p = 1.6 atm
Answer» <P>2 litre 5 litre 10 litre 4 litre Solution :Given that `K_p = 1.6 atm` `V_1 = 20 L, V_2 = ?, T_1 = 400 K , T_2 = 400K` `K_P = P_(CO_2)` `:. P_(CO_2) = 1.6 atm` `P_1 = 0.4 atm` `P_2= 1.6 atm` `(P_1V_1)/T_1= (P_2V_2)/T_2` `:. V_2 = (P_1V_1)/T_1 xx T_2/P_2 =(0.4 atm xx 20L)/(cancel 400K) xx (cancel400K)/(1.6atm) = 5L`

Discussion

50812.	A 20 litre container at 400 K contains CO_(2) (g)at pressure 0.4 atm and an excess of SrO (neglect the volume of solid SrO). The volume of the container is now decreased by moving the movable piston fitted in the container . The maximum volume of the container when pressure of CO_(2) attains its maximum value, will be ("Given that " : SrCO_(3)(s) hArr SrO (s) + CO_(2) (g)K_(p) =1*6 atm)
Answer» 5 litre 10 litre 4 litre 2 litre Solution :MAXIMUM pressure of `CO_(2)` = Pressure of `CO_(2)` at equilibrium For the reaction `SrCO_(3) (s) HARR SRO (s) + CO_(2) (g) ` `K_(p) = p_(CO_(2))= 1.6 " ATM = maximum pressure of "CO_(2)` Volume of container at this stage `V=nRT/P""` ...(i) INTIALLY, no . of moles of the gas (n) ` (PV)/(RT)= (0.4 xx20)/(RT)= 8/(RT)` As no. of moles of the gas will remain constant , therefore, at equilibrium also, `n=8/(RT)` Putting this value in eqn. (i), `V= 8/(RT)*(RT)/P = 8/P= 8/1.6=5L`

Discussion

50813.	A 20 liter container at 400 contains CO_(2)(g) at pressure 0.4 atm and an excess of SrO (neglect the volume of solid SrO). The volume of the container is now decreased by moving the movable piston fitted in the container. The maximum volume of the container, when pressure of CO_(2) attains Its maximum value will be : Given that : SrCO_(3)(S)hArrSrO(S)+CO_(2)(g) K_(p)=1.6 atm
Answer» 2 litre 5litre 10litre 4 litre Answer :B

Discussion

50814.	A 2.0 L container at 25^(@)C contains 1.25 mol of oxygen and 3.3mol of carbon. (a) What is the initial in the flask ? (b) If carbon and oxygen and oxygen react as completely as possible to form CO, what will be the final pressure in the container ?
Answer» Solution :(a) V=2.0 L, T=298 K, n=1.25 mol (because only GAS will exert pressure). Hence, PV=nRTor`P=(NRT)/(V)=((1.25" mol")(0.0821" L atm "K^(-1)mol^(-1))(298 K)/(2.0" L")=15.3" atm"` (B) The reaction will be :`C(s)+(1)/(2)O_(2)(g) to CO(g)` As 1 mol of C reacts with `(1)/(2)` mol of `O_(2)`, the limiting reactant will be 1.25 mol of `O_(2)`. Thus, `(1)/(2)` mol of `O_(2)` produces CO=1 mol. `:.1.25` mol of `O_(2)` will produce CO=2.50 mol. Now, n(GASEOUS)=2.50 mol. Hence, `P=(nRT)/(V)=((2.50" mol")(0.0821" L atm "K^(-1)mol^(-1)(298"K"))/(2.0" L ")=30.6" atm "`.

