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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Gold occurs as face centred cube and has a density of 19.30kg dm^(-3). Calculate atomic radius of gold. (Molar mass of Au=197) |
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Answer» Molar MASS=`197g"mol"^(-1)` Avogadro constant `=N_(A)=6.022xx10^(23)"mol"^(-1)` Atomic radius of Au=? If fcc unit cell, there are 8 ATOMS of Au at 8 corners and 6 atoms at 6 face centres. Number of Au atoms in the unit cell `=(1)/(8)xx8+(1)/(2)xx6` 4atoms Mass of 1Au atom `(197)/(6.022xx10^(23))=3.271xx10^(-22)g` `therefore "Mass of 4 Au atoms"=4xx3.271xx10^(-22)g` `therefore "Mass of unit cell"=1.308xx10^(-22)g` `""=1.308xx10^(-24)kg` Densidy of the unit cell `=("Mass of unit cell")/("Volume of unit cell")` `therefore d=(1.308xx10^(-21))/(a^(3))` `therefore (a^(3)=1.308xx10^(-21))/(d)=(1.308xx10^(-24))/19.3` `=6.77xx10^(-26)dm^(3)` `=6.77xx10^(-23)cm^(3)` `therefore a=(6.77xx10^(-23))^(1//3)=(67.77xx10^(-24))^(1//3)` `""=4.077xx10^(-8)cm` If r is the radius of Au atom, then for fcc unit cell, `r=(a)/(2sqrt2)` `=(4.077xx10^(-8))/(2sqrt2)=1.442xx10^(-8)cm=144.2cm` |
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| 2. |
Gold numbers of some colloids are : 0.005-0.01, Grum arabic : 0.15-0.25,Oleate : 0.04-1.0, Starch:15-25. Which among these is a better protective colloid? |
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Answer» Gelatin i.e., The smaller the value of gold number of lyobhilic sol, the GREATER is the protective action. Hence, gelatin will be better protective colloid. |
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| 3. |
Gold numbers of some colloids are : Gelatin : 0.005 - 0.01 , Gum Arabic : 0.15 - 0.25, Oleate : 0.04 - 1.0, Starch : 15 - 25. Which among these is a better protective colloid ? |
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Answer» Gelatin |
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| 4. |
Gold numbers of protective colloids, A, B, C and D are respectively 0.50, 0.01, 0.10 and 0.005. The correct order of the stability of colloids is ....... |
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Answer» `BLT D LT A lt C ` |
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| 5. |
Gold numbers of protective coloids A,B, C and D are 0.50,0.01, 0.10 and 0.005 reprectively. The correct order of their protective powers is |
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Answer» `A lt C lt B lt D` D `(0.005) gt B (0.001) gt C (0.05) gt A (0.50)` or `A lt C lt B lt D.` |
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| 6. |
Gold numbers of four protective colloids A, B, C and D are 0.5 , 0.01, 0.1 and 0.005 respectively. Arrange them in the correct order of their protective power. |
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Answer» Solution :LESSER the value of gold number , higher will be the PROTECTIVE power. The order of protective poweris :`D gt B gt C gt A`. |
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| 7. |
Gold number of haemoglobin is 0.03. Hence, 100 mL of gold sol will require haemoglobin so that gold is not coagulated by 1 mL of 10% NaCl solution: |
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Answer» `0.03 MG` weight of Hb in mg `=0.30` |
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| 8. |
Gold number of few colloids are given below: Gelatin = 0.005"" Strach = 25 . Egg albumin = 0 .08 ""Gum arabic =0.10 Whichis the best protective colliad ? |
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Answer» Gelatin `therefore ` Gelatin is the best protective colloid. |
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| 9. |
Gold number of gum arabic is 0.15. The amount of gum arabic required to protect 100 mL of red gold sol from coagulation by 10 mL of 10% NaCl solution is |
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Answer» `0.15` milimoles `=0.15xx10=1.5mg` (NOTE that for 10 mL of gold sol, 1 mL of `10%` mL of `10%` NaCl sol. Is required). |
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| 10. |
Gold number of a lyophilic solution is such property that |
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Answer» The LARGER its value, the GREATER is the peptising power |
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| 11. |
Gold number is associated with: |
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Answer» Only lyophobic COLLOIDS |
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| 13. |
Gold number is associated only with............ |
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Answer» LYOPHOBIC COLLOIDS |
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| 14. |
Gold number is: |
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Answer» The number of mg of LYOPHILIC colloid whichshould be added to 10 ml of ferric hydroxide sol so as to prevent its coagulation by the addition of 1 ml of 10 % sodium chloride solution |
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| 15. |
Gold number gives: |
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Answer» AMOUNT of GOLD PRESENT in a collodd |
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| 16. |
Gold number gives............... |
