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.
| 951. |
Q15. Calculate C-Cl bond enthalpy from the reactionGiven ΔΗ0[CHI = 140KJ mol-1AHO[C-Cl] = 240KJ mol-1AHO[H-Cl] = 430 KJm01-1 |
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Answer» sorry question is misprinted....we have to find bond enthalpy of Cl-Cl |
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| 952. |
Cl97)Which of the following is Homomer ofClCH3 Cl HC CCICH98 Which one of the following structues i n c, |
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Answer» A and C are homomers as in A , the ch3 is in wedge position and cl is in dash position. and C is just the same structure with change in conformation. |
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| 953. |
49. Adiabatic process involves:(BHU 1999)(a) Δυ-0(b) Δν0 |
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Answer» d...Anadiabatic process(Constant heat capacities) is one for which there is no heat transfer between the system and surroundings, i.e dQ=0. ... |
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| 954. |
Explain the following:(i) Isobaric processQ.8(ii) Adiabatic process |
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Answer» Anisobaric processis a thermodynamicprocessin which the pressure remains constant. This is usually obtained by allowing the volume to expand or contract in such a way to neutralize any pressure changes that would be caused by heat transfer Anadiabatic processis one that occurs without transfer of heat or mass of substances between a thermodynamic system and its surroundings. |
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| 955. |
Fig.8.21and BC II Veaertices D, E and F respectively (see Fig. 8.22).and Δ DEF, AB DE, AB 11 DE, BC EFEF. Vertices A, B and C are joined toShow thatquadrilateral ABED is a parallelogramquadrilateral BEFC is a parallelogram(i)(ii) AD I CFand AD CF(x) quadrilateral ACFD is a parallelogramFig. 8.22(V) AC-DF(vi) Δ ABC Δ DEF. |
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| 956. |
ी ही 5है ि न्/ 2 oo Yoo lmown abouf fho /D“fl“w"qfl« |
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Answer» Plywoodis a material manufactured from thin layers or "plies" of woodveneerthat are glued together with adjacent layers having their wood grain rotated up to 90 degrees to one another. Allplywoodsbind resin and wood fibre sheets (cellulose cells are long, strong and thin) to form a composite material. |
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| 957. |
15-WHAT IS NORMALLY STANDSUP AND GROWS DOWN |
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Answer» Plants normally stand up but its roots grow down. Please hit the like button if this helped you. |
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| 958. |
А.А.O Fine DUPSE NAME OFD CH3-CH-CH2=CH2 2-chino butaneсен -- 4 -сб/СИ,В) С) 3 - #1, -су - CH2 – СНАCitace(9) CL-CH, CEC - сну -ctBn -В сне си - снасы-4 с.-С -СН-С.#2 |
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Answer» (1) 2-chloro Butane |
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| 959. |
. A stone-is dropped into a quiet lake and waves move in circles at the s5 cm/s. At the instant when the radius of the circular wave is 8 cm, how fast isthe enclosed area increasing? |
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| 960. |
निम्नलिखित अभिक्रिया तृतीयक ब्यूटिल एथिल ईथर बनाने के लिए उपयुक्त नहीं है।CH,CH,C,HONa + CH-C-Cl ---> CH,-C-OC,H,CH,(1) इस अभिक्रिया का मुख्य उत्पाद क्या होगा?(ii) तृतीयक-ब्यूटिलएथिल ईथर बनाने के लिए उपयुक्त अभिक्रिया लिखिए?CH, |
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| 961. |
CH,CH(Cl)CH(Br)CH,ICH),CCH-CCIC,(1) CH,CI,(u) CHCI, |
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Answer» Could u please post the question in English? |
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| 962. |
C6H5-CH=CH-CH2-Cl |
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Answer» IUPAC NAME: 1-chloro-3-phenyl prop-1-ene Please hit the like button if this helped you |
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| 963. |
12.Complete the following reactions-6) CH, CH OH Conc H,so,(ii) CH,COOH+ NaHCO(iii) CH, +CL sunlight |
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Answer» Like if you find it useful |
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| 964. |
What type of graph is shown below?A Point-slope plotB. Line graphC Scatter plotD Bar graph |
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Answer» its a scatter plot graph |
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| 965. |
Kinetic energy of molecule is directly proportional to(a) temperature(b) pressure(c) Both (a) & (b)(d) Atmospheric pressure |
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Answer» The averagekinetic energyof a collection of gas particles isdirectly proportionalto absolute temperature only. |