Discussion

50815.	A 2.0 g sample containing Na_2CO_3 and NaHCO_3 loses 0.248 g when heated at 300° C, the temperature at which NaHC03 decomposes to Na_2CO_3, CO_2 and H_2O. What is the percentage of Na_2CO_3 in the mixture ?
Answer» SOLUTION :`Na_2CO_3` does not decompose at the given temperature. It is only `NaHCO_3` which decomposes at this temperature. The corresponding equation is: `underset(168.016 g) (2NaHCO_(3)) overset(300^(@) C) to underset(44.01 g)(Na_(2)CO_(3)) + underset(18.016 g)(CO_(2)) + H_(2)O` `CO_(2)` and `H_(2)O` scape as gases and cause a loss in mass. TOTAL mass of `CO_2 + H_2O` (loss in mass) =44.01 + 18.016 = 62.026 g As according to the above equation,`therefore 0.248` g mass is LOST by `NaHCO_3 = 168.016` g `therefore 0.248`gmasswillbelostby`NaHCO_(3) = (168.016)/(62.026) xx 0.248 = 0.672 g` Since, the total mass of the sample is 2.0 g, the mass of `Na_2CO_3` PRESENT in it = `2.0 - 0.672 = 1.328 g` `therefore` PERCENTAGE of `Na_(2)CO_(3) = (1.328)/2 xx 100 = 66.4` Hence, the given sample contains `66.4% Na_(2)CO_(3)`.

Discussion

50816.	A 2 lit vessel is filled by 1 mole of each gas A "&" B. If K_(C) for reaction A(g)hArrB(g) is 1.5 at temp. T. [Atomic mass of A is 40 & B "is" 20].
Answer» `[A]` VS TIME is graph `I` `[B]` vs time is graph `I` `[A]` vs time is graph `II` `[B]` vs time is grap `II` Solution :N//A

Discussion

50817.	A 2 lit solution contains 0.04 mol of each of [CO(NH_(3))_(5)SO_(4)]Br and [CO(NH_(3))_(5)Br]SO_(4). To 1 lit of this solution, excess of AgNO_(3) is added. To the remaining solution of excen of BaCl_(2) is added. The amounts of precipitated salts, respectively, are
Answer» 0.01 mol & 0.01 mol <BR>0.01 mol & 0.02 mol 0.02 mol & 0.01 mol 0.02 mol & 0.02 mol SOLUTION :Given that 2L solution contains 0.04 mol each of `[CO(NH_(3))_(6)SO_(4)]Br and [CO(NH_(3))_(5)Br]SO_(4)`. Therefore, the concentration is 0.02 mol `"LIT"^(-1)`. `underset("0.02 mol")([CO(NH_(3))_(5)SO_(4)]Br)underset("0.02 mol0.02 mol")(rarrBr+[CO(NH_(4))_(5)SO_(4)]^(+))` `underset("0.02 mol")([CO(NH_(3))_(5)Br]SO_(4))underset("0.02 mol ")(rarrSO_(4)^(-2))+underset("0.02 mol")([CO(NH_(3))_(5)Br]^(+2))` `underset(("EXCESS"))(AG^(+))+underset((0.02))(Br^(-))rarrunderset(0.02)(AgBr)` `underset(("excess"))(Ba^(+2))+underset((0.02))(SO_(4)^(-2))rarrunderset(0.02)(BaSO_(4))`

Discussion

50818.	A 2-L flask contains 1.6 g of methane and 0.5 g of hydrogen at 27^(@)C. Calculate the partial pressure of each gas in the mixture and hence calculate the total pressure.
Answer» SOLUTION :`n_(O_(2))=(70.6)/(32)=2.21, n_(NE)=(167.5)/(20)=8.375, x_(o_(2))=(2.21)/(2.21+8.375)=0.21, x_(Ne)=0.79`

Discussion

50819.	A 2 L flask containing nitrogen at 60 cm pressure is connected to a 4 L flask containing carbon monoxide at 48 cm pressure. If the temperature is kept constant, calculate the final pressure of the mixture.
Answer» SOLUTION :52 CM

Discussion

50820.	(a) 2 g of metal carbonate were dissolved in 50 mL of N HCl. 100 mL of 0.1 N NaOH were required to neutralise the resultant solution. Calculate the equivalent mass of the metal carbonate. (b) How much water should be added to 75 mL of 3 N HCl ot make it a normal solution ?
Answer» SOLUTION :N//A

Discussion

50821.	A 1800 gm mixture of anhydrous CuSO_(4)(s) and its hydrated form [CuSO_(4).5H_(2)O(s)] undergoes 20% loss in mass on heating. Mole fractrion of CuSO_(4) in mixture is (Atomic mass of Cu = 64) :
Answer» `(3)/(40)` `(4)/(9)` `(5)/(9)` `(1)/(2)` ANSWER :C