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Answer» the AMOUNT of gold present in the colloid |
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| 17. |
Gold number gives ……………….. |
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Answer» the amount of gold present in the colloid |
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| 18. |
Gold is soluble in ……………………… |
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Answer» conc. `HNO_(3)` |
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| 19. |
Gold is heavier than aluminium. If we put a 100 g biscult of gold in water taken in a measuring cylinder or we put a 100 g aluminium bar in the measuring cylinder, will the rise in level of water be same or different in the two cases ? Give reason. |
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Answer» |
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| 21. |
Gold is extracted using : |
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Answer» AMALGAMATION PROCESS |
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| 22. |
Gold has a close- packed structure which can be viewed as spheres occupying 0.74 of the total volume . If the density of gold is 193 g//c c, calculate the apparent radius of a gold ion in the solid. |
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Answer» |
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| 23. |
Gold has a close-packed structure which can he viewed as spheres occupying 0.74 of the total volume. If the density of gold is 19.3 g/cc, calculate the apparent radius of a gold ion in the solid. |
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Answer» Solution :Gold has a close-packed structure with a packing fraction value of 0.74. This SHOWS that it has a face-centred cubic cell. The number of IONS in a face-centred unit cell is 4. Now, density `=("mass of unit cell")/("VOLUME of unit cell")` or `19.3 = (4 xx (197) xx 1.66 xx 10^(-24))/a^(3) , a = 4.07 xx 10^(-8)` CM. In a face-centred cubic cell, radius `=(sqrt(2)a)/4 = (sqrt(2) xx 4.07 xx 10^(-8))/4 = 1.439 xx 10^(-8) cm` |
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| 24. |
Gold is extracted by hydrometallurgical process based on its property |
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Answer» of being electropositive `4Au+8KCN+2H_(2)O+underset(air)(O_(2))to` `4K[Au(CN)_(2)]+4KOH` `2K[Au(CN)_(2)]+Znto2Au+K_(2)[Zn(CN)_(4)]` |
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| 25. |
Gold extracted using |
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Answer» AMALGAMATION process |
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| 26. |
Gold dissolves in cyanide solution the presence of air to form [Au(CN)_(3)] which is stable in a cons solution. Au_((s))+CN_((s))^(-)+O_(2)+H_(2)O_(2) hArr [Au(CN)_(2)]_((aq))^(-)+OH_((aq))^(-) Aquaregia a 3 : 1 mixture of conc. HC and HNO_(3) was developed by the alchemists as a means to dissolve gold. The process is actually a Redox reaction. Au_((s))+NO_(3(aq))+Cl^(-) hArr AuCl_(4(aq))^(-)+NO_(2(g)) Gold is too noble to react with HNO_(3) However gold does react with a waregia becuase the complex AuCl_(4)^(-) forms Au_((aq))^(+3)+3e^(-) to Au_((s)) E^(theta)=15V to I AuCl_(4(aq))^(-)+3e^(-) to Au_((s))+4Cl_((q))^(-)E^(0)=1V to 2 The function of HC is to provide C what is the purpose of the Cr in the above reaction select your choice from the following. |
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Answer» It is an OXIDISING agent `:. Cl^(-)` can be used as a complexing agent. |
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| 27. |
Gold exhibits the variable oxidation states of: |
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Answer» `+2, +3` |
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| 28. |
Gold dissolves in cyanide solution the presence of air to form [Au(CN)_(3)] which is stable in a cons solution. Au_((s))+CN_((s))^(-)+O_(2)+H_(2)O_(2) hArr [Au(CN)_(2)]_((aq))^(-)+OH_((aq))^(-) Aquaregia a 3 : 1 mixture of conc. HC and HNO_(3) was developed by the alchemists as a means to dissolve gold. The process is actually a Redox reaction. Au_((s))+NO_(3(aq))+Cl^(-) hArr AuCl_(4(aq))^(-)+NO_(2(g)) Gold is too noble to react with HNO_(3) However gold does react with a waregia becuase the complex AuCl_(4)^(-) forms Au_((aq))^(+3)+3e^(-) to Au_((s)) E^(theta)=15V to I AuCl_(4(aq))^(-)+3e^(-) to Au_((s))+4Cl_((q))^(-)E^(0)=1V to 2 Calculate the formation constant approximately, of Auct at 25^(@)C |
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Answer» `10^(5)` |
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| 29. |
Gold dissolves in cyanide solution the presence of air to form [Au(CN)_(3)] which is stable in a cons solution. Au_((s))+CN_((s))^(-)+O_(2)+H_(2)O_(2) hArr [Au(CN)_(2)]_((aq))^(-)+OH_((aq))^(-) Aquaregia a 3 : 1 mixture of conc. HC and HNO_(3) was developed by the alchemists as a means to dissolve gold. The process is actually a Redox reaction. Au_((s))+NO_(3(aq))+Cl^(-) hArr AuCl_(4(aq))^(-)+NO_(2(g)) Gold is too noble to react with HNO_(3) However gold does react with a waregia becuase the complex AuCl_(4)^(-) forms Au_((aq))^(+3)+3e^(-) to Au_((s)) E^(theta)=15V to I AuCl_(4(aq))^(-)+3e^(-) to Au_((s))+4Cl_((q))^(-)E^(0)=1V to 2 How many grams, approximately of NACN are needed to extract 20g of gold from are? |
