InterviewSolution
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.
| 92301. |
30 g of urea (M = 60 g "mol"^(-1)) is dissolved in 846 g of water. Calculate the vapour pressureof water for this solution if vapour pressure of pure water at 298 K is 23.8 mm Hg. |
|
Answer» Solution :The solution MAY be taken as dilute solution. Applying the RELATIONSHIP : `(p_1^0 - p_1)/(p_1^0) = (w_2 XX M_1)/(M_2 xx w_1)` ` 1 - (p_1)/(p_1^0) = (w_2 xx M_1)/(M_2 xx w_1)` Substituting the values of the above EQUATION, we have ` 1- (p_1)/(23.8) = (30 xx 18)/(60 xx 846)` ` 1 - (p_1)/(23.8) = 0.0106 ` `(p_1)/(23.8) = 1- 0.0106` `(p_1)/(23.8) = 0.9894" or" p_1 = 0.9894 xx 23.8` `p_1 = 23.55 mg Hg ` |
|
| 92302. |
3.0 g of non-volatile solute when dissolved in 1 litre water, shows s osmotic pressure of 2 atmosphere at 300 K. Calculate the molecular mass of th solute. (R=0.082 L K^(-1)mol^(-1)). |
|
Answer» `M_(B)=(W_(B)xxRxxT)/(pixxV)=((3.0g)XX(0.082"L atm K"^(-1)mol^(-1))xx(300K))/((2 atm)xx(1L))=20.0.116 g.` |
|
| 92303. |
3.0 g of H_(2) react with 29.0 g of O_(2) to form H_(2)O. (i) which is the limiting reagent? (ii) Calculate the maximum amount of H_(2)O that can be formed. (iii) Calculate the amount of the reactant left unreacted. Molecular mass of H_(2)=2.016. |
|
Answer» Solution :`UNDERSET(2xx2.016=4.032g)(2H_(2))+underset(32g)(O_(2))rarrunderset(2XX(1.016+16)=36.032)(2H_(2)O)` `"3 g of "H_(2)" require "O_(2)=(32)/(4.032)xx3=23.8g` Thus, `O_(2)(29g)` is present Hence, `H_(2)` is the limiting reactant. `H_(2)O" FORMED "=(36.032)/(4.032)xx3g=26.8g` `O_(2)" left unreacted "=29-23.8=5.2g` |
|
| 92304. |
30 g of CH_(3)MgBr reacts with excess of CH_(3)CH_(2)CH_(2)OH forming xg of a gas. What isthe value of x? |
|
Answer» Solution :`underset((=119g))(CH_(3)MGBR)+CH_(3)CH_(2)underset(OH)underset(|)(CH_(2))rarrunderset((=16g))(CH_(4)(g))+CH_(3)CH_(2)underset(OMgBr)underset(|)(CH_(2))` `119g` of `CH_(3)MgBr` EVOLVE `CH_(4)=16 g` 30 g of `CH_(3)` MgBr evolve `CH_(4)=(30g)XX((16g))/((199g))=4g.` `:.` Value of x=4 |
|
| 92305. |
30 g. of acetic acid is dissolved in 1 dm^(3) of a solvent. The molality of the solution will be (Givent, density of solvent =1.25 g cm^(-3) |
|
Answer» `0.40` |
|
| 92306. |
30 cc of (M)/(3) HCl, 20 cc of (M)/(2) HNO_(3) and 40 cc of (M)/(4) NaOH solutions are mixed and the volumes was made up to 1dm^(3). The pH of the resulting solution is |
|
Answer» 2 Total milliequivalent of `OH^(-) = 40 xx (1)/(4) = 10` Thus, milliequivalents of `H^(+)` LEFT `= 20 - 10 = 10` Total volume of solution `= 1 dm^(3) = 1000 ml` `:. [H^(+)] = (10)/(1000) = 10^(-2)` `rArr PH = -log[H^(+)] = -log [10^(-2)] = 2`. |
|
| 92307. |
30 cc of M/3 HCl, 20 cc of M/2 HNO_(3) and 40 cc of M/4 NaOH solutions are mixed and the volume wasmade up of 1 dm^(3) . The pH of the resulting solution is |
|
Answer» 2 `=20xx1/2+30xx1/3=20` Total milliequivalents of `OH^(-)=40xx1/4=10` Thus, milliequivalents of `OH^(-)=40xx1/4=10` Thus, milliequivalents of `H^(+)`left `=20-10=10` Total VOLUME of solution `=1dm^(3)=1000` ml `:.[H^(+)]=10/1000=10^(-2)` `impliespH=-LOG[H^(+)]=-log[10^(-2)]=2` |
|
| 92309. |
3 xx 10^(3) kg of acetic acid is added to 500 cm^(3) of water. If 25% acetic acid is dissociated in water, what will be DeltaT_(f)K_(f)=1.86 K//m, d of H_(2)O = 1g cm^(-3). |
|
Answer» |
|
| 92310. |
3 Which is the common oxidation state of the first transition series of elements |
