InterviewSolution
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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.
| 2151. |
Element x has 12 protons in its nucleus .To which group in the peiodic tabel would it belong? |
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Answer» IVA(14) Its ELECTRONICCONFIGURATION =2,8,2 `THEREFORE` X belongs to IIA (2) group since there are tow ELECTRONS in the valence shell |
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| 2152. |
Element X has 12 neutorns in its nucleous .To which group in the periodic table would it belongs? |
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Answer» 1 |
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| 2153. |
A substance is a form of matter that has a ……….. composition. |
| Answer» SOLUTION :definite | |
| 2154. |
A students wants to prepare a colloidal solution with gas phase and liquid medium. Which of the following should be choose ? (i) Shaving cream (ii) Face cream (iii) Soap (iv) Pumice |
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Answer» (i) and (II) |
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| 2155. |
Element A,B and C with atomic number a,a+1 and z+2 respecitively form positvie ions having equal numberof electons compare and contrast the size of the give reasons |
| Answer» Solution :As the positive ions have equal NUMBER of ELECTONS A,B and C must be ioselectronic and form unipositive dipositive and tripositive ions .with the increase in nuclear charge the effective nuclear force of attraction on the outermost eletrons increasetherefore the ionic size decrease from `A^(+),B^(+2)` and `C^(+3)` | |
| 2156. |
Electrons have negligible mass and charge . |
| Answer» Solution :Correct STATEMENT : ELECTRONS have NEGLIGIBLE MASS and have negative change . | |
| 2158. |
Electrons are ……….. Charged particles. |
| Answer» SOLUTION :NEGATIVELY | |
| 2159. |
A student was aked a prepare a 20% ( volume by volume ) solution of ehtanol in water . How should the studentdo it ? |
| Answer» SOLUTION :20ML ETHANOL + 80ML WATER | |
| 2160. |
Electrons and Protons |
Answer» SOLUTION :
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| 2161. |
Electronic spectra can distinbuish isobars but not isotopes. Justify. |
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Answer» Solution :(i) FUNDAMENTAL particle responsible for the spectra (ii) relation between fundamental particle and structure of spectra. (iii) difference in number of the above PARTICLES between ISOBARS and isotopes (iv) EFFECT of this on structure of spectra. |
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| 2162. |
Electronic configurations of elements P, Q, Rand S are given below. (These are not actual symbols). P - 2, 2 Q - 2, 8, 2 R - 2, 8, 5 S - 2,8 Which are those included in the same group? |
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Answer» <P> |
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| 2163. |
Electronic configurations of elements P, Q, Rand S are given below. (These are not actual symbols). P - 2, 2 Q - 2, 8, 2 R - 2, 8, 5 S - 2,8 Which among these elements are included in the same period? |
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Answer» <P> |
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| 2164. |
A student prepared a 2.5% ( by mass ) solution of glucose in water . Which of the following represents the correct composition of the solution ? |
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Answer» 2.5 g GLUCOSE + 100 g WATER |
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| 2165. |
Electronic configurations of elements P, Q, Rand S are given below. (These are not actual symbols). P - 2, 2 Q - 2, 8, 2 R - 2, 8, 5 S - 2,8 Which among them is a noble gas? |
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Answer» |
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| 2166. |
A student need 3.011 xx 10^(23) atoms of each magnesium and carbon. Then how much will he take both the elements by weight? |
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Answer» 24 g Mg and 6g C Mole of 6 g C=`(6)/(12)` =0.5 |
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| 2167. |
Electronic configurations of elements P, Q, Rand S are given below. (These are not actual symbols). P - 2, 2 Q - 2, 8, 2 R - 2, 8, 5 S - 2,8 To which group and period does the element R belong? |
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Answer» |
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| 2169. |
A student mixed egg albumen with water and stirred well. He observed that ______ . |
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Answer» a TRANSPARENT solution was formed |
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| 2170. |
Electron affinity values of noble gases are zero wheres they are negative for alkaline earth metals how do you account for this ? |
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Answer» Solution :(i) valence shell electronic CONFIGURATION of noble gases and alkalin earth metals (ii) relation between valence shell electronic configuration and stability in both the cases (iii) TYPES of forcesexisting between an electorn going to be added to a noble GAS (iv) energy change involved when an electron is added to a noble gas ELEMENT |
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| 2171. |
