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.
| 7801. |
Structure of `N_(2)O` is `N = N = O`. Calculate bond enthalphy of `N = N` bond in `N_(2)O`. Given : `{:(DeltaH_(f,N_(2)O)^(@)=100 "kJmol"^(-1)" "BE_(N=N)=950 kJ mol^(-1)),(BE_(N=0)=600 "kJmol"^(-1)" "BE_(O=O)=500 "kJmol"^(-)):}` and resonance energy of `N_(2)O = - 100 kJ mol^(-1)`A. `500 kJ mol^(-1)`B. `450 kJ mol^(-1)`C. `400 kJ mol^(-1)`D. `420 mol^(-1)` |
| Answer» Correct Answer - C | |
| 7802. |
The electron gain enthalpy (in kJ/mol) of fluorine, chlorine, bromine and iodine, respectively are : (1) – 333, – 349, – 325 and – 296 (2) –296, – 325, – 333 and – 349 (3) – 333, – 325, – 349 and – 296 (4) –349, – 333, – 325 and – 296 |
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Answer» Answer is (1) – 333, – 349, – 325 and – 296 Order of electron gain enthalpy (magnitude) is Cl > F > Br > I Answer is (1) Electron gain enthalpy of Cl > F > Br > I |
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| 7803. |
Electrolysis of copper sulphate solution is performed with copper electrodes: write the equation for the reaction at the anode and cathode |
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Answer» At Cathode — Cu+2(aq) + 2e- → Cu(s) (Reduction) At Anode — \(2\bar OH+\overset{-}{\longrightarrow}\) H2O + 1/2 O2 + 2e- (Oxidation) |
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| 7804. |
Beryllium and magnesium do not give colour to flame whereas other alkaline earth metals do so, why? |
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Answer» Due to the small size, the ionisation enthalpies of Be and Mg are much higher than those of other alkaline earth metals. This means that the valence electrons in beryllium and magnesium are more tightly held by the nucleus. Therefore, they need large amount of energy for excitation of electrons to higher energy levels. Since such a large amount of energy is not available in bunsen flame, therefore, these metals do not impart any colour to the flame. |
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| 7805. |
Explain why alkali and alkaline earth metals cannot be obtained by chemical reduction methods? |
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Answer» Alkali and alkaline earth metals are themselves very strong reducing agents and reducing agents stronger than them are not easily available. Therefore, these metals cannot be obtained by chemical reduction methods. |
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| 7806. |
Choose the false statements and rewrite them. (a) Alkali metals possess both +1 & +2 oxidation states. (b) Lithium is found to be the strongest reducing agents among the alkali metals. (c) Manufacturing of rayon is known as viscose process. (d) Washing soda is used to remove temporary hardness of water. |
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Answer» (a) False. Alkali metals possess only +1 oxidation state. (b) True (c) True (d) False. Washing soda is used to remove the permanent hardness. |
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| 7807. |
Alkali metal halides are all high melting, colourless crystalline solids. 1. Write any other physical property of alkalimetal halides. 2. How alkali metal halides are prepared? |
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Answer» 1. All the alkali metal halides are soluble in water except LiF. They have high negative enthalpies of formation. The melting and boiling points always follow the trend: fluoride > chloride > bromide > iodide. 2. Alkali metal halides are prepared by the reaction of the appropriate oxide, hydroxide or carbonate with aqueous hydrohalic acid. |
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| 7808. |
(a) Write the scientific name of slaked lime. (b) What is its role in white wash? (c) What do you mean by slaking of lime? (d) What happens when slaked lime is treated with dry chlorine? |
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Answer» (a) Calcium hydroxide (b) It is disinfectant hence it is used in white wash. (c) Calcium hydroxide is obtained by adding water to quick lime. This process is called slaking of lime. (d) Calcium hydroxide reacts with dry chlorine to form calcium hypochlorite, a constituent of bleaching powder. 2Ca(OH)2 + 2Cl2 → CaCl2 + CaOCl2 + 2H2O |
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| 7809. |
Which of the alkali metals is having least melting point? Justify.(a) Na (b) K (c) Rb (d) Cs |
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Answer» (d) Cs. As the atomic size of the metal increases, the strength of metallic bonding decreases and hence its melting point decreases. Since the size of Cs is the largest, therefore its melting point is the lowest. |
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| 7810. |
Listen the chemical reaction,1. Write the common name of the reactant. 2. What is the temperature corresponding to A? 3. What is B (product)? Write its chemical formula. |
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Answer» 1. Gypsum 2. 393 K 3. Plaster of Paris (CaSO4 . ½H2O) or (CaSO4) .H2O |
