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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.
| 90151. |
A current of 0.2 ampere is passed for 600 sec. through 100 mL of 0.1M NaCl. If hydrogen and chlroine gas is produced at the cathode and anode respectively, calculate |
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Answer» Solution :Anode : `2CL^(-) rarr Cl^(2)+2e` Cathode : `2H^(-)+2e rarr H_(2)` `[H^(+)]` of `H_(2)O` are used up at cathode `= [OH^(-)]` of `H_(2)O` left free `:.` EQ. of `Cl^(-)` used = Eq. of `[H^(+)]` used = Eq, of `[OH^(-)]` left `= (i.t)/(96500) = (0.2 xx 600)/(96500) = 1.24 xx 10^(3)` `:.` Mole of `{OH^(-)]` left `= 1.24 xx 10^(-3)` `:.` MOLARITY of `[OH^(-)] = (1.24 xx 10^(-3))/(100//1000) = 1.24 xx 10^(-2)` Also Eq. of`NaCl` left = Initial eq.-Eq. of `Cl^(-)` lost `= 0.1 xx 0.1 - 1.24 xx 10^(-3)` `= 8.76 xx 10^(-3) =`Mole of `Cl^(-)` `M_(NaCl) = (8.76 xx 10^(-3))/(100) xx 1000` `= 8.76 xx 10^(-20) M` |
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| 90152. |
A current of 0.25 amp is passed through CuSO_(4) solution for 45 minutes. Calculate the mass of copper deposited on the cathode (At.wt of Cu = 63.) |
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| 90153. |
A current of 0.20A is passed for 482 s through 50.0mL of 0.100 M NaCl. What will be the hydroxide ion concentration in the solution after the electrolysis? |
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Answer» 0.0159M The reaction at cathode is `2H_(2)O+2e to H_(2)+2OH^(-)` 1F `=1" of mol "OH^(-)` `9.99 xx 10^(-4)F=9.99 xx 10^(-4)" mol of OH"` Concentration of `OH^(-)` in solution After electrolysis `=(9.99 xx 10^(-4)"mol")/((50)/(1000)L)=0.0199M` |
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| 90154. |
A current of 0.20 A is passed for 48.2 s through 50.0 mL of 0.100 M NaCl. What will be the hydroxide ion concentration in the solution after the electrolysis? |
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Answer» 0.0159 M |
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| 90155. |
A current of 0.1A was passed for 4hr through a solution of cuprocyanide and 0.3745 g of copper was deposited on the cathode. Calculate the current efficiency for the copper deposition. (Cu GAM 63.5 or Cu-63.5) |
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Answer» `79%` `therefore % "EFFICIENCY" =0.3745/0.4738xx100=79%` |
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| 90156. |
A current of 0.0965 amp is passed for 1000 seconds through 50 mL of 0.1 M NaCl. If the only reactions are reduction of H_(2)O to H_(2) at the cathode and oxidation of Cl^(-) to Cl_(2) at the anode, what will be the average concentration of OH^(-) in th e final solution ? |
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Answer» Solution :Mole of ELECTRIC charge `= (0.0965 XX 1000)/(96500) = 0.001 F`. `therefore` equivalent of `OH^(-)` LIBERATED = 0.001. Mole of `OH^(-)` liberated = 0.001. `therefore` concentration of `OH^(-)` in mole per litre `= (0.001"( mole)")/(0.05 "(litre)")` = 0.02 M |
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| 90157. |
A current of 0.1 A was passed for 2 hr through a solution of cuprocyanide and 0.3745 g of copper was deposited on the cathode. Calculate the current efficiency for the copper deposition. (Cu = 63.5) |
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Answer» `79%` |
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| 90158. |
A current of 0.193 amp is passed through 100 ml of 0.2M NaCl for an hour. Calculate pH of solution after electrolysis if current efficiency is 90%. Assume no volume change. |
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| 90159. |
A current is passed through two valtameters connected in series. The first voltameter contains XSO_4(aq) while the second voltameter contains Y_2SO_4(aq).The relative masses of X and Y are in the ratio of 2:1.The ratio of the mass of X liberated to the mass of Y liberated is : |
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Answer» `1:1` |
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| 90160. |
A current is passed through two cells connected in series. The first cell contains X(NO_(3))_(3)(aq) and the second cell contains Y(NO_(3))_(2)(aq). The relative atomic masses of X and Y are in the ratio 1:2, what is the ratio of the liberated mass of X to that of Y? |
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Answer» SOLUTION :if atomic MASS of X=a, then that of Y=2a. As eq.WT=atomic mass/valency, Eq. wt. of X=a/3, Eq. wt. of Y=2 a/2=a. `("Mass of "X)/("Mass of "Y)=("Eq. wt. of "X)/("Eq. wt. of Y")=(a//3)/(a)=(1)/(3)=1:3` |
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| 90161. |
A current is passed through two cells connected in series. The first cell contians X(NO_(2))_(2)(aq). The relative atomic masses of X and Y are in the ratio 1:2. What is the ratio of the liberated mass of X to that of Y ? |
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Answer» `3:2` |
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| 90162. |
A current is passed through two cells connected in series. The first cell contains X(NO_(3))_(3)(aq) and the second cell contains Y(NO_(3))_(2)(aq). The relative atomic masses of X and Y are in the ratio 1:2. what is the ratio of the liberated mass of X to that of Y? |
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Answer» Solution :Suppose atomic mass of X=a Then atomic mass of Y=2A EQ. wt. of `X=(a)/(3)`, Eq wt. of `Y=(2a)/(2)=2`(`because` VALENCY of X=3, Valency of Y=2) Applying Faraday's second LAW of electrolysis: `("Mass of X")/("Mass of Y")=("Eq. wt. of X")/("Eq. wt. of Y")=(a//3)/(a)=(1)/(3)=1:3` |
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| 90163. |
A current is passed through two cells connected in series. The first cell contains X(NO_(3) )_( 3) (aq) and the second cell contains Y(NO_(3) )_( 2) (aq). The relative atomic masses of X and Y are in the ratio 1: 2. What is the ratio of liberated mass of X to that of Y? |