Discussion

50822.	A 1.50gm samples of type metal (an alloy of S_(n), Pb, Cu and Sb) is dissolved in nitric acid and metastannic acid, H_(2)SnO_(3) precipitate. This is dehydrated by heating to Tin(IV) oxide, which is found to weigh 0.50gm. What percentage of tin was in the original type metal sample? (Sn= 119)
Answer» 0.3333 0.2627 0.2938 0.5254 Answer :B

Discussion

50823.	A 15dm^(3) cylinder contains hydroen gas at 5 atm at 27^(@)C, How many balloons of capacity of 1.5 dm^(3) at 1 atm and 27^(@)C can be filled using the gas available from the given cylinder ?
Answer» ANSWER :40

Discussion

50824.	A 1.250 g sample of octane(C_(8)H_(18)) is burnedin excess of oxygen in a bomb calorimeter. The temperatre of the calorimeter risesfrom 294.05 K to 300.78K. If heat capacityof the calorimeter is 8.93kJ//K, find the heat transferred to the calorimeter. Also calculate the enthalpy combustion of the sample of octane.
Answer» SOLUTION :Rise in temperature ` = 300.78 - 294.05K = 6.73K ` Heattransferredto the calorimeter `= ` Heat capacity of the calorimeter `xx` Rise in TEMP. Molar mass of `= ( 8.90kJ K^(-1)) ( 6.73K) = 60.1 kJ` `C_(8) H_(18) = 8 xx 12 + 18 = 114 G mol^(-1)` `:. `ENTHALPY of combustion`= ( 60.1)/( 1.250 ) xx 114 kJ mol^(-1)=5481.1kJ mol^(-1)`

Discussion

50825.	A 1.20 gm sample od cocaine, [alpha]_(D)=-16^(@) was dissolvedin 7.50 mL of cholroform and placed in polarimeter tube having a path length of 5.00 am calculate the observed rottion.
Answer» `+1.3^(@)` `-2.6^(@)` `+2.6^(@)` `-1.3^(@)` SOLUTION :N//A

Discussion

50826.	A 1.2 g mixture of Na_(2)CO_(3) and K_(2)CO_(3) was dissolved in water to form 100 cm^(3) of a solution. 20 cm^(3)of this solution required 40 cm^(3) of 0.1 N HCl for neutralisation. Calculate the mass of Na_(2)CO_(3) and K_(2)CO_(3) in the mixture.
Answer» SOLUTION :N//A

Discussion

50827.	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 SULPHUR acid Answer :A::B

Discussion

50828.	A 1.10 g sample of copper ore is dissolved and the Cu^(2+)is treated with excess KI. The liberated I_2requires 12.12 mL of 0.10M Na_(2) S_(2)O_3solution for titration. Find the % of copper by mass in ore.
Answer» Solution :`Cu^(2+) +erarrCu^(+), 2I^(-) rarrI_(2)+2E` Meq.of `Cu^(+2)` = Meq.of liberated `I_2` = Meq.of `Na_(2)S_2O_3` `= 12.12 xx0.1 XX1 =1.212` `:.(W_(cu^(2+)))/(63.6//1)xx1000=1.212` ` :. W_(cu) =W_(cu^(2+))=0.077` `( :. Cuoverset(H_2SO_4)rarrCuSO_(4))` `:. %Cu=(0.077)/(1.10)XX100 =7%`

Discussion

50829.	A 1.0g sample of pyrolusite ore containing MnO_2 (MW = 87)was dissolved in a concentrated HCI solution and liberated Cl_2(g) was passed through a concentrated KI solution releasing I_2 . If the liberated iodine required 16mL of 1.25 M sodium thiosulphate (Na_(2)S_(2)O_3)solution, mass percentage of MnO_2in the given sample is
Answer» ` 43.5% ` `87%` ` 21.75 %` `0.875%` Solution :No.of eqts. `MnO_2` = No.of m.eqts .of `CI_2=` = No.of eqts .KI = No.of m.eqts. hypo No.of m.eqts of `MnO_(2)= 16x×1.25 = 16xx5/4=20` No.of milli MOLES of `MnO_(2)=10` Weight of `MnO_(2)=10`Weight % of `MnO_(2) =10xx87xx10^(-3)=0.87g` Weight % of `MnO_(2)=(0.87)/1 xx100 =87%`