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| 7811. |
A piece of metallic sodium is added to liquid ammonia. 1. What is the observation? 2. What is the reason for this? 3. What happens when the solution is kept for some time? 4. What happens if the solution is concentrated? |
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Answer» 1. The solution turns into blue. 2. The blue colour of the solution is due to the presence of solvated electrons. 3. On standing, the solution slowly liberates hydrogen resulting in the formation of amide. 4. When the concentration of the solution increases the colour turns bronze and the solution becomes diamagnetic. |
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| 7812. |
1. What is the name of the method that is used in manufacturing of sodium hydroxide? 2. Explain the method. 3. Write the equations of the reactions involved in this process. |
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Answer» 1. Castner-Kellner method 2. The Castner-Kellner cell consists of a large tank. The bottom of the tank is filled with mercury. The tank is made into 3 compartments by 2 partitions. The outer compartments are filled with NaCl solution and the middle compartment is filled with a very dilute solution of NaOH. The mercury layer of the bottom of cell in the outer compartments acts as cathode and the mercury in the middle compartments acts as anode. As a result of electrolysis Na and Cl2 are produced in the outer compartments. Na gets coated with mercury, and sodium amalgam is produced in the outer compartments. Due to the action of eccentric wheels at the bottom of the tank, the tank can be tilted. As a result of this Na is carried into the middle compartment and is then allowed to react with water in the middle compartment. As a result of this reaction NaOH is produced. When the concentration of NaOH in the middle compartment is raised to 40%, it is taken out from the tank. On cooling, crystals of NaOH separate out. 3. NaCl → Na+ + Cl- At cathode: Na+ + e- → Na Na + Hg → Na/Hg (Sodium amalgam) At anode: Cl- → Cl + e- Cl + Cl → Cl2 (g) |
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| 7813. |
Carbon dioxide A. Is carried as carboxyhaemoglobin on the haemoglobin molecule.B. Uptake by the blood increases its oxygen-binding power. C. Uptake by the blood leads to similar increases in H+ and HCO3- ion concentrations. D. Stimulates ventilation when breathed at a concentration of 20 per cent. E. Content is greater than oxygen content in arterial blood. |
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Answer» A. False It is carried as carbamino hemoglobin; carboxy hemoglobin is the combination of hemoglobin with carbon monoxide. B. False It decreases as the oxygen dissociation curve is shifted to the right. C. False HCO3- ions increase more; H+ ions are largely buffered by hemoglobin. D. False Breathing 20 per cent CO2 causes respiratory depression; stimulation of respiration by CO2 is maximal when breathed at concentrations of around 5 per cent. E. True CO2 content in arterial blood is about 500 ml/l; O2 content is about 200 ml/l. |
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| 7814. |
Mg2C3 on hydrolysis gives ........ |
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Answer» Mg2C3 on hydrolysis gives Propyne. |
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| 7815. |
The raw materials required for the manufacture of cement clinker are ......... |
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Answer» limestone & clay |
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| 7816. |
In portland cement what enhances the settling time?(a) CaSO4, 1/2 2H2O(b) CaSO4, 2H2O(c) CaCO3(d) CaSO4 |
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Answer» (b) CaSO4, 2H2O Gypsum (CaSO4, 2H2O) is usually added to prevent early hardening and increase the settling time |
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| 7817. |
Portland cement contains 'X' to enhance the setting time. What is 'X'?(A) CaSO4. 1/2 H2O(B) CaSO4.2H2O(C) CaSO4(D) CaCO3 |
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Answer» Correct option is (B) CaSO4.2H2O Gypsum (CaSO4.2H2O) is used to enhance setting time in portland cement. |
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| 7818. |
Origin of replication of DNA in E. coli is shown below, Identify the labelled parts (i),(ii), (iii) and (iv)A. (i)- discontinuous synthesis , (ii)- continuous synthesis (iii) 3’ end (iv) 5’end B. (i)- continuous synthesis , (ii)- discontinuous synthesis (iii) 5’ end (iv) 3’end C. (i)- discontinuous synthesis, (ii)- continuous synthesis (iii) 5’ end (iv) 3’end D. (i)- continuous synthesis , (ii)- discontinuous synthesis (iii) 3’ end (iv) 5’end |
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Answer» Correct option is D. (i)- continuous synthesis , (ii)- discontinuous synthesis (iii) 3’ end (iv) 5’end |
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| 7819. |