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| 966. |
Calculate the mole fraction of benzene in solution containing 30% bymass in carbon tetrachloride. |
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| 967. |
Calculate the mole fraction of Benzene (C6H6) in carbon tetra chloride (CCl4) solution containing 30% by mass of Benzene |
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| 968. |
Qul:31 A Solution containing 1.gg Pu loom & KCL| (M= 74.5 g mol-1) O P O p potonic with asolution Containing 39 pe looml of weeaAssume that Both the solutions have| same temperature |
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Answer» 1 Secondary School Chemistry 5 points A solution containing 1.9g per 100 mL of KCl (M=74.5) is isotonic with a solution containing 3 g per 100 mL of urea ( M= 60). Calculate degree of dissociation of KCl solution. Assume that both the solution have same temperature. Ask for details Follow Report byAditisharma025513.03.2019 solution Log into add a comment Answers  techtro Ambitious Isotonic solutions have same osmotic pressure if osmotic pressure of KCl solution is to calculate the degree of dissociation of KCl solution. Check below solution  2.3 3 votes THANKS 0 Comments Report Log into add a comment kobenhavn Ambitious Answer: 25% Explanation: Isotonic solutions are those solutions which have the same osmotic pressure. If osmotic pressures are equal at the same temperature, concentrations must also be equal.  for non electrolytes such as urea  for electrolytes such as   = osmotic pressure C= concentration R= solution constant T= temperature i= vant hoff factor For urea solution: 3 g of urea is dissolved in 100 ml of solution. For solute A: 1.8 g of A is dissolved in 100 ml of solution.    As      0.25 0 0    Total moles after dissociation = thus    Thus degree of dissociation is 25%. |
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| 969. |
The number of mole of oxygen in one litre of aircontaining 21% oxygen by volume, under standardconditions, is(1) 0.0093 mole(2) 2.10 moles(3) 0.186 mole |
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| 970. |
If 100 ml of 1.0 M NaOH solution is dilutedto 1.0 L, the resulting solution contains-(A) 1 mole of NaOH(B) 0.1 mole of NaOH(C) 10.0 mole of NaOH(D) 0.05 mole of NaOH |
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Answer» According to M1V1=M2V2hence M1=1.0M V1=100ml V2=1000mlhenceM2=M1V1/V2=1*100/1000=0.1 mole of NaOH option b |
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| 971. |
Calculate the boiling point of an aqueous solution containing 1of MgBr, in 200 g of water. (Atomic mass : Mg 24, Br 80)(K, for H20 0.52 K kg mol-1).0.50 g2 |
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Answer» DTf= kf*m dTf= 0.512*(10.50/184)1000/200=0.146’C dTf=freezing pt. of pure water- Freezing pt. of solution 0.146=0’C-Freezing pt. of solution Freezing pt. of solution(containing MgBr2)=-0.146’C |
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| 972. |
The number of moles of oxygen in 1L of aircontaining 21% oxygen by volume, understandard conditions, is(A) 0.0093 mole(B) 2.10 moles(C) 0.186 mole(D) 0.21 mole9.[AIPMT 1995 |
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Answer» One litre of gas consist of 1/22.4 moles. And if 21% of it is oxygen then it means Moles of oxygen = (1/22.4)*(21/100) = 0.009375 moles of oxygen |
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| 973. |
Time taken for an electron to complete onerevolution in the Bohr orbit of hydrogen atom is ____(A)4π^2mr^2/nh(B) nh/4π^2mr(C)nh/4π^2mr^2(D) h/2πmr |
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| 974. |
How many structural isomers can you draw for pentane?What are the two properties of carbon which lead to the huge numberof carbon compounds we see around us?1.2.3. What will be the formula and electron dot structure of cyclopentane? |
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| 975. |
hat are two properties of carbon which lead to the huge number of carbon compounds we seearound us? |
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| 976. |
What are the two properties ofcarbon which lead to the hugenumber of carbosee around us?n compounds we |
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Answer» 1)Tetravalency of carbon, which makes it capable of making bond with 4 other atoms or radicle ....2)Catenation,by which carbon has the capability to bond with other carbon atoms.In other words it is the self linkage property of carbon to make bond with other carbon atoms. |
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| 977. |
okSECTION-B22. Why is it advised to rub magnesium ribbon with sand paper before burning it in air?23. A thermometer has 20 equal divsiom |