A student heats a beaker containing ice and water. He measures the temperature of the content of the beaker as a function of lime. Which of the following would correctly represent the result ? Justify your choice . |
| Answer» Solution :We know that the melting point of ice and the FREEZING point of water in pure states are both zero. This means that at this temperature, both are present. Upon heating, the temperature would actually not change but heat energy supplied would be abosrbed by the ice as latent heat of fusion. This would result in the melting of ice. The temperature would remain zero degree till the whole of ice has melted. Further heating would INCREASE the temperature of water till it STARTS boiling at `100^(@)C`. The curve (d) gives the correct REPRESENTATION. | |
| 2172. |
Electrolysis principle is not used in the following processes : |
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Answer» plating of SILVER on copper |
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| 2173. |
Effervescence is observed when the water is warmed.What is the reason behind it ? |
| Answer» SOLUTION :DUE to the evolution of dissolved gases, like, oxygen, carbon dioxide, etc., effervescence is OBSERVED. SOLUBILITY of a gas in liquid DECREASES with the increase in temperature. | |
| 2174. |
A straight chain saturated hydrocarbon wihth vapour density 43 on heating in absence of air gives X and y.y on treatment with bromine water gives Z identify X,Y and Z. |
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Answer» Solution :(i) calculation of molecular weight of hydro- carbon . (ii) identification of molecular formuja of hydro carbon based on given data (iii) identification of PRODUCTS namely X and Y of HYDROCARBON upon cracking (iv) identification of Z from BROMINATION of X |
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| 2175. |
During the preparation of artifical diamond can molten copper be used instead of molten iron? Justify. |
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Answer» Solution :(i) CHANGE involved on solidification (ii) arrangement of carbon ATOMS in diamond (III) CONDITIONS required for the preparationof diamond (iv) comparison of VOLUME changes during solidification of iron and copper. |
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| 2176. |
A statue coated with chemical substance X on long exposure to polluted atmosphere becomes black. This colour can be restored by treatment with H_2O_2. Identify element X and also the types of chemical changes involved. |
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Answer» Solution :(i) metal PRESENT in paints which GETS darkened (black) on long exposure to polluted AIR (ii) nature of `H_2O_2` (iii) CHANGE in the colour of the compound when washed with `H_2O_2` |
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| 2178. |
(a) state the postulates stated by neils bohr in order to overcome the objection as raised against Ruthford's model of atom. (b) summarise the rules for the distrubution of elctrons in the first eighteen elements |
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Answer» SOLUTION :(a) For ANSWER CONSULT SECTION 4.5 (B) For answer consult section 4.9 |
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| 2179. |
(a) State six postulates of Dalton's atomic theory. (b) A 0.24g sample of compound of carbon and oxygen on analysis was found to contain 0.096 g of carbon and 0.144 g of oxygen. Find the percentage composition of the compound by weight. |
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Answer» SOLUTION :(a) For the POSTULATES of Daltaon's atomic THEORY, cosult seciton. `(b) "PERCENTAGE of carbon"=("MASS of carbon")/("Mass of compound")xx100=((0.096g))/((0.24g))xx100=40%` `"Percentage of oxygen"=("Mass of oxygen")/("Mass of compound")xx100=((0.144g))/(0.24g)xx100=60%` |
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| 2180. |
During the manufacture of wrought iron, cast iron is taken in a reverberatory furnace and stirred at high temperature and thus process is called ______________ |
| Answer» SOLUTION :N/A | |
| 2181. |
You are given two liquids, on a solution and the other a compound. How will you distinguish the solution from the compound ? |
| Answer» SOLUTION :(a) TRY to separate the constituents present in both by some SUITABLE PHYSICAL method. Separation is possible in case of a solution but not in a compound. | |
| 2182. |
During summer days, water kept in an earthen pot (pitcher) becomes cool because of the phenomenon of : |
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Answer» diffusion |
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| 2183. |
A stable unipositive ion of an element contains three fully filled orbits.What is the atomic number of the element ? |
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Answer» SOLUTION :(i) MAXIMUM NUMBER of electrons present in an orbit (Bohr -Bury SCHEME). (iv) electronic configuration of neutral atom (V) 37 |
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| 2184. |
During respiration, glucose and oxygen enter our body cells and waste products carbon dioxide and water leave the body cells by the process of : |
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Answer» effusion |
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| 2185. |
During calcination, carbonate ores are converted to _________. |