Transcription unit is represented in the diagram given below.Identify site (i), factor (ii) and Enzyme (iii) responsible for carrying out the process. A. (i) Promoter Site, (ii) Rho factor (iii) RNA polymerase B. (i) Terminator Site, (ii) Sigma factor (iii) RNA polymerase C. (i) Promoter Site, (ii) Sigma factor (iii) RNA polymerase D. (i) Promoter Site, (ii) Sigma factor (iii) DNA polymerase |
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Answer» Correct option is C: (i) Promotor Site, (ii) Sigma factor (iii) RNA polymerase |
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| 7820. |
Assertion: Adenosine deaminase (ADA) deficiency can be corrected by gene therapy. Reason: Bone narrow transplantation cannot completely cure ADA deficiency. a. Both assertion and reason are true, and reason is the correct explanation of assertion. b. Both assertion and reason are true, but reason is not the correct explanation of assertion. c. Assertion is true but reason is false d. Both assertion and reason are false. |
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Answer» b. Both assertion and reason are true, but reason is not the correct explanation of assertion. |
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| 7821. |
Which one of the following is active in treatment of sewage? (a) Zooplankton (b) Fishes (c) Bacteria (d) All of these |
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Answer» Bacteria is active in treatment of sewage. |
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| 7822. |
In order to form a dinucleotide during DNA synthesis which functional group at 3' must be free? A. Methyl group B. Phosphate group C. Carboxylic acid D. Hydroxyl |
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Answer» Correct option is D. Hydroxyl |
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| 7823. |
Which of the following is not a function of placenta? A. secretes relax in B. facilitates removed of Co2 and waste products C. secretes oxytocin D. supplies oxygen and nutrients |
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Answer» Correct option is C. secretes oxytocin |
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| 7824. |
Deficiency of Adenosine deaminase can be overcomed by- (a) Gene therapy (b) Antibiotics (c) Human growth hormone (d) Producing interferon |
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Answer» Deficiency of Adenosine deaminase can be overcomed by Gene therapy. |
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| 7825. |
Which one of the following is an incorrect statement with regard to pedigree analysis? A. It verifies that DNA is the carrier of genetic information. B. It helps to understand whether the trait depicted in the chart is dominant or recessive. C. It confirms that the trait is linked to one of the autosome. D. It helps to trace the inheritance of a specific trait. |
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Answer» Correct option is A. It verifies that DNA is the carrier of genetic information. |
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| 7826. |
A person suffering from colourblindness can not differentaite between- (a) Red and yellow colour (b) Green and blue colour (c) Red and green colour (d) Any colour |
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Answer» A person suffering from colourblindness can not differentaite between Red and green colour. |
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| 7827. |
Select the incorrect statement regarding pedigree analysisA. Solid symbols show unaffected individualsB. Proband is the person from which case history startsC. It is useful for genetic counsellorsD. It is an alalysis of traits in several generations of a family |
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Answer» Correct Answer - A In pedigree analysis, solid symbol shows affected individuals. |
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| 7828. |
Red green colourblindness is a sex linked trait. Which of the given statements is not correct regarding colourblindness ?A. It is more common in males than in femalesB. Homozygous recessive condition is required for the expression of colourblindness in females.C. Males can be carriers of the traitD. Colourblind women always have colourblind father and always produce colorblined son. |
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Answer» Correct Answer - C Since colourblindness is a sex-linked recessive trait and males just have one X chromosome, they can never be the carriers. Males will always express the disease/phenotype. |
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| 7829. |
The possibility of a female becoming haemophilic is extremely rare because mother of such a female has to be at least `overset("(i)") "____"` and father should be `overset("(ii)") "____"`A. (i) haemophilic, (ii) carrierB. (i) carrier, (ii) haemophilicC. (i) haemophilic, (ii) normalD. (i) haemophilic, (ii) haemophilic |