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Answer» When it is exposed to air it starts reacting with oxygen and forms metal oxides. With Mg the oxide layer forms a white coating on the outside of the metal thal will slowdown burning. The ribbons should be sanded with the sand paper to remove the layer. hope this answer helps you please like the solution 👍 ✔️ |
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| 978. |
Count the number of H-atoms in 1 gram of H2 gas..... |
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| 979. |
what is the weight of 0.2mole of H2 gas? |
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Answer» Weight = moles × moles mass = 0.2 ×2 = 0.4 |
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| 980. |
SECTION-Anon-metal 'X' is polyatomic and has atomicity 4. Identifyn atom of element 'X' has three orbits around its nucleus. Wholding capacity?Define Health00 L E |
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Answer» A non - metal with atomicity 4 is Phosphorus.The number of electrons in an orbit is 2n^2. Therefore, an element with three orbits will have 2(3)^2 = 18 electrons.Health is a state of complete physical, mental and social well-being and not merely the absence of disease or infirmity. |
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| 981. |
A non-metal X is polyatomic and has atomicity 4.Identity "XAn atom of element 'X' has three orbits around its nucleus. What is its maximum electronholding capacity?Define Health. |
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Answer» Atomicity tells us the number of atoms present in a molecule of an elementHere X has atomicity 4.So, phosphorus is the required element as it also has atomicity 4. |
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| 982. |
100 mL of O2 gas diffuses in 10 s. 100 mL of gas 'Xdiffuses in 't' second. Gas 'X' and time 't' can be(1) H2, 2.5s(3) CO, 10 s(2) SO2 16 s(4) He, 4s |
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| 983. |
100 mL of O2 gas diffuses in 10 s. 100 mL of gasdiffuses in 't' second. Gas 'X' and time 't' can be(1) H2, 2.5s(3) CO, 10 s(2) SO2, 16 s(4) He, 4 s |
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| 984. |
100 mL of O2 gas diffuses in 10 s. 100 mL of gas 'Xdiffuses in 't' second. Gas 'X' and time 't' can be(1) H2, 2.5 s(3) CO, 10 s(2) SO2, 16 s(4) He, 4 s |
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Answer» (1/10) / (1/t) = sqrt(Mx / 32)whereMx = (32/100)*t^2 one equation with 2 unknown...? It's a diophantine equation with several solutions:t.......Mx5.......815.....7220....12825....20030....288 |
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| 985. |
21.The maximum number of molecules are presentIAIPMT 2004]In(A) 15 L of N2 gas at STP(B) 5L of N2 gas at STP(C) 0.5 g of H2 gas(D) 10 g of 02 gas |
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Answer» (i) Number of moles in 15 L of N2gas at STP. = (15 L / 22.4 L) = 0.67 moles (ii)Number of moles in 5 L of N2gas atSTP. = (5 L / 22.4 L) = 0.22 moles (iii) Number of moles in0.5 g of H2gas = (0.5 g / 2 g mol-1) = 0.25 moles (iv) Number of moles in10 g of O2gas = (10 g / 32gmol-1) = 0.312 moles Hence, as we know that 1 mole of any gas has Avogadro number of molecules (6.022 x 1023) in it, we can say that the higher number of moles i.e. in 15 L of N2gas will contain maximum number of molecules, as it has higher number of molar quantity. |
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| 986. |
gThe maximum number of molecules is present in5 L of H2 gas at S.T.PC) 0.5 g of H2 gasB) 5 L of N2 gas at S.T.P.D) 10 g of O2 gas |
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Answer» 1 mole of H2 has volume → 22.4 L so, in 15L → moles = 15/22.4 = 0.67 mol similarly for 5L , mole → 5/22.4 = 0.22 mol 0.5g H2 has mole = 0.5/2 = 0.25 mol and 10g Of O2 has mol = 10/32 = 0.33 so, option A |
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| 987. |
lAir (1 atm.)A gas is filled in the glass bulb of capacity 10 litre as shown infigure at 300 K.Density of glycerine 2.72 g/ml.Density of mercury 13.6 g/ml.Now, select the correct statement(s) among the following:(A) pressure of the gas is 3.28 atm(B) pressure of the gas is 0.342 atm(C) moles of the gas in the bulb are 0.34(D) moles of the gas in the bulb are 1.3310 lit.2.5 mGlicerine |
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| 988. |
34 11. 10 dm3 of N2 gas and 10 dm of gas X at the same temeperature contain the same number of molecalet.?isISCOB) CO2C) H2D) NO |
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| 989. |
10. A gas at 2c has pressure 2 aim. Caicuiate the temperature at which the pressure of gasbecomes dotibles Rir same oomaine¡whch its enclosed. |
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| 990. |
Gas laws and ideal gas aoA sample of gas occupies 10 L under a pressureof 1 atm. What will be its volume if the pressureis increased to 2 atm? Assuming that temperatureof the gas sample does not change |
