NTPC ACT 2018 Memory-Based Questions | NTPC Assistant Chemist Trainee Recruitment 2026 | Part 1

[00:00:00]Hello, everyone. Welcome back to Chemistry with the I shall back to N 2 V C Assistant Chemistry trainer 2018 exam water Chemistry subject may with the memory based and model question given a minute and about 10 minutes total 50 question Chemistry say and 10 question the executive aptitude test water with me say I much bigger than you So that a question what about that blood of P upon V into P P not minus P versus P upon P not the slope at intercept hello that I can I and the mono layer volume kitna over VM you find out connect earlier water to other update equation Korea Korea the equation come formula check P upon P not minus P into one upon V A D equal C minus one upon C V M into P upon P not plus one upon C V M is equation for other straight line equation say compare correct to you okay yeah why yeah yeah slope M yeah X and the intercept C to slope the value that I can I slope 0.40 C minus one upon C V M and intercept hello yeah 0.10 equation say intercept kitna hai one upon C V M I got home slope upon intercept calculate correct calculate that it 0.40 divided by 0.10 = c - 1 upon CVM into CVM divided by 1 = c - 1 next c - 1 = either I got four c - 1 = 4 c = five So I'm I'm going to be in the budget.

[00:02:32]So 1 upon CVM 0.10 into five equal 1 upon VM VM equal 2.0 centimeter cube gram inverse so answer is that option A correct answer 2.0 centimeter cube gram inverse Next other I'm taking question two which one of the following pairs of nucleoids are isotones? isotones isotones guy same number of neutron same number of neutron Okay, neutron density because that neutron equal mass number minus atomic number Okay.

[00:03:39]14 C6 is going to be 15 over 14 minus 6 equal 8 and 15 7 nitrogen is going to be 15 over 15 minus 7 8 So mass number but atomic number defined hey mass number of a same same over to get over isobar mass number defined hey atomic number same hey isoto okay next question assertion a molecule lacking any plane of symmetry cannot exist as a major compound your policy major compound does not have to possess a plane of symmetry it only needs to be a chiral due to an internal symmetry element that symmetry element may be plane of symmetry be also your most common okay center of symmetry and improper rotation axis SN hence the absence of a plane of symmetry alone does not rule out a major compound other plane of symmetry present in here so you put me over something key or major compound may over I got sent I got molecule the center of symmetry present in here or improper axis of plane of symmetry present in here to be all molecule major compound over okay so over Okay.

[00:05:51]So, assertion false.

[00:05:56]Next, a meso compound is optically inactive because it possesses internal compensation arising from molecular symmetry. Internal compensation compensation plane of symmetry, center of symmetry, improper axis of symmetry. So, correct answer is D. A is false, but R is true. Okay?

[00:06:22]Next question.

[00:06:24]According to Fajan's rule, arrange the following compound in the increasing order of covalent character.

[00:06:32]So, according to Fajan's rule, covalent character increases with the polarizing power increases.

[00:06:52]Polarizing power of the cation increases.

[00:06:55]Polarizing power increases when the cation has smaller ionic size.

[00:07:08]Okay. Next, higher positive charge.

[00:07:19]So, polarizing power of cation increases when smaller ionic charge and higher positive charge.

[00:07:26]So, magnesium 2 plus the charge that I had So, it's a polarizing power substitute that over polarizing power substitute that over magnesium 2 plus then over smaller ionic Then lithium plus then sodium plus then.

[00:07:50]Potassium plus potassium plus the polarizing power subsidy come over. So hence the covalent character.

[00:07:58]Covalent character color guy over.

[00:08:07]KCL got subsidy come over.

[00:08:10]Next.

[00:08:11]NaCl.

[00:08:13]Next.

[00:08:14]LiCl.

[00:08:16]Next mgcl2.

[00:08:21]MgCl2. Okay.

[00:08:23]Option B correct answer.

[00:08:27]Next question.

[00:08:29]Which one of the following nuclei is expected to have a nuclear spin quantum number. Equal to zero. I equal zero. I equal zero over two.

[00:08:40]In a man.

[00:08:42]Inactive over. Okay.

[00:08:45]13 C guy get night.

