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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.
| 8751. |
17 gof NH_(3) gaswill occupy a volume of _____________ cm^(3)at NTP. |
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| 8754. |
16 grams of NaOH is dissolved in 100 grams of water at 25^@ C to form a saturated solution. Find the mass percentage of solute and solvent.Mass of the solute (NaOH) = 16 g Mass of the solvent H_2 O = 100 g |
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Answer» Solution :(i) Mass percentage of the solute Masspercentageof solute =`("Mass of the solute")/("Mass of the solute +Mass of THESOLVENT") XX 100` `=( 16 xx 100 )/( 16 +100 )=(1600 )/( 116)` Masspercentageof the solute`= 13.79 % ` (II )Mass percentageof solvent =100 - ( Masspercentageof thesolute ) `=100-13.79 %= 86.21 %` |
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| 8755. |
16 g of an ideal gas SO_(x) occupies 5.6 L at STP. What is its molecular mass ? What is the value of X ? |
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| 8756. |
15 ml of water and10 ml of Sulphuric acid are mixed in a beaker. (i) State the method that should be followed with reason. (ii) What is the process called ? |
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Answer» Solution :The ACID must be added slowly to WATER with constant stirring to PREVENT the mixture to be splashed out. The reaction is highly exothermic, therefore glass container may also break due to excessive heating. (ii) The PROCESS is called DILUTION. |
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| 8757. |
1.5 g of solute is dissolved in 15 g of water to form a saturated solution at 298K. Find out the solubility of the solute at the temperature. |
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Answer» Solution :Massof thesolute= 1.5 g Massof THESOLVENT =15G Solubilityof thesolute `= ("Mass of thesolute ")/("Massof thesolvent ") XX 100` Solubilityof thesolute `=(1.5 )/( 15 ) xx 100 =10 g` |
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| 8759. |
14.3 g of Na_(2)CO_(3) xH_(2)O is dissolved in water and the volume is made up to 200 mL. 20 mL of this solution required 40 mL of (N)/(4) HNO_(3) for complete neutralisation. Calculate x |
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Answer» SOLUTION :`V_(1) N_(1) = N_(2) V_(2)` `20 (N_(1)) = 40 XX 0.25` `N_(1) = 0.5 N` Normality of `Na_(2)CO_(3) XH_(2)O` solution is 0.5 `0.5 = (14.3)/("Eq.WT") xx (1000)/(200)` Eq. Wt `= (14.3 xx 5)/(0.5)` Eq. Wt = 143 But equivalent weigth `Na_(2) CO_(3) XH_(2)O` is `(106 + x(18))/(2) = 143` `:. x = (286 - 106)/(18)` `x= 10` `:. Na_(2) CO_(3) . 10H_(2)O` |
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| 8760. |
10ml of water and 5ml of sulphuric acid are to be mixed in a beaker. (i) State the method that should be followed. (ii) Why should this method be followed. (iii) What is this process called? |
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Answer» Solution :(i) The ACID must SLOWLY be added to water. (ii) The MIXTURE may SPLASH out causing burns, as lot of heat is generated in this process. (III) Dilution of the acid |
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| 8761. |
10g of argon is compressed isothermally and reversibly at a temperature of 27^(@)C from 10 L to 51. Calculate q, w, Delta U and Delta H. |
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Answer» Solution :`Q = -2.303 nRT log V_(2)//V_(1) = -2.303 xx 10//40 MOL xx 2 "cal" K^(-1) mol^(-1) xx 300K xx log 5//10 = - 103.635` Cal For isothermal compression `Delta U = 0` `W = Delta U -q = 0- (-103.635) = + 103.635` Cal Also when temperature is CONSTANT, PV = constant, `Delta H = Delta U + Delta (PV) = 0+ 0 =0` |
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| 8762. |
100% pure ethanol is known as……………….. |
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Answer» POWER alcohol |
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| 8764. |
100 ml of water can dissolve 36g of NaCl at 25^@ C to attain saturation. |
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| 8765. |
100 mL of 0.05 M aqueous CuSO_(4) solution is subjected to electrolysis. Calculate the quantity of electricity required for the deposition of entire copper at the cathode. |
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Answer» 0.005 F `rArr W_(Cu) = (0.05 xx 63.5)/(10) = 0.3175 G` NO. of EQUIVALENTS `= (0.3175)/(31.75) = 0.01` `:.` No of FARADAYS is 0.01 F |
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| 8766. |