Discussion

50830.	A 10gmixture of Cu_(2)S and CuS was treated with 200 mL of 075M MnO_(4)^(-)in acid solution producing SO_(2), CU^(2+) and Mn^(2+). The SO_(2) was boiled off and the excess of MnO_(4)^(-). The SO_(2) was boiled off and the excess of MnO_(4)^(-) was titrated with 175 mL of 1M Fe^(2+) solution. Caculaate % fo CuSin original mixture.
Answer» ANSWER :`57.94%`

Discussion

50831.	A 10cm column of air is trapped by a colume of Hg 8cm long is capillary tube horizontally fixed as shown below at 1atm pressure Calculate the length of air column when the tube is fixed at same temperature (a) Vertically with open end up (b) Vertically with open end down (c ) At 45^(@) from horizontal with open end up .
Answer» ANSWER :`9.05 CM 11.18cm 9.3 cm`

Discussion

50832.	A 100mL solution containing 0.4 M As_(2)S_(3), 5M NaOH and 6M H_(2)O_(2) are reacted to form AsO_(4)^(3)" and "SO_(4)^(2-) as product. If final solution is allowed to stand for some time, what maximum volume of O_(2) at 1 atm and 273Kcan be obtained by decomposition of H_(2)O_(2)?
Answer» 112mL 224mL 336mL 448mL Answer :d

Discussion

50833.	A 100mL solution containing 0.4 M As_(2)S_(3), 5M NaOH and 6M H_(2)O_(2) are reacted to form AsO_(4)^(3)" and "SO_(4)^(2-) as product. What may be the correct coefficient of As_(2)S_(3),H_(2)O_(2) and NaOH respectively in a balanced reaction?
Answer» 1,14,12 1,12,14 1,28,20 1,28,18 Answer :a

Discussion

50834.	A 100mL mixture containing 72% of CH_(4) by volume and the rest an unknown gas 'X' was kept in a vessel Due to a very fin e crack the mixture effused out 21mL of the mixture was lost and the remaining mixture contained 68.35% of methane by volume Calculate molecular mass of gas X All the measurement are made at same temperature and pressure .
Answer» ANSWER :87.1

Discussion

50835.	A 100 watt bulb emits monochromatic light of wavelength 400 nm.Calculate the number of protons emitted per second by the bulb.
Answer» ANSWER :`2.012 XX 10^(20) s^(-1)`

Discussion

50836.	A 100 watt bulb emits monochromatic light of wavelength 400 nm. Calculate the number of photons emitted per second by the bulb.
Answer» Solution :Power of the BULB = 100 WATT `= 100 J s^(-1)` Energy of one photon, `E = hv = (hc)/(lamda)((6.626 xx 10^(-34)Js) xx (3 xx 10^(8) ms^(-1)))/(400 xx 10^(-9)m) = 4.969 xx 10^(-19) J` `:.` NUMBER of photons emitted `= (100 Js^(-1))/(4.969 xx 10^(-19)J) = 2.012 xx 10^(20) s^(-1)`

Discussion

50837.	A 100wartbulbemitsmonochromaticlight ofwavelength 400 bn, Calculatethe numberofphotonsemittedper secondby the bulb.
Answer» SOLUTION :Powerof the bulb - 100 walt= 10 `js^(-1)` Energyof one photonE `hv = (hc )/(lambda)= (6.626xx 10^(34) Js xx3.0xx 10^(8) MS^(-1))/( 400 xx 10^(9) m)` Noof photon4.9695 `xx 10^(19)` energyfrom 1photon 100J senergyfromhowmanyphoton Numberof photonsemited `=(1xx 100 Js^(-1))/(4.9695 xx 10^(19) J)= 2.0123 xx 10^(20) s^(-1)`

Discussion

50838.	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 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//A