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Answer» Correct Answer - B Haemophilia is genetically due to the presence of a recessive sex linked gene h, carried by X-chromosome. A female becomes haemophilic only when both its X-chromosomes carry the gene `(X^(h)X^(h))` However. Such females generally die before birth because the combination of these two recessive alleles is lethal. A female having only one allele for normal blood clotting present on the other X-chromosome is dominant. Such females are known as carriers. In case of males, a single gene for the defect is able to express it self as the Y-chromosome is devoid of any corresponding allele (X^(h)Y)`. The possibility of human female becoming haemophilic is extremely rare because she has to be homozygous recessive for the trait, i.e., her father must be haemophilic and mother must be atleast a carrier. |
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| 7830. |
What are the preparations are to be done before waxing. |
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Answer» 1. Position the client according to the area being waxed, taking into account the mobility and comfort of the client. 2. Protect the client’s clothes with towels and disposable paper. 3. When waxing the eye brows, apply eye pads to protect the eyes and petroleum jelly to control the amount of hair being removed. 4. Remove all Jewellery in the areas to be waxed. |
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| 7831. |
Write down the guidelines for conducting a skin sensitivity test before waxing. |
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Answer» Guidelines for conducting a skin sensitivity test before waxing. 1. Sensitivity test should be performed on the forearm as is free from hair, clean and dry. 2. Perform the test 24 hours before the treatment. 3. Check the wax for operating temperature by applying the wax on to the forearm. 4. Observe for any reactions on the skin after removing the wax. |
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| 7832. |
On the sensitive and thicker skin of the client, after investigating, what would you recommend for her? |
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Answer» It is advised to give microdermabrasion or ultrasonic exfoliation treatment on sensitive and thicker skin to remove dead skin cells to even out the skin tone. This helps to stimulate circulation of the blood and make the skin smooth. Procedure: 1. Apply water solution on the skin to loosen surface dead skin cells. 2.Use an ultrasonically activated spatula, lift dead cells of the skin and infuse water solution deep into the skin and hydrate new cells. Contra Indication: 1. Skin diseases 2. Acute infectious diseases 3. Pregnancy |
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| 7833. |
Why is the poem called March? |
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Answer» The poem describes nature as seen in the month of march, so it as called ‘March’. It is spring time. |
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| 7834. |
Name the following:A mineral that is required for keeping our bones healthy. |
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Answer» The minerals that are required to keep our bones healthy are calcium and vitamin D. We need sufficient calcium to keep our bones healthy and vitamin D helps to absorb the calcium. |
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| 7835. |
What is the difference between paneer, dahi and cheese? |
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Answer» Self-incompatibility is a spread mechanism in flowering plants that prevents in breeding and promotes out crossing and response in genetically controlled. In favor cross pollination the in to different male and female gametes. |
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| 7836. |
State two advantages of parallel circuit over the series circtuit. |
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Answer» (i) Each appliance gets the required amount of electric current according to its resistance. (ii) If one appliance is switched on/off others are not affected |
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| 7837. |
Methane gas is also called as …… (a) Marsh gas (b) Mass gas (c) Molecular gas(d) Model gas |
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Answer» (a) Marsh gas |
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| 7838. |
A: Highly electropositive metals are extracted by electrolysis of their fused salts. R: Highly electropositive metals cannot be reduced by chemical reduction methods.A. If both (A) and ( R) are correct and ( R) is the correct explanation of (A).B. If both (A) and ( R) are correct, but ( R) is not the correct explanation of (A).C. If (A) is correct, but ( R) is incorrect.D. If (A) is incorrect, but ( R) is correct. |
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Answer» Correct Answer - D Correct assertion : Highly electropositive metals cannot be extracted by electrolysis of their fused salts. |
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| 7839. |
The pair, in which ions are isoelectronic with Al3+ is: (A) Br and Be2+(B) Cl- and Li+(C) S2- and K+ (D) O2-and Mg2+ |