10.0 g of CaCO_3on heating gave 4.4 g of CO_2 and 5.6g of CaO. The observation is in agreement with the |
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Answer» law of constant COMPOSITION mass of REACTANT = 10.0 g mass of products = mass of CaO + mass of `CO_2` = 5.6 +4.4 = 10.0 g As mass of reactant = mass of products. The OBSERVATION is in agreement with law of conservation of mass. |
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| 8767. |
10 mol of Zn react with 10 mol of HCl. Calculate the number ofmoles of H_(2)produced. |
| Answer» ANSWER :A | |
| 8768. |
10 mL of a solution of NoOH is found to be completely neutralised by 8 mL of a given solution of HCl. If we take 20 mL of the same solution of NaOH, the amount of HCl solution (the same solution as before ) is required to be neutralised . It will be |
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Answer» 4 mL |
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| 8769. |
10 mL of a solution of NaOH is found to be completely neutralised by 8 mL of a given solution of HCl. If we take 20 mL of the same solution of NaOH, the amount HCl solution (the same solution as before) required to neutralise it will be |
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Answer» Solution :The amount of HCl required will increase proportionately because the concentrations of both the NAOH and HCl solutions remain the same. The amount of HCl required will be ` 8 xx (20)/(10) = 16 mL` THUS option (d) is correct. |
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| 8770. |
10 mL of 1 N HCl, 25 mL of 2 N H_(2)SO_(4) and 40 mL of X N HNO_(3) are mixed and made up to 2000 mL. 100 mL of this solution required 30 mL of NaOH taken from a solution containing 4 g of NaOH in 250 mL of solution. What is the value of X ? |
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Answer» Solution :Normality of NaOH `N_(1) = (4)/(10) xx (1000)/(250)` `N_(1) = 0.4 N` `:. N_(1) V_(1) = N_("mix") V_("mix")` `0.4 xx 30 = N_("mix") xx 100` `N_("mix") = 0.12 N` `((10 xx 1) + (25 xx 2) + (40 xx X))/(2000) = 0.12 N` `240 = 60 + 40x` `180 = 40x` `x = (180)/(40) = 4.5` `N_(HNO_(3)) = 4.5 N` |
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| 8771. |
(1) The probability of finding an electron is maxium at certain places in space (2) Energy change takes place only during excitation or deexcitation of electron (3) Electron has both particle and wave nature (4) Angular momentum of the electron revolving in different ellipticals orbits is quantised Arrange the above statements of different models or principals in the chronological order , which ultimately led to the development of modern structure of the atom |
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Answer» 1,2,3,4 Angular momentum of the electrons revolv-ing in different ELLIPTICAL ORBITS is quantised. Electron has both particle and wave nature. ltbrrgt The PROBABILITY of FINDING an electron is maxi- mum at certain places in space. |
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| 8772. |
Write whether the statement is true or false?1 mole of C_(12)H_(22)O_(11)contain 22 hydrogen atoms. |
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| 8774. |
1 mole of any substance contains ……………..molecules. |
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Answer» `6.023xx10^(23)` |
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| 8775. |
1 mole of an ideal gas expand isothermally and reversibly from a pressure of 10 atm to 1 atm at 300K. Calculate the height to which an object of 50 kg can be lifed by this expansion. |
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Answer» Solution :`w_(("exp")) = -2.303nRT` `LOG P_(i)//P_(f) = - 2.303 xx 1 MOL xx 8.314 JK^(-1) mol^(-1) xx 300K xx log 10//1 = 5.74 xx 10^(3)J` Now, mgh `= 5.74 xx 10^(3) J or 50 kg xx 9.81 MS^(-2) xx h = 5.744 xx 10^(3)J` `:. h = 11.7m` |
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| 8776. |
1 mole of a substance (atoms ,ions,molecules) is equal to |
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Answer» `6.022 XX 10^23` ATOMS |
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| 8777. |
1 mole of gas at STP is equal to ..... L |
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Answer» 11.2 L |
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| 8779. |
{:(1. "Galvanisation",,(a) "Noble gas elements"),(2."Calcination ",,"(b) Coating with Zn"),("3. Redox reaction",,"(c) Silver-tin amalgam"),("4. Dental filling",,"(d) Alumino thermic process"),("5.Group 18 elements",,"(e) Heating in the absence of air"):} |
| Answer» Solution :1-(b), 2-(E ), 3-(d), 4-(c ), 5-(a) | |