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Answer» When temperature is constant , gas law states that P1*V1 = P2*V2 => 10*1 = 2*V2 so, V2 = 10/2 = 5L. Thank you so much |
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| 991. |
Q.10 Using the equation of state PV niRT show that at a given temperature, the density of the gas is proportionai to the gaspressure P |
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Answer» The equation of state is given by, pV = nRT ……….. (i) Where, p → Pressure of gas V → Volume of gas n→ Number of moles of gas R → Gas constant T → Temperature of gas From equation (i) we have, p = n RT/V Where n= Mass of gas(m)/ Molar mass of gas(M) Putting value of n in the equation, we have p = m RT/ MV ------------(ii) Now density(ρ) = m /V ----------------(iii) Putting (iii) in (ii) we get P = ρ RT / M OR ρ = PM / RT Hence, at a given temperature, the density (ρ) of gas is proportional to its pressure (P) |
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| 992. |
current through the coil is doubled and the radius of the coil is halved? |
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Answer» At the center of a current carrying coil, the magnetic field intensity is directly proportional to the current and inversely proportional to the radius of the coil.So, if the current is doubled and radius is halved, then the value of B increases by 4 times the initial value. |
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| 993. |
&+’ -3:-2(0.01 नारे=T a2 X" +x—-6 |
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| 994. |
ung dcius, () twWhy NaOH and KOH pellets should not be left exposed to air? |
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Answer» Sodium Hydroxide and potassium hydroxide are hygroscopic and readily absorbs carbon dioxide from the air and forms Sodium Carbonate and potassiym carbonate respectively, that's why, they should not be left exposed in the air |
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| 995. |
nt fiuno Should ba lotpont intheungo-02856 X298, 15 x 0.112O SI8S |
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Answer» Least precise term i.e. 0.112 is having 3 significant digits. ∴ There will be 3 significant figures in the calculation. thanks least number after decimal is two hence there will be two significant figure in the calculation |
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| 996. |
Calculate K, for the reaction,CH, (g) + H2 (g)aio i#-100 kJ ınol-1 at 25°C.19C,H, (g) |
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Answer» Thefree energychange of the reaction in any state, ΔG (when equilibrium has not been attained) is related to the standardfree energychange of the reaction, ΔG0(which is equal to thedifferenceinfree energiesof formation of the products and reactants both in their standard states) according to the equation. use value of R as 8.3145temp: 273+25 = 298Kand the value of free gibb's energy given and using the equation given above, solve the question. |
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| 997. |
31. ग्लिसरॉल है एक : . R(3) प्राइमरी अल्कोहल (8) सेकेण्डरी अल्कोहल ()(020-टर्शियरी अल्कोहल (0) ट्राइहाइडिक अल्कोहल 15, सा! |
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Answer» primary alcoholप्राइमरी अल्कोहल your wrong answer |
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| 998. |
Assuming that sea water is a 3.50 wt% aqueous solution of NaCl. What is the molality of sea water? |
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Answer» Assuming 100 g of sea water,it contains 3.5 g of Nacl.No of moles of Nacl=3.5/58.4=0.06 molesso for 100 gm of sea water,no of moles is 0.06so molality=0.06/100=0.0006let me know if this right and it is not,I'll revise it for you. |
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| 999. |
Which of the following is the best conductor ofElectricity?(a) Ordinary water (b) Sea water(c) Boiled water d) Distilled water48. |
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Answer» This is becausesalt wateris agood conductor of electricity.Saltmolecules are made of sodium ions and chlorine ions. (An ion is an atom that has anelectricalcharge because it has either gained or lost an electron.). These ions are what carryelectricity throughwater. Like my answer if you find it useful! |
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| 1000. |
ctnwhich.isitetow-seawater orfreshwater why3 In which water is it easier to swim, sea water or fresh water. Why? |
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Answer» It is easier to swim in sea water than in a river because, the sea water contains salt which increases the density of water and also increases its upthrust so, the body sinks less in it and we swim easily |
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