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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 50251. |
A gas absorbs 100 J of heat and is simultaneously compressed by a constant external pressure of 1.5 atm from a volume of 8.0L to 2.0L. The change in internal energy for the gas in Joules is (1L - atm = 101.32 J)) |
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Answer» `-1011.9` |
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| 50252. |
A Galvanic cell consits of three compartment as shown in figure. The first compartment contain ZnSO_4(1M) and III compartment contain CuSO_4(1M). The mid compartment contain NaNO_3 (1M). Each compartment contain 1L solution: E_(Zn^(2+)//Zn)^(@)=-0.76, E_(Cu^(2+)//Cu)^(@)=+0.34 The concentration of SO_4^(2-) ions in III compartment after passage of 0.1 F of charge will be: |
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Answer» 1.05M |
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| 50253. |
A Galvanic cell consits of three compartment as shown in figure. The first compartment contain ZnSO_4(1M) and III compartment contain CuSO_4(1M). The mid compartment contain NaNO_3 (1M). Each compartment contain 1L solution: E_(Zn^(2+)//Zn)^(@)=-0.76, E_(Cu^(2+)//Cu)^(@)=+0.34 The concentration of NO_3^- in mid compartment after passage of 0.1 F of charge will be: |
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Answer» 0.95M |
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| 50254. |
A Galvanic cell consits of three compartment as shown in figure. The first compartment contain ZnSO_4(1M) and III compartment contain CuSO_4(1M). The mid compartment contain NaNO_3 (1M). Each compartment contain 1L solution: E_(Zn^(2+)//Zn)^(@)=-0.76, E_(Cu^(2+)//Cu)^(@)=+0.34, The concertation of Zn^(2+) in first compartment after passage of 0.1 F charge will be: |
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Answer» 1M |
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| 50255. |
(A): Gallium is used as a thermometric liquid (R):Gallium has wide liquid range of temperature |
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Answer» A and R are TRUE, and R is the correct explanation of A |
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| 50256. |
A gas absorbs 250 J of heat and expands from 1 litre to 10 litre at constant temperature against external pressure of 0.5 atm . The values of q, w and DeltaE will be respectively. |
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Answer» `10 CM ^(3) , 500J` |
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| 50257. |
A g of a metal displaces VmL of H_(2) at NTP Eq.wt. of metal, E is (are): |
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Answer» `E=(Axx1.008xx22400)/("Vol.of" H_(2)"displaced"xx2)` |
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| 50258. |
a' g KHC_(2)O_(4) required to reduce 100 mL of 0.02 M KMnO_(4) in acid medium and 'b' g KHC_(2)O_(4) neutralises 100 mL of 0.05 M Ca(OH)_(2) then : |
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Answer» a= b |
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| 50259. |
A fundamental goal of thermodynamics is the ____of the reaction |
| Answer» SOLUTION :PREDICTION of SPONTANEITY | |
| 50260. |
A fuel will have a larger fuel value if one gram of it one burning give more of |
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Answer» `CO_(2)` |
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| 50261. |
A fuel has same knocking property as a mixture of 70% iso octane (2,2,4 trimethyl pentane) 30%n-heptane by volume the octane number of the fuel is |
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Answer» 70 |
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| 50262. |
A fruity smell is produced by the reaction of C_(2)H_(5)OH with |
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Answer» `PCl_(5)` |
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| 50263. |
A fruity smell is obtained by the reaction of ethanol with |
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Answer» `PCl_(5)` |
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| 50264. |