[00:08:48]Half.

[00:08:50]17 over I get night.

[00:08:53]5 by 2.

[00:08:54]And 15 nitrogen guy.

[00:08:57]1 by 2. 16 over I.

[00:09:00]Zero over two key.

[00:09:03]16 8.

[00:09:05]Is guy.

[00:09:07]Proton B 88 neutron B.

[00:09:10]88.

[00:09:13]A nucleus.

[00:09:16]With an.

[00:09:20]Even.

[00:09:21]Number of.

[00:09:25]Proton.

[00:09:27]And.

[00:09:30]An even number of.

[00:09:34]Neutrons.

[00:09:38]Even.

[00:09:40]even nucleus has I equal zero and NMR inactive.

[00:09:51]So, correct answer option D.

[00:09:55]Next question.

[00:09:57]The mass of copper obtained from 100 g of pure CuSO4 is approximately atomic mass is 63.5 and sulfur 32 and O 16.

[00:10:10]So, CuSO4 car molar mass kitna hoga? 63.5 plus sulfur 32 then oxygen car 16 into four.

[00:10:23]Total hoga 159.5 g g mol inverse.

[00:10:30]Okay?

[00:10:32]In 159.5 g of CuSO4 is my copper kitna hai?

[00:10:42]63.5 g I got 100 g of CuSO4 hoga the mass of copper kitna hoga?

[00:10:57]63.5 divided by 159.5 into 100 g a calculation karne ke baad aayega 39.81 g approximately 39.8 g. Okay? Option B correct answer.

[00:11:20]Next question.

[00:11:21]For a gas obeying the van der Waals equation, the ratio PCVC RTC is the value nikalna padega.

[00:11:30]So, for a van der Waals gas ke liye, PC ka value kitna hota hai? A by Okay.

[00:11:53]equal Okay.

[00:12:21]Next question they collected. The common name of methanoic acid is B, formic acid. Okay. Next question, balance the following disproportionation reaction. Okay.

[00:12:39]P4 plus 3OH- plus 3H2O H2 PO2- plus PH3 So, A value here one B 3 C 3 D 3 E 1 So, A plus B plus C plus D plus E equal 1 plus 3 plus 3 plus 3 plus 1 equal 11 Answer A, correct answer. Next question, a solution is prepared by dissolving 110 g of urea molar mass 60 g mol inverse in 1000 g of water. If the density of the solution is 1.15 g ml inverse, what is the molarity of the solution?

[00:13:39]So, calculate moles of urea.

[00:13:45]It now will be 110 g divided by 60 g mol inverse.

[00:13:53]Okay.

[00:13:54]Calculate karne ke baad I got 1.8 33 mol.

[00:14:01]Next, solution ka mass kitna hoga? Mass of solution.

[00:14:07]Solution mein kya kya hai? Urea hai 110 g plus water hai 1000 g.

[00:14:17]Equal 1100 10 g.

[00:14:23]Ab hum volume nikalenge. Solution ka volume kitna hoga? Guess where I got density equal mass upon volume.

[00:14:38]So, volume equal mass upon density.

[00:14:47]Density kitna hai?

[00:14:52]1.15 g ml inverse.

[00:14:58]Or mass where 1110 g.

[00:15:04]So, calculate karne ke baad I got 965.2 ml. Isko agar liter mein convert kare, 0.9652 liter.

[00:15:17]Okay.

[00:15:18]Abhi molarity nikalna hai. Molarity equal guy.

[00:15:23]Molarity equal moles of solute divided by volume of solution in liter.

[00:15:36]Volume of solution in liter, okay?

[00:15:43]Moles of solute 1.8 33 1.833 divided by 0.96 52 liter a mole.

[00:15:57]So, calculate about I got 1.

[00:16:02]91 molar.

[00:16:06]So, molar equal mole liter inverse.

[00:16:11]Okay, answer I got 1.91 molar answer B.

[00:16:19]Next question.

[00:16:21]Next question.

[00:16:23]The correct order of CF bond length in the following compound is So, the first compound is CH3 C F F F, okay?

[00:16:42]Next compound is CH3 C H F F Next about CH3 C H H F This is the distance here.