| Answer» ANSWER :A | |
| 92311. |
3-Phenylpropene on reaction with HBr gives (as a major produt) |
|
Answer» `C_(6)H_(5)CH_(2)CH(BR)CH_(3)` `C_(6)H_(5)CH_(2)CH=CH_(2)+HBr to C_(6)H_(5)CH_(2)-underset("Br ")underset("| ")("C ")H=CH_(3)` |
|
| 92312. |
3-Pentanol is produced by which of the following reactions? 1. CH_(3)CH_(2)Br overset(Mg)underset("Ether")to overset(CH_(3)CH_(2)OH)to overset(H^(oplus)underset(H_(2)O)to 2.H-=CH overset(NaNH_(2))to overset(CH_(3)CH_(2)CHO)to overset(H^(oplus))underset(H_(2)O)to 3.(CH_(3)CH_(2))_(2)CO overset(LiAlH_(4))underset("Ether")to overset(H^(oplus))underset(H_(2)O)to |
|
Answer» (i) only |
|
| 92313. |
3-Phenylpropene on reaction with HBr gives (as a major product) |
|
Answer» `C_6H_5CH_2CH(Br)CH_3` `C_6H_5CH=CH-CH_3+HBr to oversetoverset(Br)(|)(C_6H_5CHCH_2CH_3)` |
|
| 92314. |
When 3-phenylpropene reacts with HBr in the presence of peroxide, the major product formed is |
|
Answer» `C_(6) H_(5)CH_(2) CH_)(2) CH_(2) Br` |
|
| 92317. |
3-ntrio 2-methyl butane -2-ol is the condensation products of |
|
Answer» ethnal and NITROMETHANE |
|
| 92318. |
Write chemical reactions of affect the 3-Nitrobromobenzene to 3-nitrobenzoic acid transformations. |
Answer» SOLUTION :
|
|
| 92319. |
3-Nitrobromobenzene to 3-nitrobenzoic acid . |
Answer» SOLUTION :
|
|
| 92320. |
3 moles of ideal gas X (C_(p,m) = (5)/(2)) and 2 molesof idealgas Y(C_(p,m) = (7)/(2)R) are takenin vessa andcompressed reversibly and adiabitically, duringthisprocess temperatureof gaseousmixture increasedfrom300 K to 400 K. Calculate change in internal energy (DeltaU) in cal of gaseous mixture (Given R = 2 cal//mol.K) |
|
Answer» `n_(1)C_(V_(1)m)DELTAT + n_(2)C_(V_(1)m)DeltaT` `=[3 XX(3)/(2)R xx 100 ]+[2 xx (5R)/(2)xx100]` `= 450 R + 500 R` `= 950 R` `= 1900 cal` |
|
| 92321. |
3 moles of ethanol react with one mole of phosphorus tribromide to form 3 moles of bromaethane and one mole of X. Which of the following is X? |
|
Answer» `H_(3)PO_(4)` |
|
| 92322. |
3 moles of CO_(2) gas expands isothermally against external pressure of 1 bar. Volume increases from 10 L to 30 L respectively. The system is in thermal contact of surroundings at temperature 15^(@)C. Entropy change in isotehrmal process is: DeltaS=2.303nR" log "((V_(2))/(V_(1))). Q. Select the correct relation: |
|
Answer» `DeltaS_("system")GT0,DeltaS_(surr).=0` |
|
| 92323. |
3 moles of CO_(2) gas expands isothermally against external pressure of 1 bar. Volume increases from 10 L to 30 L respectively. The system is in thermal contact of surroundings at temperature 15^(@)C. Entropy change in isotehrmal process is: DeltaS=2.303nR" log "((V_(2))/(V_(1))). Q. If CO_(2) behaves like an ideal gas, then entropy change of system (DeltaS_("system")) will be: |
|
Answer» `+27.4JK^(-1)` |
|
| 92324. |
3 moles of an ideal gas is compressed from 30 dm^3 to20 dm^3 against a constant pressure of 3.039 xx 10^5Nm^-2. The work done in calories is ___________ . (1 J = 0.239 cal) |
|
Answer» `+30.39 CAL` |
|
| 92325. |
3 moles of A and 4 moles of B are mixed together and allowed to come into equilibrium according to the following reaction :A(g)+4B(g)hArr 2C(g)+3D(g)When equilibrium is reached, there is 1 mole of C. The equilibrium extent of the reaction is |
|
Answer» `1//4` |
|
| 92326. |