(A) Friedel-Crafts reaction between benzene and acetic angydride in presence of anhydrous AICI_(3) yields acetophenone and not polysubstitution product. ( R) Acetophenone formed poisons the catalyst, preventing further reaction. |
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Answer» If both (A) and ( R) are CORRECT and ( R) is CORRCT EXPLANATION of (A). |
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| 50265. |
A free radical is aspecies with an unpaired valence electron |
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Answer» |
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| 50266. |
A formula analogous to Rydberg formula applies to the series of spectral ines which arise from transition from higher energy level to the lower energy level of hydrogen atom. A muonic hydrogen atom is like a hydrogen atom in which the electron is replaced by a heavier particle,t he 'muon'. the mass of the muon is about 207 times the mass of an electron, while the charge remains same as that of the electron. Rydberg formula for hydrogen atom is: (1)/(lamda)=R_(H)[(1)/(n_(1)^(2))-(1)/(n_(2)^(2))](R_(H)=109678cm^(-1)) Q. Distance between first and third Bohr orbits of muonic hydrogen atom will be: |
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Answer» `(0.529)/(207)xx2`Ã… |
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| 50267. |
A formula analogous to Rydberg formula applies to the series of spectral ines which arise from transition from higher energy level to the lower energy level of hydrogen atom. A muonic hydrogen atom is like a hydrogen atom in which the electron is replaced by a heavier particle,t he 'muon'. the mass of the muon is about 207 times the mass of an electron, while the charge remains same as that of the electron. Rydberg formula for hydrogen atom is: (1)/(lamda)=R_(H)[(1)/(n_(1)^(2))-(1)/(n_(2)^(2))](R_(H)=109678cm^(-1)) Q. Angular momentum of 'muon' in muonic hydrogen atom may be given as: |
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Answer» `(h)/(PI)` |
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| 50268. |
A formula analogous to Rydberg formula applies to the series of spectral ines which arise from transition from higher energy level to the lower energy level of hydrogen atom. A muonic hydrogen atom is like a hydrogen atom in which the electron is replaced by a heavier particle,t he 'muon'. the mass of the muon is about 207 times the mass of an electron, while the charge remains same as that of the electron. Rydberg formula for hydrogen atom is: (1)/(lamda)=R_(H)[(1)/(n_(1)^(2))-(1)/(n_(2)^(2))](R_(H)=109678cm^(-1)) Q. |
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Answer» `+(13.6)/(207)EV` |
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| 50269. |
A formula analogous to Rydberg formula applies to the series of spectral ines which arise from transition from higher energy level to the lower energy level of hydrogen atom. A muonic hydrogen atom is like a hydrogen atom in which the electron is replaced by a heavier particle,t he 'muon'. the mass of the muon is about 207 times the mass of an electron, while the charge remains same as that of the electron. Rydberg formula for hydrogen atom is: (1)/(lamda)=R_(H)[(1)/(n_(1)^(2))-(1)/(n_(2)^(2))](R_(H)=109678cm^(-1)) Q. Radius of first Bohr orbit of muonic hydrogen atom is: |
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Answer» `(0.259)/(207)`Ã… |
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| 50270. |
A formula analogous to Rydberg formula applies to the series of spectral ines which arise from transition from higher energy level to the lower energy level of hydrogen atom. A muonic hydrogen atom is like a hydrogen atom in which the electron is replaced by a heavier particle,t he 'muon'. the mass of the muon is about 207 times the mass of an electron, while the charge remains same as that of the electron. Rydberg formula for hydrogen atom is: (1)/(lamda)=R_(H)[(1)/(n_(1)^(2))-(1)/(n_(2)^(2))](R_(H)=109678cm^(-1)) Q. Energy of first Bohr orbit of muonic hydrogen atom is: |
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Answer» `-(13.6)/(207)EV` |
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| 50271. |