[00:17:04]This is the order to calculate the order.

[00:17:07]So, CF bond length depends on the electron density around the central carbon atom.

[00:17:14]This is the central carbon atom here.

[00:17:16]This may the other electron density will be higher.

[00:19:12]>> regulates fight or flight response.

[00:19:28]Okay.

[00:19:30]Next, thyroxine hormone regulates metabolism.

[00:19:42]Thyroxine regulate metabolism.

[00:19:46]And antidiuretic hormone your vasopressin B or promote reabsorption of water from the kidney tubules and collecting ducts. So, correct answer, option C.

[00:20:03]Next question, the potential difference between the two electrodes of an electrochemical cell under open circuit conditions when no current is drawn from the cell is known as Okay.

[00:20:20]Electrode potential, electrode potential voltage potential of a single electrode relative to a reference electrode.

[00:20:46]Okay.

[00:20:47]Cell phone voltage under load voltage when current is flowing it is usually less than the electromotive force due to internal resistance.

[00:21:30]Overpotential again, extra potential required to drive an electrode reaction at a desired rate.

[00:21:59]Okay.

[00:22:01]Next, electromotive force of the cell.

[00:22:04]Electromotive force is the maximum potential difference between the electrodes of a cell when no current flow through the external circuit.

[00:22:16]Or open circuit condition may no current flows.

[00:22:26]Okay. When no current is drawn from the cell, so correct answer C.

[00:22:32]Electromotive force of the cell.

[00:22:35]Next question, the oxidation state of iodine in I2, ICL, and HIO4.

[00:22:43]Oxidation state I2 kitna hai?

[00:22:46]Elemental iodine exists in it is free state. So, oxidation state of I equal zero. Okay.

[00:23:00]ICL I and Cl may I N C L may chlorine is more electronegative than iodine. So, chlorine equal minus one.

[00:23:17]X plus minus one equal zero.

[00:23:20]X equal plus one. So, I didn't go oxidation state I C L make it my plus one.

[00:23:26]Next H I O 4 H I O 4 I double bond O double bond O O H double bond O to I didn't go oxidation state of X 1 X plus three double bond O here to three into minus two and one O H here one into minus one equal zero X minus six minus one equal zero X equal plus seven. So, oxidation state keep my zero plus one and plus seven. Zero plus one and plus seven option A correct answer.

[00:24:18]A chiral alcohol reacts with S O C L 2 without pyridine.

[00:24:24]The product is formed with job we chiral alcohol reacts with final chloride in absence of pyridine the reaction proceeds through the S N I substitution nucleophilic internal mechanism, okay.

[00:24:42]S O C L 2 plus pyridine over the S N 2 mechanism follow for the to is make it over inversion S N 2 make it over inversion.

[00:24:58]S O C L 2 pyridine or product over retention.

[00:25:12]Okay?

[00:25:14]So, correct answer option A, retention of configuration.

[00:25:20]Is may at first formation of an alkyl chlorosulfite intermediate over heat chloride ion attacks from the same side.

[00:25:31]Is stereochemistry retained over?

[00:25:36]Phosphorus pentachloride dissociate to the extent of 8% according to the equilibrium.

[00:25:43]If the equilibrium constant KP equal 1.6 the total pressure of the system is So, degree of dissociation equal 8% equal 8 by 100 equal 0.

[00:26:06]08 Okay?

[00:26:09]Equilibrium reaction guy PCL5 gas equilibrium PCL3 gas plus CL2 gas Is color initial pressure P over in the local zero?

[00:26:28]at equilibrium Is color P into 1 minus alpha alpha degree of dissociation?

[00:26:35]Is color P alpha is color P alpha.

[00:26:39]So, KP get now over.

[00:26:41]P alpha into P alpha divided by P into 1 minus alpha equal alpha square into P divided by 1 minus alpha.

[00:26:55]So, KP value 1.6 alpha value 0.08 I square into P or niche 1 minus 0. 08 to calculate total pressure I got 230 ATM.

[00:27:15]Okay.

[00:27:16]Answer B correct answer.