3 mole of A and 4 moles of B are mixed together and allowed to come into equilibrium according to the following reaction 3A_((g))+4B_((g))hArr2C_((g))+3D_((g)) |
|
Answer» Solution :`3A+4BhArr2C+3D` `impliesK_(1)=([C]^(2)[D]^(3))/([A]^(3)[B]^(4))""...(i)` When moles of `C=1` the equation becomes `3/2A+2BhArrC+3/2D` `impliesK_(2)=([C][D]^(3//2))/([A]^(3//2)[B]^(2))""...(ii)` On comparing Eq. (i) with Eq. (ii), we GET `K_(2)^(2)=K_(1)^(2)or K_(2)=(K_(1))^(1//2)` HENCE, equlibrium extent`=1/2` |
|
| 92327. |
3 mof of mixture of NaHCO_(3) and Na_(2)CO_(3) is strongly heated at 200°C, when 1 mol of CO2 is obtained. What is the weight of residue? What is molar percentage of NaHCO_(3) and Na_(2)CO_(3) in the mixture? |
| Answer» SOLUTION : RESIDUE : 2 MOLE, | |
| 92328. |
3-Methylpent-2-ene on reaction with HBr in presence of peroxide forms an addition product. The number of possible stereoisomers for the product is |
|
Answer» TWO . With two chiral carbon atoms in the PRODUCT, total number of STEREOISOMERS are `=2^(n)=2^(2)=4`. |
|
| 92329. |
3-methylpentan-3-ol will be prepared from |
|
Answer» ETHYL formate and methyl magnesium bromide `{:(""CH_(2)H_(5)""CH_(2)CH_(3)),(""|""|),(CH_(3)-C-OHgBrunderset(-MG(OC_(2)H_(5))Br)toCH_(3)C=O),(""|),(""OC_(2)H_(5)):}` `{:(""CH_(2)CH_(3)),(""|),(overset(C_(2)H_(5)MgBr)toCH-C-OHgBr),(""|),(""CH_(2)-CH_(3)):}` `{:(""CH_(2)CH_(3)),(""|),(overset(H.OH)toCH_(3)-C-OH),(""|),(""CH_(2)CH_(3)),("""3-methylpentan-3-ol"):}` |
|
| 92330. |
3-Methylpent-2-ene on reaction with HBr in presence of peroxide forms an addition product. The number of possible steroisomers for the product is: |
|
Answer» TWO |
|
| 92331. |
3-menthyl-pent-2-ene on reaction with HBr in presence of peroxide forms an addition product. The number of possiblestereoisomersfor the product is |
|
Answer» six |
|
| 92332. |
3-methylbut-1-eneon reaction with HBr gives (as major product) ? |
|
Answer» `(CH_(3))_(2)CHCH_(2)CH_(2)BR`
|
|
| 92333. |
3 - methyl cyclohexene on oxidation will give |
|
Answer»
|
|
| 92334. |
3-Methyl butane-2-ol on heating with HI gives |
|
Answer» 2-iodo-3-methylbutane |
|
| 92335. |
3-methyl but-1-ene on HBO reaction gives |
|
Answer» 3-methyl butan-2-ol |
|
| 92336. |
3-methyl-2-butanol |
Answer» SOLUTION :
|
|
| 92337. |
3-methyl-2 butanol on treatment with HClgives predominantly |
|
Answer» 2, 2-dimethylpentane |
|
| 92338. |
3-methyl-2-butanol on treatment with HCI gives predominantly : |
|
Answer» 2-chloro-2-methylbutane |
|
| 92339. |
3-Hydroxy butanal is formed when (X) reacts with (Y) in dilute (Z) solution. What are(X),(Y) and(Z)? |
|
Answer» `{:(ULX,ulY,ULZ),(CH_3CHO,(CH_3)_2CO,NAOH):}` |
|
| 92340. |
3- hydroxy 2- methyl pentanal is formed when Xreact with Y in dilute Z solution. What are X,Y,Z |
|
Answer» `{:(""X,""Y,""Z),(CH_(3) - CHO,CH_(3) - CHO,NaOH):}` |
|
| 92341. |
3 gram of non-volatile solute in a 1000 cm^(3) of water shows an osmotic pressure of 2 bar at 300K. Calculate the molar mass of the solute (R=0.0853" L bar "K^(-1)mol^(-1)). |
| Answer» SOLUTION :`37.35" G "MOL^(-1)`. | |
| 92342. |
3 gram of activated charcoal was added to 50 mL of acetic acid solution (0.06N) in a flask. After an hour it was filtered and the strength of the fitrate was found to be 0.042 N. The amount of acetic acid adsorbed (per gram of charcoal) is : |
| Answer» ANSWER :A | |
| 92343. |