| 8781. |
1 faraday = _____ coulombs |
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Answer» 10000 |
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| 8782. |
(1) As long as an electron revolves in a particular orbit , the electron does not lose its energy Therefore, these orbits are called stationary orbits and the electrons are said to be in station-ary energy state (2) Each orbit or shell is associated with a definite amount of energy . Hence , these are also called energy levels An electron jumps from a lower energy level to a higher energy level , by absorting energy . It jumps from a higher energy to a lowerlevel by emmitting energy in the form of electromagnetic radiation Electrons move around the nucels in specified circular paths called orbits or shells or energy levels and are designated as K,L,M and N shells respectively Arrange the above postulates of Bohr's theory in a correct sequence |
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Answer» 4,1.3,2 Each orbit or shell is associated with a definite 'amount of energy. Hence, these are also CALLEDENERGY levels.' (iii) .As long as the electron revolves in a particu - lar orbit, the electron does not lose its energy Therefore, these orbits are 'called stationary orbits and the electrons are said to be in sta- tionary energy states. An electron jumps from a lower to a higher energy level by absorbing energy. It jumps from a higher to a lower energy level by emit- ting energy in the form of ELECTROMAGNETIC RADIATION. |
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| 8784. |
0.5 moles of CO thken in a 2 L flask is maintained at 750 k in the presence of a catalyst so that the following reaction can take place: CO+2H_(2) hArr CH_(3)OH. When hydrogen is introduced, the pressure of the system in increased to 23.629 antm from 15.129 atm at equilibrium and 0.08 moles of gaseous product, methanol is formed. Clculate k_(C). |
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Answer» Solution :`CO+2H_(2) hArr CH_(3)OH` Initial moles `0.5 a0` At EQUILBRIUM 0.5=x a-2x x Given that x=0.08 Moles of CO=0.5-0.08=0.42 Moles of `H_(2)` =a-0.16 Moles of `CH_(3)OH=0.08` No. of moles of `H_(2)` can be calculated by using PV =nRT Pressure of hydrogen =23.629-15.129=8.5 ATM `n=(PV)/(RT)=(8.5xx2)/(0.082xx750)=0.276(a)` `therefore ` Number of moles of `H_(2)` at equilibrium =0.276-0.16=0.116 `K_(c)=(0.08//2)/(((0.166)/(2))^(2)xx(0.42)/(2))=56.66 l^(2) "mol"^(-1)` |
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| 8785. |
0.5 moles of a salt contains '3N' oxygen atoms. Identify theformula of thesalt. |
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Answer» `MXO_(3)` `:.` 1 MOLE of salt contains = `(1 xx3N)/(0.5)` = 6 N oxygen atoms ` :." The salt is "M(XO_(3))_(2)`. |
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| 8786. |
0.5 mole of triatomic gas contains ____________________atoms. |
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Answer» |
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| 8787. |
0.1225 g of potassium chlorate decomposes completely. Write the possible chemical equation andthe amount of oxygen liberated. |
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Answer» `4KClO_(3) to 4KCl + 6O_(2),0.048` g |
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| 8788. |
__________triad does not follow Dobereiner's law of triad. |
| Answer» Solution :Mg, Ca, Sr | |
| 8789. |
……………… technique used to renovate Pamban bridge . |
| Answer» SOLUTION : PROTECTIVE COATING | |
| 8791. |
__________ resembles alkali metals as well as halogens. |
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Answer» LITHIUM |
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| 8792. |
_________ reactionsare unidirectonal . |
| Answer» SOLUTION :IRREVERSIBLE | |
| 8793. |
…………..Property , which predicts the nature of bonding between the atoms in a molecule . |
| Answer» SOLUTION : ELECTRONEGATIVITY | |
| 8794. |
_____ process dependson the preferentialwettability of the orewith oil and gangue particles. |
| Answer» SOLUTION :FROTH FLOTATION | |
| 8795. |
…………. Period is the longest peroid and it contains ………… elements. |
| Answer» SOLUTION : SEVENTH , 32 | |
| 8799. |
…………….. Is used in making manhole covers and drain pipes and ……………… is used in making transmission cables and T.V towers . |
| Answer» SOLUTION :PIG IRON , STEEL | |
| 8800. |
………………………… is used in making anchors and electromegnets . |
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Answer» Steel |
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