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Answer» 20 g `2Na[Au(CN)_(2)]^(-)+ZN to Na_(2)[Zn(CN)_(4)]+2Au` `196......2xx197` `x......20` `169xx20=x xx 2xx197` `:.x=(196xx20)/(2xx197)=10 gm` |
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| 30. |
Gold dissolves in aquaregia forming |
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Answer» `AU(NO_3)_2` |
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| 31. |
Gold dissolves in aqua-regia forming |
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Answer» CHLOROAURIC acid |
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| 32. |
Gold dissolves in aqua-regia forming : |
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Answer» AURIC chloride |
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| 33. |
Gold dissolves in a aqua-regia forming: |
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Answer» AURIC chloride |
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| 35. |
Gold crystallizes in a face centred unit cell. Its edge length is 0.410nm. The radius of gold atom is |
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Answer» 0.205 nm ` :. r = ( 0.410)/( 2 xx 1.414) = 0.145 `nm |
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| 36. |
Gold can exhibit the oxidation states |
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Answer» `I and +II` |
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| 37. |
Gold biscults are available in the market which look exactiy similar to gold but actually they are not of pure gold (but of gold called fool's gold). How will check it by some simple physical property? Density of pure gold is well known to be 19.3 g cm^(-3). |
| Answer» SOLUTION :We can find out the exact MASS and VOLUME* of the given gold bisult and then calculate its density. If it is not made of PURE gold, its density will come out to be DIFFERENT from that of pure gold. | |
| 38. |
Gold (atomic radius = 0.144 nm) crystallize into face centered unit cell, then what is the edge length of unit cell ? |
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Answer» 0.4574 NM |
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| 39. |
Gold [atomic radius = 0.144 nm] crystallises in face-centred unit cell. What is the length of a side of the cell ? |
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Answer» Solution :For FCC LATTICE `a=2sqrt2, R` SUBSTITUTING the values, we GET `a=2xx1.414xx0.144=0.407nm`. |
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| 40. |
Gold (atomic radius = 0.144 nm) crystallises in a face-centred unit cell. What is the length of a side of the cell ? |
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Answer» SOLUTION :RADIUS of gold atom r = 0.144 nm In a face-centred unit CELL. ` 4r = sqrt(2)a` `therefore a = 2sqrt(2)r` `therefore a = 2 XX 1.414 xx 0.144` nm `therefore` a = 0.407 nm |
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| 41. |
Gold (at. Mass 197 g mol^(-1))crystallises in cubic closest packed structures (the face-crntred cubic) and has a density of 19.3 g//cm^(3) . Atomic radius is |
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Answer» `144.17` pm `a=3sqrt((197xx4)/(6.023xx10^(23)xx19.3))` `a=407.8xx10^(-10)cm=407.8` pm `r=(407.8)/(SQRT(8))=144.18`pm |
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| 42. |
Gold and silver are extracted from their repective ores by |
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Answer» Calcination |
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| 43. |
Gold and silver are called noble metals, because: |
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Answer» They do not NORMALLY react |
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| 44. |
Go through the following graph and answer the following questions. Which of the following reaction is true? |
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Answer» REDUCTION of calcined/roasted haematite ore to PIG IRON in blast furnance takes place in the lower temperature range and in the higher temperature range by CO and C respectively |
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| 45. |
Going from fluorine to chlorine, bromine and iodine, the electronegativity |
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Answer» Increases |
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| 46. |
Go through the following graph and answer the following questions. To make the following reduction process spontaneous, temprature should be : ZnO+C rarr Zn+CO |
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Answer» `LT 1000^(@)C` |
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| 47. |
Go through the following graph and answer the following questions. At what approximate temperature, zinc and carbon have equal affinity for oxygen. |
| Answer» Answer :a | |
| 48. |
Glyptal polymer is obtained from glycerol by reacting it with |
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Answer» Malonic acid |
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| 49. |
Glyptals are chiefly employed in |
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Answer» Toy MAKING |
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