A forensic chemist needed to determine the conventration of HCN in the blood of a suspected homicle victimand decided to titrate a diluted sample fo the blood with iodine, using the reaction, HCN_((aq.)) + I_(3)^(-) rarr ICN_((aq.)) + 2I_((aq.))^(-) + H_((aq.))^(+) Adiluted blood sample of volume15mL was titrated to the stoihometric point with 5.21mL of an I_(3)^(-) aolurion. The molarity of I_(3)^(-) in the solution was determinedby titrating it against arsenci (III) oxide, which in solution formsarenious acid, H_93)As)_(3). It was foundthat 10.42 mL of thetri-iodide solution was neededto reach teh stoichmetric point witha 10mL sample of 0.1235 M H_(3) AsO_(3) in the reaction. H_(3)AsO_(3(aq)) +v I_(3(aq))^(-) + H_(2)O_((l)) rarr H_(3)AsO_(4(aq)) + 3I_((aq.))^(-) + 2H_((aq.))^(+) (a) What is the normally of tri-iodie ions in the initlal solution? (b) What is the molar concentration of HCN in the blood sample? |
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Answer» |
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| 50272. |
(A) : For n = 3, 1 may be 0,1,2 and .m. may be 0, (+-2, +-1 and 0) (R) : For each value of n there are 0 to (n-1) possible values of 1 and for each value of l values of .m. are -1….0….+1 |
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Answer» Both (A) and (R) are true and (R) is the CORRECT EXPLANATION of (A) |
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| 50273. |
(a)for any mutielectron specie, compare the energies of following orbitals. Justify [ External magnetic & electric fields are absent ] Orbitals to be compared 3d_(xy),2s,4p_(x),4p_(y) (b)compare the "average distance " of the following orbital fromthe nucles. Also commemt on their 'chances of closeness to the nucleus'if they have same averge distance"Orbitals to be compared 1s,2s,3s,3p_(x),3d_(xy). |
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Answer» Solution :(a)` 2slt3d_(xy) lt4p_(x)=4p_(y)` justification compare` (N+l)&if(n+l)` same then 'n' (b) Average distance (on the basis of n) ` 1slt2slt3s=3p_(x)=3d_(xy)` CLOSENESS comparison 3S will be more CLOSE as compare to `3p_(x)` as compared to `3d_(xy)` |
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| 50274. |
(A): For a process to be spontaneous, the change in free energy must be negative. (R): The change in entropy for a process must always be positive if it is spontaneous. |
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Answer» Both A and R are true and R is thecorrect explanation |
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| 50275. |
A fluorine disposal plant was constructed to carryout the reactions: F_(2)+2NaOHrarr1/2O_(2)+2NaF+H_(2)O 2NaF+CaO+H_(2)OrarrCaF_(2)+2NaOH As the plant operated, excess lime was added to bring about complete precipitation of the fluoride as CaF_(2). Over a period of operation, 1900 kg of fluorine was fed into a plant and 10,000 kg of lime was required. What was the percentage utilisation of lime? [At, mass F=19], [Lime=CaO] |
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Answer» |
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| 50276. |
A flask of 1L having NH_(3)(g) at 2.0atm and 200K is connected with another flask of vol 800ml having HCl(g) at 8atm & 200K through tube of negligible volume. The two gases reacts to form NH_(4)Cl(s) with evolution of 43 KJ/mol heat. If heat capacity of at constant volume is 20J/K/mol and neglecting heat capacity of flask and solid, volume of solid formed and its pressure (R = 0.08). The final pressure in the flask is |
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Answer» 5.375KJ `{:("INITIAL mole", (1 xx 2)/(0.08 xx 200), (8 xx 0.8)/(0.08 xx 200)0,0),(,= 0.125,=0.4,),("FINAL moles",0,0.275,0.125):}` `:.` heat product = `0.125 xx 43 = 5.375 KJ` |
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| 50277. |
A flask of 1.5 Lcapacity contains 400 mg of O_(2) and 60 mg of H_(2) at 100^(@)C. Calculate the total pressure of the gaseous mixture. If the mixture is permitted to react to form water vapour at 100^(@)C, what materials will be left and what will be their partial pressures? |