Discussion

50839.	A 100 ml mixture of Na_(2)CO_(3) and NaHCO_(3) is tittrated against 1 M HCl. If v_(1)L and v_(2)L are consumed when phenolphthalein and methyl orange are used as indicators respectively in two separate titrations, which of the following is true form molarities in the original solution.
Answer» molarity of `Na_(2)CO_(3)=20v_(1)` molarity of `NaHCO_(3)=10(v_(2)-2v_(1))` molarity of `Na_(2)CO_(3)=10(v_(2)+v_(1))` molarity of `NaHCO_(3)=10(v_(2)-v_(1))` Solution :When HPh USED `(1)/(2)xx` Eqts of `Na_(2)CO_(3)` = Eq of HCl `(1)/(2)xx(100)/(1000)xxM_(Na_(2)CO_(3))xx2=V_(1)xx1` `M_(NC_(2)CO_(3))=10V_(1)......(1)` When meOH used Eqts of `NC_(2)CO_(3)+` Eqts of `NaHCO_(3)` = Eq of HCl `((100)/(1000)xxM_(Na_(2)CO_(3))xx2)+((100)/(1000)xxM_(NCHCO_(3))xx1)=V_(2)xx1....(2)` On solving (1), (2) `M_(NaHCO_(3))=10(V_(2)-2V_(1))`

Discussion

50840.	A 1.00 m^2 column of air extending from the earth's surface through the upper atmosphere has a mass of about x producing an atmospheric pressure. Then 'x' is
Answer» 10,300 KG 10,300 GM 1030 gm 1030 MG ANSWER :A

Discussion

50841.	A 10.0 g piece of gallium (m=69.7) at 25.0^(@) c is placed in 10.0 g of H_(2)O"at" 55.0^(@)c what is the final temperature when this system comes to equilbium ? (Assume the specific heat capacity of liquid Ga is the same as that of solid Ga =0.37jg^((-1))"^(@)C^(-1)).
Answer» `35.0^(@)C` `38.1^(@)C` `41.8^(@)C` `52.6^(@)`C Answer :d

Discussion

50842.	A 100 meter hollow tube of uniform thickness has two open ends X and Y. Ammonia gas is sent into the tube from end X and hydrogen chloride from end Y, simultaneously. At what distance from end 'X', the gases first meet to form a smoke ring?
Answer» SOLUTION :Two gases diffuse in to the pipe from their respective ends. A, B and MEET at a point .O. ACCORDING to Grahm.s law `(r_(1))/(r_(2)) = sqrt((M_(2))/(M_(1))) = sqrt((36.5)/(17)) = 1.47` The RATIO of rate of diffusion is also equal to the ratio of distance travelled. `(1.47)/(1) = (l)/(100-1),` SOlving l = 59.5 The two gases first meet at a distance from end X = 59.5meter

Discussion

50843.	A 10-L vessel filled with O_(2) at 30 K is connected to an open limb manometer containing glycerine. The level in the open limb was found to be higher than the other limb by 50 cm. Calculate the number of moles of the gas in the vessel (Given d_("glycerine")=2.72 g mL^(-1), d_(Hg)=13.6 g mL^(-1))
Answer» Solution :`(hxxdxxg)Hg=(hxxdxxg)_("GLYCERINE")` `hxx13.6=50xx2.72"or"h=10 CM` `P_(gas)=(76+10)_(cm)=86 cm` `PV=nRT :. (86)/(76)xx10=n0.0821xx300"or"n=0.46" mole"`.

Discussion

50844.	A 10L cylinder has helium at 8 atm and 32^@C. How many balloons of 2L each at 2atm and 32^@C can be filled with the gas available from the cylinder.
Answer» Solution :The PVof balloon ` = 2 xx 2 = 4` lit-ATM Gas from the cylinder is available to fill in the BALLOONS, until pressureof gas in cylinder drops from 8 atm to 2 atm. The PV of helium available ` = 10 xx (8-2) = 60` lit-atm Number of balloons filled ` = ("PV of the gas available")/("PV of the balloon")= (60)/(4) = 15`

Discussion

50845.	A 10 g sample of oxygen gas is taken in a container of volume 1 litre and is found to exert a pressure of 3 bar. Which of the following options is correct regarding speed of the molecuels?
Answer» All the molecules are MOVING at a same SPEED which is equal to 310 m/sec. `U_("avg")=300m//"sec"` `U_("mps")=300xxsqrt((2)/(5)` m/sec. `U_("mps")=310` m/sec. Answer :C