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Answer» Correct option is (D) O2--and Mg2+ Isoelectronix species have same no. of electrons Al+3, O2-, Mg+2 all have 10 electrons. |
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| 7840. |
The humidity at dry-bulb temperature of 23^oC is 30% and the humidity at dry-bulb temperature of 28^oC is 25%, what is the slope of adiabatic cooling line?(a) -0.01(b) -0.02(c) -0.03(d) -0.05 |
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Answer» Right option is (a) -0.01 Best explanation: Slope of adiabatic cooling line = (0.3 – 0.25)/(23 – 28) = -0.01. |
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| 7841. |
A conductor has `14.4 xx 10^(-19)` coulombs positive charge. The conductor has : (Charge on electron `= 1.6 xx 10^(-19)` coulombsA. 9 electron in excessB. 27 electron in shortC. 27 electrons in excessD. 9 electrons in short |
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Answer» Correct Answer - D Positive charge shows the deficiency of electrons Number of electrons `= (14.4 xx 10^(-19))/(1.6 xx 10^(-19))=9`. |
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| 7842. |
According to Hardy Schultz rule, correct order of flocculation value for Fe(OH)3 sol is : (1) K2CrO4 > K3[Fe(CN)6] > KNO3 > KBr = AlCl3 (2) K3[Fe(CN)6] > K2CrO4 > KNO3 = KBr = AlCl3 (3) K3[Fe(CN)6] < K2CrO4 < KNO3 = KBr = AlCl3 (4) KNO3 > KBr = K2CrO4 > AlCl3 = K3[Fe(CN)6] |
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Answer» Answer: (3) K3[Fe(CN)6] < K2CrO4 < KNO3 = KBr = AlCl3 According to hardy-schultz rule, Coagulation value or flocculation value ∝ 1/Coagulation power |
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| 7843. |
Find how many odd factors are there in the number 1240.1. 22. 43. 54. 16 |
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Answer» Correct Answer - Option 2 : 4 Given: number 1240 Formula used: Odd factors of pa × qb = (a +1)(b + 1), Where p and q should be odd prime numbers Calculation: Factor of 1240 = 23 × 51 × (31)1 ⇒ Odd number are 51 × (31)1 ⇒ Odd factors of pa × qb ⇒ (a +1)(b + 1) ⇒ (1 + 1)(1 + 1) = 4 ∴ 1240 has 4 odd factors |
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| 7844. |
What is power (in watt) of a light source emitting `3.125 xx 10^(18)` photons per second, while the wavelength of light is `620 nm` ? (use `hc = 12400 eVÅ`) |
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Answer» Correct Answer - `1` Let the power be `x` wants. No. of photons emitted per second `xx` Energy of `1` photon `=` Total energy `:. 3.125 xx 10^(18) xx (12400)/(6200) = (x)/(1.6 xx 10^(-19)` `:. x = 1` |
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| 7845. |
A. 7.5 moles B. 0.2 molesC. 0.6 moles D. 0.4 moles |
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Answer» Here is Reaction with Balance 10FeC2O4 + 6KMnO4 + 24H2SO4 ----> 3K2SO4+ 6MnSO4 + 5Fe2(SO4)3 + 24H2O + 20CO2 So we see that 6 moles of KMnO4 is required to oxidize 10 moles of FeC2O4 Then, 1 mole of FeC2O4 would be oxidized by = ? 6/10=0.6 |
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| 7846. |
Write the complete symbol for the atom with the given atomic number (Z) and atomic mass (A):(i) Z = 17, A = 35 (ii) Z = 92, A = 233 (iii) Z = 4, A = 9 |
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Answer» (i) 3517Cl (ii) 23392U (iii) 49Be |
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| 7847. |
The metal which gives photoelectrons most easily is (a) Lithium (b) Sodium(c) Calcium (d) Cesium |
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Answer» The metal which gives photoelectrons most easily is Cesium. |
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| 7848. |
Bohr’s orbits are called stationary states because (a) Electrons in them are stationary (b) Their orbits have fixed radii (c) The electrons in them have fixed energy (d) The protons remain in the nuclei and are stationary |
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Answer» (c) The electrons in them have fixed energy |
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| 7849. |
In Leibig’s method. 0.24 g of organic compound on combustion with dry oxygen produced of 0.62 g of CO2 and 0.11 g of H2O. Determine the percentage composition of the compound. |
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Answer» Mass of organic compound = m = 0.24 g Mass of carbon dioxide formed = 0.62 g Mass of water formed = 0.11 g Percentage of carbon = 12/4 x 0.62/0.24 x 100 = 70.45 Percentage of hydrogen = 2/4 x 0.11/0.24 x 100 = 5.09 Percentage of oxygen = [100 – (70.5 + 5.0)] = 24.46 |
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| 7850. |
The values of dissociation constant of some acids (at 25°C) are given below. The strongest acid in water is:(a) 1.4 x 10–2 (b) 1.6 x 10–4(c) 4.4 x 10–10 (d) 4.3 x 10–7 |
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Answer» Correct option: (a) 1.4 x 10–2 Explanation: More is Ka, lesser is pKa(pKa = – log Ka), more is acidic strength. |
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