| Answer» SOLUTION :N/A | |
| 2186. |
A stable dipositive ion and a dinnegative ion are isoelectrnic with an octet configuaration in the second shell of their atoms. Identify of the preceding and succeeding elements and write their electronic configurations. |
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Answer» Solution :LET the dipositive ion be `X^(+2)` and dinegative ion is `Y^(-2)`. The octet in the SECOND shell `to 2,8` The number of electrons in `X=10+2=12` The number of electrons in `Y=10-2=8` X is magneshium with an ELECTRONIC configuration 2, 8,2,Y is oxygen with a congfiguration 2,6. The element preceding magnesium is sodium and the succeeding one is aluminium. They have eletronic configuration `2,8,1, and 2,8,3`, respectively. The elements preceding and succeeding for oxygen are nitrogen and fluorine which have electronic configuration 2,5 and 2,7, respectively. |
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| 2187. |
During an experiment the students were asked to prepare a 10% (mass/mass) solution of sugar in water. Ramesh dissolved 10 g of sugar in 100 g of water while Sarika prepared it by dissolving 10g of sugar in water to make 100 g of the solution. (a)Are the two solutions of the same concentration? (b) Compare the mass % of the two solutions. |
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Answer» Solution :(a) No, the two solutions have DIFFERENT concentrations. (B) Mass % of solution prepared by Ramesh = `("Mass of sugar")/("Mass of solution")xx 100` `=(10)/((10 + 100)) xx 100 = (10)/(110) xx 100 = 9.09%` Mass % of solution prepared by Sarika = `("Mass of sugar")/("Mass of solution")xx 100 = (10)/(100) xx 100 = 10.0%` |
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| 2188. |
A solution of sodium sulphate is made in 100 g of water of mass by mass percentage 20. Calculate the mass of sodium sulphate in this solution. |
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Answer» Solution :Suppose the mass of solute (sodium sulphate) = x g The mass of solvent (WATER) = 100 g `THEREFORE` The mass of solution = (x + 100)g `therefore` Mass percentage of solution = `("Mass of solute")/("Mass of solution")xx100` `20=x/((x+100))xx100` `20(x+100)=100X` `20X+2000=100x` 2000 = 100x - 20x 2000 = 80x `2000/80=x` 25 = x x = 25g 25 g of sodium sulphate would be ADDED in the solution. |
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| 2189. |
During a chemical reaction ____ takes place between the reactant molecules. |
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| 2190. |
Due to which property does the aroma of hot food reaches us some metres away? |
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Answer» The INTERMOLECULAR force of atrraction is different in different matters. |
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| 2191. |
A solutionis madeby dissolving50 g of glucose in 250 g of water calculate the concentration of the solution in mass percentage |
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Answer» |
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| 2193. |
A solution is a ______ mixture composed of two or more substances. |
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| 2194. |
Draw the structure of oxygen and sulphur atoms . |
Answer» SOLUTION :
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| 2195. |
Draw the energy profile diagram and also calculate the activation energy of the forward reaction for a reaction A+B_(to)^(larr)C+D. Energy released is equal to twice the activation energy of the forward reaction and E_(a) of the backward reaction is 120 kJ/mole. |
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Answer» Solution :(i) comparision of energies of reactants and PRODUCTS (II) determination of empirical formula (iii) relation between molecular weight and VAPOUR density (iv) determination of molecular formula (v) 40 kj/mole |
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| 2196. |
Homogeneous mixture of two or more than twocomponents is called |
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Answer» |
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| 2197. |
Draw the electron distribution diagram for the formation of Carbon dioxide (CO_(2)) molecule |
Answer» SOLUTION :
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| 2198. |
Draw the atomic structure of sodium atom on the basis of Bohr's model. |
Answer» SOLUTION :
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| 2199. |
A solution contains 60 g of sugar in 340 g of water. Calculate the concentration in terms of mass by mass percentage of the solution. |
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Answer» Mass of WATER (solvent) = 340 g Mass of solution = (60 + 340) g = 400 g Mass percentage of solution = `("Mass of solute")/("Mass of solution")XX100` = `60/400xx100=15%` |
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| 2200. |
Draw energy profile diagrams for exothermic and endothermic reactions. |
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Answer» Solution :(i) Energy charges during the formation of a product. (II) COMPARISON of energy of the REACTANT and product in CASE of endothermic and exothermic REACTIONS. (iii) relation among the energy of the reactant, product and activation energy. |
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