Discussion

50846.	A 1.0 g sample of KCIO_3 was heated under such conditions that a part of it decomposed according to the equation: (i) 2KCIO_3to 2KCI+3O_2 and the remaining underwent a change according to the equation (ii) 4KCIO_3to 3KClO_4+ KCI. If the amount of oxygen evolved was 146.8 mL at S.T.R, calculate the percentage by weight of KClO_4 in the residue.
Answer» Solution :SUPPOSE, X g of `KClO_3` decomposed according to the equation: `underset(2 xx 122.55 g)(2KClO_(3)) to underset(2 xx 74.55 g)(2KCl) + underset(3 xx 22.4 L "at S.T.P.")(3O_(2))` Therefore, `1-x`g of `KClO_(3)` decomposed according to the equation, `underset(4 xx 122.55 g)(4KClO_(3)) to underset(3 xx 138.55 g)(3KClO_(4)) + underset(74.55 g)(KCl)` From these equations, it is CLEAR that oxygen is evolved only in the first type of decomposition. `therefore 2 xx 122.55 g` of `KCIO_3` on decomposition gives `O_2 = 3 xx 22.4` L `therefore` x gm of `KCIO_3` on decomposition will give: `O_(2) = (3 xx 22.4)/(2 xx 122.55) xx x` L at S.T.P. `therefore` The percentage of `KClO_(4)` in the residue `=0.394/0.790 xx 100 = 49.9`

Discussion

50847.	A 1.0 g sample of H_(2)O_(2) solution containing 'x' per cent by weight requires x mL of KMnO_(4) solution for complete oxidation under acidic conditions. Calculate the normality of the KMnO_(4) solution.
Answer» SOLUTION :N//A

Discussion

50848.	A 10 cm column of air is trapped by a column of mercury, 8 cm long, in a capillary tubehorizontally fixed as shown below, at 1 atm pressure. When the tube is held atcertain angle theta^@ with open end up, the weight of Hg is borne partially by the gas. Vertical Height of Hg is a measure of additional pressure on gas. When the tube's held at 45^@ with the horizontal with open end up the length of air column is
Answer» `(76 xx 10)/(76 + 8/(sqrt2)) cm` `(76 xx 8/(sqrt2))/(76 + 10) cm` `76 xx 8/(sqrt2)` `10 xx 8/(sqrt2)` SOLUTION :At `45^@C`, pressure on GAS = `76 + h COS THETA` `= 76 + 8/(sqrt(2))` `P_1l_1 = P_2l_2 , 76 xx 10 = (76 + 8/(sqrt2))l_2`.

Discussion

50849.	A 10 cm column of air is trapped by a column of mercury, 8 cm long, in a capillary tubehorizontally fixed as shown below, at 1 atm pressure. When the tube is held atcertain angle theta^@ with open end up, the weight of Hg is borne partially by the gas. Vertical Height of Hg is a measure of additional pressure on gas. The length of air column, when the tube is fixed vertically at the same temperature with open end up is .
Answer» `(76 XX 84)/10 cm` `(76 xx 10)/84 cm` `(84 xx 10)/76 cm` `(76 xx 10)/(84) cm` SOLUTION :Boyle.s LAW , `P_1 V_1 = P_LV_L (or) P_1L_1 = P_2l_2` `implies 76 xx 10 - (76_1 + 8) l_2 implies l_2 = (76 xx 10)/(84) cm`

Discussion

50850.	A 10 cm column of air is trapped by a column of mercury, 8 cm long, in a capillary tube horizantally fixed as shown below, at 1 atm pressure. When the tube is held at certain angle theta^(@), with open end up, the weight of Hg is borne partially by the gas. Vertical Height of Hg is a measure of additional pressure on gas: The length of air coloumn.when the tube is fixed vertically with opend end down at same temperature is
Answer» `(76 XX 68)/10 ` 0 cm `(76 xx 10)/(84) ` `(76 xx 10)/84 ` Solution :`underset(("horrizontal"))(p_1l_1) = underset(("VERTICALLY down"))(p_2l_2)` `76 xx 10 = (76 - 8) l_2 implies l_2 = (76 xx 10)/(68) cm`.

Discussion

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