[00:27:20]Next question in the parallel first order reaction A to B and A to C the concentration of product at a given time found to be B equal to mole liter inverse and C equals 6 mole liter inverse. The ratio of K1 and K2 a parallel first order reaction A K1 K2 So parallel first order reaction may concentration of B divided by concentration of C equal K1 upon K2 Okay.

[00:28:04]Concentration of B 2 mole liter inverse concentration of C 6 mole liter inverse equal K1 upon K2 So K1 upon K2 value 1 1 by 3 1 is to 3 Correct option answer B Next question. Consider the following statements regarding the Lewis concept of acid and bases. Statement one, every Lewis acid is an electron pair acceptor, but not every Lewis acid is a Bronsted acid.

[00:28:50]The Bronsted acid guy proton or H+ donor.

[00:29:03]All Lewis acid guy electron pair acceptor. So, substance like BF3, AlCl3, Fe3+ are Lewis acid because they accept electron pair. Although they do not donate H+.

[00:29:20]So, statement one true.

[00:29:24]Next statement, statement two, BF3, AlCl3, Fe3+ behave as Lewis acid because they can accept an electron pair from suitable donors.

[00:29:38]Okay. BF3 has an incomplete octet and accept a lone pair.

[00:29:48]AlCl3 also AlCl3 has only six electrons around aluminum and accept an electron pair.

[00:30:00]And Fe3+ accept lone pair from the ligands to form coordination complexes. Hence, all behave as Lewis acid.

[00:30:14]So, all act as Lewis acids.

[00:30:20]So, a statement two also correct. Both statement one and statement two are true. Both statement one and statement two are true.

[00:30:31]And it's going to be answer of option A.

[00:30:34]Statement two is the correct explanation of statement one. So, why is statement two the correct explanation. Statement two provides example that illustrate the fact state in statement one that Lewis acid are electron pair acceptor and need not be Bronsted acid.

[00:30:57]On bromination of toluene using BR2FE, the methyl group activates the benzene ring primarily because of The CH3 group is an electron donating group.

[00:31:17]It activates the benzene ring mainly through hyperconjugation. You have the three hydrogen here.

[00:31:24]Yeah, hyperconjugation will participate participate three alpha hydrogen.

[00:31:30]Or benzene ring will activate it.

[00:31:33]This increase the electron density at the ortho and para position making electrophilic substitution reaction such a bromination faster.

[00:31:43]Minus I effect Alkyl group have plus weak pi plus I effect. Okay.

[00:31:54]Minus M effect The methyl group cannot show a minus M effect because it has no lone pair or pi bond.

[00:32:05]Pi bond be negative. So, minus I effect may participate negative.

[00:32:10]On electromeric effect you negative.

[00:32:12]Electromeric effect operates only in the presence of an attacking reagent and it is not possible for activation.

[00:32:21]Answer will be B, hyperconjugation.

[00:32:26]Next question, which one of the following is an aromatic amino acid?

[00:32:35]C NH2 COOH hydrogen hydrogen You would like glycine.

[00:32:48]Next alanine case structure yoga other amino acid basic structure drop in What's the weather is fine CS3 Lego?

[00:33:04]Do you over?

[00:33:05]alanine Is it about female alanine is the structure yoga?

[00:33:14]Is fine CS2 whiskey sub engine ring?

[00:33:19]at the show girl Do you over female?

[00:33:29]alanine Are you aromatic with a cookie split engine being?

[00:33:35]Yeah, aromatic so correct answer.

[00:33:38]C I'll tell you the structure yoga.

[00:33:42]basic structure drop on the keyboard CWH H NH2 What's the weather is fine CH?

[00:33:54]CS3 CS3 over Do You over value to your aromatic name over.

[00:34:02]Do important point.

[00:34:05]other amino acid cup Joe 20 to 21 to amino acid a Is the structure case you got lucky?

[00:34:15]Tricks a lot is topic with you today to comment garden. I had my question.

[00:34:22]1 to 20 is video make over here.

[00:34:27]Okay.

[00:34:29]other Your part one over.

[00:34:33]I got up up I got I got part two video tell you.

[00:34:40]The video go like garden now share garden and subscribe garden and comment be.

[00:34:48]Part two video.

[00:34:50]Comment garden.

[00:34:51]Thank you so much video last video tell you.