3 gm of copper metal in cathode is deposited on electrode when 3 ampere current is pass for 2 hours through aqueous solution of CuSO_(4). Then what is the efficiency of current ? (Atomic weight of Cu=63.5 gm/mol) |
|
Answer» 0.33 `=3xx7200` `=(3xx7200)/(96500)=(3xx72)/(965)F`. . . (i) MOLE=`("weight")/("MOLECULAR weight")=(3gm)/(63.5" gm mole"^(-1))` `Cu_((aq))^(2+)+2e^(-) to Cu_((S))` So, 2F gives 1 mole Cu . . . (Cathodic reaction) So, `(3xx72)/(965)` F Gives Theoretical mole of Cu `=(3xx72)/(965)F((1" mole")/(2F))=(3xx36)/(965)` So, efficiency `=("PRACTICAL value")/("theoretical value")=(3)/(63.5)xx(965)/(3xx36)=0.4221` `=42.21%` |
|
| 92344. |
3g of urea is dissolved in 45g of H_(2)O. The relative lowering in vapour pressure is |
|
Answer» 0.05 `(p_(A)^(@)-p_(A))/(p_(A)^(@))=x_(B)` where `x_(B)` is the mole fraction of solute (UREA) `x_("urea")=(n_("urea"))/(n_("urea")+n_(H_(2)O))~~(n_("urea"))/(n_(H_(2)O))` (for dilute solutions) `n_("urea")=(3)/(60)=0.05` `n_(H_(2)O)=(45)/(15)=2.5` `n_("urea")=(0.05)/(2.5)=0.02` |
|
| 92345. |
3 g of facetic acid is added to 250 mLof 0.1 HCl and the solutions made up to 500mL.To 20 mLof this solutions 5mLof 5M NaOHis added . The pH of the solutions is b/wn& n+ 1value of 'n' is :[Given : pKa of acetic acid = 4.75 molar mass of acetic acid =60 g/mol ] Neglect any changes in volumes. |
|
Answer» |
|
| 92346. |
3 g of activated charcoal was added to 50 mL of acetic acid solution (0.06N) in a flask. After an hour it was filtered and the strength of the filtrate was found to be 0.042N. The amount of acetic acid absorbed (per gram of charcoal) is |
|
Answer» 18 mg FINAL m moles of `CH_(3)COOH=0.043xx50` Hence, MASS of `CH_(3)COOH` adsorbed per gram of charcoal`=((0.06-0.042)xx50xx10^(-3)xx60xx10^(3))/(3)=18mg` |
|
| 92347. |
3 g of activated charcoal was added to 50 mL of acetic acid solution (0.06N) in a flask. After an hour it was fitered and the strength of the filtrate was found to be 0.042N. The amount of acetic acid adsorbed (per gram of charcoal) is: |
|
Answer» 18 mg `= (0.06 XX 50)/(1000) = 3 xx 10 ^(-3)` Moles of acetic acid after ADSORPTION `= (0.042 xx 50)/(1000) = 2.1 xx 10 ^(-3)` Moles of acetic acid adsorbed `= 3.0 xx 10 ^(-3) -2.1 xx 10 ^(-3)` `= 0.9 xx 10 ^(-3)` Mass of acetic acid `=0.9 xx 10 ^(-3) xx 60 = 54 xx 10 ^(-3) g` Amount of acetic acid adsorbed PER gram of charcoal `= (54 xx 10 ^(-3))/(3) = 18 xx 10 ^(-3) g` `or =18` mg |
|
| 92348. |
3 g of activated charcoal was added to 50 mL of acetic acid solution (0.06N) in a flask. After an hour it was filtered and the strength of the filtrate was found to be 0.042N. The amont of acetic acid absorbed (pr gram of charcoal) is |
|
Answer» Solution :Milleq of acetic acid present initially `=0.06xx60=3` millieq of acetic acid left after adsorption `=0.042xx50=2.1` `THEREFORE` millieq of acetic acid adsorbed by 3 g of charcoal `=3-2.1=0.9meq=0.9xx60 mg=54mg` `therefore` Amount of acetic acid adsorbed per GRAM of charcoal `=54/3=18` mg |
|
| 92349. |
3 g of activated charcoal was added to 50 ml . Of acetic acid solution (0.06 N ) in a flask . After an hour it was filtrate was found to be 0.042 N . The amount of acetic acid adsorbed ( per gram of charcoal )is |
| Answer» Solution :m.e of `CH_(3)COOH` adsorbed = `50(0.6 -0.042)` | |
| 92350. |
3 g of a hydrocarbon on combustion in excess of oxygen produces 8.8g of CO_(2) and 5.4 g of H_(2)O. The data illustrates the law of: |
|
Answer» CONSERVATION of mass |
|