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Answer» <P> Solution :`n_(oO_(2))=(400xx10^(-3))/(32)=0.0125``n_(H_(2))=(60xx10^(3))/(2)=0.03` LTBR. Here `O_(2)`. Is the limiting reagent and is completely consumed. Total moles `n=0.0125+0.03=0.0425mol` total pressure `P=(0.0425xx0.082xx373)/(1.5)=0.867atm` `underset(0.005" mol")underset(0.03" mol")(2H_(2)(g))+underset(-)underset(0.0125" mol")(O_(2)(g))tounderset(0.025" mol...After reaction")underset(-"...initial")(2H_(2)O(g))` here `O_(2)` is the limiting reagent and is completely consumed. Now `p_(H_(2)O)=(0.867xx0.025)/(0.0425)=0.509atm` |
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| 50278. |
A flask is initially filled with pure N_2O_3(g) having pressure 2 bar and following equilibria are established. N_2O_3(g)hArrNO_2(g)+NO(g)""K_(P_1)=2.5bar 2NO_2(g)hArrN_2O_4(g)""K_(P_2)=? If at equilibrium partial pressure of NO(g) was found to be 1.5 bar,then: |
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Answer» EQUILIBRIUM parial pressure of `N_2O_3(g)` is 0.5 bar. |
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| 50279. |
A flask contains a mixture of isohexane and 3-methylpentane. One of the liquids boils at 63^(@)C while the other boils at 60^(@)C. What is the best way to separate the two liquids and which one will be distilled out first? |
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Answer» Fractional distillation, ISOHEXANE |
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| 50280. |
A flask contains 1gm, of H_2 , 2gms. of Ne and 1.6 gms. of O_2 at a pressure of 2 atmoshperes at 27^@C. Calculate the partial pressures of each gas and the volume of the flask. |
| Answer» SOLUTION :1.54atm, 0.301 ATM, 0.154atm, 8lit | |
| 50281. |
A flask contains 1.2 inoles of methane gas at 25^@C and 74.6mm pressure. If 4gm of the same gas is sent into the flask, what would be its pressure ? |
| Answer» SOLUTION :90.14 MM | |
| 50282. |
A flask containing 12 g of a gas of relative molecular mass 120 at a pressure of 100 atm was evacuated by means of a pump unitl the pressure was 0.01 atm at the same temperature. What is the approximate number of molecules left in the flask ? |
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Answer» <P> Solution :Initially,`n_(1)=(12)/(120)` mole=0.1 mole, `P_(1)`=100 atm, `T_(1)`=T, `V_(1)`=VAfter evacuation,`n_(2)`=? `P_(2)=0.01`atm, `T_(2)=T, V_(2)=V` ApplyingPV=nRT As V and T an constant, `(P_(1))/(P_(2))=(n_(1))/(n_(2))` `(100)/(0.01)=(0.1)/(n_(2))"or"n_(2)=(0.1xx0.01)/(100)=(10^(-1)xx10^(-2))/(10^(2))=10^(-5)` mole No. of MOLECULES left `=(10^(-5))(6.02xx10^(23))=6.02xx10^(18)` |
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| 50283. |
A flask A contains 0.5 mole of oxygen gas. Another flask B contains 0.4 mole of ozone gas Which of the two flasks contains greater number of oxygen atoms. |
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Answer» SOLUTION :Flask A: 1 mole of oxygen gas = `6.023xx10^(23)` molecules `:.`0.5 mole of oxygen gas = `6.023xx1O^(23)xx0.5` molecules The number of atoms in flask A = `6.023xx10^(23)xx0.5xx2` = `6.023xx10^(23)` atoms. Flask B:1 mole of ozone gas = `6.023xx10^(23)` molecules `:.` 0.4 mole of ozone gas = `6.023 XX 10^(23)xx0.4` molecules The number of atoms in flask B = `6.023xx10^(23)xx0.4xx3` = `7.227xx10^(23)` atoms. `:.` Flask B CONTAINS a great number of oxygen atoms as compared to flask A. |
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| 50284. |
A flask containing 250 mg of air at 27^(@)Cis heated till 25.5% of air by mass is expelled from it. What is the final temperatuer of the flask ? |
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Answer» Solution :Let the volume of flask be V. Let FINAL temeperature be T(K). Mass of air expelled at `T(K)=(250x25)/(100)=62.5mg` Mass of air contained in flask `=250-62.5=187.5mg` now, volume of total air (250 mg) at higher temperature `V_(2)=(Vxx250)/(187.5)ML` Now `(V_(2))/(T)=(V)/(300)` or `T=(V_(2)xx300)/(V)=(Vxx250xx300)/(187.5xxV)=400K` or `127^(@)C` |
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| 50285. |
A five litre flask contains 3.5gm of N_2 3g of H_2 and 8g of O_2 at 27^@C. The total pressure exerted by the mixture of these gases is |
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Answer» |
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| 50286. |
a fischer projection of (2R,3S)-2,3-butanediol is |
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Answer»
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| 50287. |
A fire work gives out crimson coloured light. It contains the salt of |
| Answer» Solution :strontium | |
| 50288. |
A fire of lithium , sodium and potassium can be extinguished by |
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Answer» `H_(2)O` |
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| 50289. |
A ferromagnetic substances becomes a permanent magnet when it is placed in a magnetic field because ___________ |
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Answer» all the DOMAINS GET oriented in the DIRECTION of magnetic field |
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| 50290. |
A feromagnetic substance becomes a permanent magnet when it is placed in a magnetic field because . ….. |
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Answer» all the DOMIANS GET oriented in the direction of magnetic field. |
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| 50291. |
(A) : Fe^(3+) (g) is more stable than Fe^(2+) (g) (R) : Fe^(3+) (g) has more number of unpaired electrons than Fe^(++) (g) |
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Answer» Both (A) and (R) are true and (R) is the CORRECT explanation of (A) |
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| 50292. |
A factory was started near a village. Suddenly villagers started feeling the presence of irritating vapours in the village and cases of headache, chest pain, cough, dryness of throat and breathing problems increased. Villagers blamed the emissions from the chimney of the factory for such problems. Explain what could have happened. Give chemical reactions for the support of your explanation |
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Answer» Solution :The symptoms observed in the villagers show that oxides of nitrogen and sulphur must be coming out of the CHIMNEY. This is due to combustion of fossil fuels like COAL, oil, natural gas, gasoline, etc. to produce high temperatures at which oxidation of atmospheric nitrogen TAKES place forming NO and `NO_2` : `N_2+O_2 overset(1200-1750^@C)hArr 2NO , 2NO + O_2 overset(1100^@C)hArr 2NO_2` `SO_2`is produced due to combustion of sulphur containing coal and fuel oil or roasting of sulphide ORES like iron pyrites `(FeS_2)` , copper pyrites `(CuFeS_2)` , etc. e.g., `Cu_2S + O_2 to 2 Cu + SO_2` |
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| 50293. |
(A) fat is digested in the intestine by emulsification. (r)Bile salts stablize the emulsion so formed. |
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Answer» IF both (A) and (R) are CORRECT and (r) is the correct EXPLANATION for (a). |
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| 50294. |
A farmer was using pesticides on his farm. He used the produce of his farm as food for rearing fishes. He was told that fishes were not fit for human consumption because large amount of pesticides had accumulated in the tissues of fishes. Explain how did this happen? |
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Answer» Solution :Pesticides from soil are transferred into the crops and from the crops these are transferred into rearing fish food. Pesticides entered into water through rearing fish food and FINALLY entered into the BODIES of the fishes. Therefore through the food chain pesticides are transferred from lower trophic level to higher trophic level. Over the TIME, the concentration of pesticides in fishes reach a level which CAUSES serious metabolic and physiological DISORDERS. |
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| 50295. |
A factory was started near a village. Suddenly villagers started feeling the presence of irritating vapours in the village and cases of headache, chest pain, cough, dryness of throat and breathing problems increased. Villagers blamed the emissions from the chimney of the factory for such problems. Explain what could have happened. Give chemical reactions for the support of your explanation. |
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Answer» Solution :The symptoms observed in a village indicate that nitrogen oxide and sulphur oxide are released from the chimnery of the factory. These are produced by the burning of fossil FUELS such as natural gas, COAL, gasoline, etc. In an AUTOMOBILE engine, at high temperature when fossil fuel burnt, `N_2` and `C_2` COMBINE to yield NO is i.e., nitric oxide. `N_2 + O_2 overset(1200- 1500^@C)to 2NO` `2NO + O_2 overset(1100^@C)to 2NO_2` `SO_2` is produced by burning of sulphur containing fossil fuel or by roasting of sulphide ores such a iron pyrites, copper pyrites etc.. `Cu_2 S + O_2 to 2CU + SO_2` |
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| 50296. |
A factory, producing methanol, is based on the reaction : CO+2H_(2)rarr CH_(3)OH ltBRgt Hydrogen and carbon monoxide are obtained by the reaction CH_(4)+H_(2)Orarr CO+3H_(2) Three units of factory namely, the "reformer" for the H_(2) and CO production, the "methanol reactor" for production of methonol adn a "separator" to separate CH_(3)OH form CO and H_(2) are schematically shown in figure. The flow of methonal from valve-3 is 10^(3)mol//sec. The factory is so designed that (2)/(3) of the CO is converted to CH_(3)OH. Assume that the feromer reaction goes to completion. CO+2H_(2)rarrCH_(3)OH Delta_(r) H=-100 R What is the flow of CO and H_(2) at valve-2 ? |
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Answer» `CO:500 mol//sec,` `H_(2):1000 mol//sec` |
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| 50297. |
A factory, producing methanol, is based on the reaction : CO+2H_(2)rarr CH_(3)OH ltBRgt Hydrogen and carbon monoxide are obtained by the reaction CH_(4)+H_(2)Orarr CO+3H_(2) Three units of factory namely, the "reformer" for the H_(2) and CO production, the "methanol reactor" for production of methonol adn a "separator" to separate CH_(3)OH form CO and H_(2) are schematically shown in figure. The flow of methonal from valve-3 is 10^(3)mol//sec. The factory is so designed that (2)/(3) of the CO is converted to CH_(3)OH. Assume that the feromer reaction goes to completion. CO+2H_(2)rarrCH_(3)OH Delta_(r) H=-100 R Amount of energy released in methanol reactor in1 minute : |
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Answer» 1200 kcal |
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| 50298. |
A factory, producing methanol, is based on the reaction : CO+2H_(2)rarr CH_(3)OH Hydrogen and carbon monoxide are obtained by the reaction CH_(4)+H_(2)Orarr CO+3H_(2) Three units of factory namely, the "reformer" for the H_(2) and CO production, the "methanol reactor" for production of methonol adn a "separator" to separate CH_(3)OH form CO and H_(2) are schematically shown in figure. The flow of methonal from valve-3 is 10^(3)mol//sec. The factory is so designed that (2)/(3) of the CO is converted to CH_(3)OH. Assume that the feromer reaction goes to completion. CO+2H_(2)rarrCH_(3)OH Delta_(r) H=-100 R What is the flow of CO and H_(2) at valve-1? |
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Answer» `CO:1500 mol//sec, H_(2):2000 mol//sec` |
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| 50299. |
(A) : F_(2)+NaOH to NaF+OF_(2)+H_(2)O is an example of redox reactions with respect to fluorine. (R ) : Fluorine shows universal oxidation state of -1 in al its compounds. |
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Answer» Both A and R are TRUE and R is the CORRECT EXPLANATION |
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| 50300. |
(a) Explain the following terms : (i) Weight percent (ii) Volume percent (iii) Molarity (iv) Normality (b) How much volume of M/2 HCI solution be diluted to obtain 100 cm^3 of a decimolar solution ? |
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Answer» Solution :(ii) The number of gram moles of HCI will remain the same on dilution. Suppose, `V cm^3` of y `M/2`HCI are required to be DILUTED. Since, Number of gram moles in a solution `=("Molarity" XX "Volume in" cm^(3))/1000` Therefore, the number of gram moles in cone, solution `=(1/2 xx v)/1000` THUS, `(1/2 xx V)/1000 =(1/10 xx 100)/(1000)` Which GIVES `V = 20 cm^(3)` Hence, `20 cm^3` of `M/2`HCI should be diluted to `100 cm^3` to obtain a decimolar solution. |
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