| Answer» ANSWER :D | |
| 90164. |
A current is passed through two cells connected in series. The first cell contains X(NO_(3))_(3(aq)) and the second cell contains Y(NO_(3))_(2(aq)). The relative atomic masses of X and Y are in the ratio 1:2. What is the ratio of liberated mass of X to that of Y. |
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Answer» SOLUTION :The oxidation STATES of X and Y are `X^(3+) and Y^(2+)` given atomic masses are in the ratio of 1:2 `therefore` Equivalent masses are in the ratio `(1)/(3):(2)/(2)` or `(1)/(3):1` or `1:3` |
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| 90165. |
A current is passed through 2 voltameters connected in series, The first voltameter contains XSO(aq.) and second voltameter contains Y_2 SO_4 (aq). The relative atomic masses of X and Y are in the ratio of 2:1. The ratio of the mass of X liberated to the mass of Y liberated is |
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Answer» `1:1` |
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| 90166. |
A current is passed through two cells connected in series. The first cell contain x(NO_(3))_(2) (aq) and the second cell contains y(NO_(3))_(2)(aq) . The relative atomic masses of x and y are in the ratio of x to that of y? |
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Answer» Solution :If atomic mass of x=a, then that of y=2a `therefore` Eq. mass `=("at. Mass")/("VALENCY")` `therefore` Eq. mass of `x=a//3`. Eq mass of `y=2a//2=a`. `("Mass of x")/("Mass of y")=("Eq. mass of x")/("E Q mass of y")` `=(a//3)/(1)=(1)/(3)= 1:3` |
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| 90167. |
A current being passed for two hour through a solution of an acid liberating 11.2 litre of oxygen at NTP at anode. What will be the amount of copper deposited at the athode by the same current when passed through a solution of copper sulphate for the same time |
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Answer» 16g |
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| 90168. |
A curent of dry air was bubbled through in a bulb containing 26.66g of a n organic substance in 200gms of water, then through a bulb at the same temperature containing pure water and finally through a tube containing fused calcium chloride. The loss in weight of water bulb is 0.0870 gms and gain in weight of CaCl_(2) tube is 2.036 gm. Calculate the molecular weight of the organic substance in the solution. |
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| 90169. |
A: Cuprite is concentrated by froth floatation process R: Cuprite is the sulphide ore. |
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Answer» If both Assertion & Reason are TRUE and the reason is the CORRECT explanation of the assertion, then mark (1). |
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| 90170. |
A cuprous ore among the following is |
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Answer» cuprite |
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| 90171. |
A cup of tea placed in the room eventually acquires a room temperature by losing heat. The process may be considered close to |
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Answer» CYCLIC process |
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| 90172. |
Assertion: [Cu(CN)_4]^(2-) is more stable than [Cu(H_2O)_4]^(2+) Reason: On heating [Cu(H_2O)_4]^(2+) with NH_3, [Cu(NH_3)_4]^(2+) is formed. |
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Answer» Both A & R are TRUE, R is the correct explanation of A |
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| 90173. |
A cubic solid is made to two elements P and Q. Atoms Q are the corners of the cube and P at the body centre. What is the formula of the compound ? What are the coordination numbers of P and Q? |
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Answer» Solution :This STRUCTURE is CsCl type structure. The FORMULA of the compound will be PQ. In this structure, P is at the body centre of the cube having Q atoms at its corners. Thus, the coordination number of P is 8. When we extend the cube in three DIMENSIONS, it can be seen that coordination number of Q is ALSO EIGHT. Thus, Formula of compound = PQ Coordination number of P = 8, Coordination number of Q = 8. |
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| 90174. |
A cubic solid is made of two elements P and Q. Atoms Q are at the corners of the cube and P at the body centre. What is the formula of the compound? |
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Answer» <P> Solution :As ATOMS P are present at the body centre, therefore, number of atoms of P in the unit cell = 1.As atoms Q are present at the 8 corners of the cube, therefore, number of atoms of Q in the unit cell = `1/8 xx 8` = 1 Ratio of atoms P:Q=1:1 HENCE, the FORMULA of the compound is PQ. |
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| 90175. |
A cubic solid is made of two elements P and Q. Atoms of Q are at the corners of the cube and P at the body-centre. What is the formula of the compound ? What are the coordination numbers of P and Q? |
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Answer» SOLUTION :Atoms Q are located at the CORNER of cube `therefore` Number of Q atoms `= ( 8 XX 1/8) = 1` per UNIT cell P atom are at the body-centre of a cube `therefore`Number of P atoms per unit cell = 1 Ratio of P atom to the Q atoms = 1:1 . `therefore` Formula of a COMPOUND = PQ The co-ordination number of P and Q are 8. |
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| 90176. |
A cubic solid is made of two elements P and Q. Atoms of Q are at the corners of the cube and P at the body-centre. What is the formula of the compound ? What are the coordination numbers of P and Q ? |
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Answer» <P> Solution :Number of atoms of Q in the unit CELL `=(1)/(8)xx8=1`Number of atoms of P in the unit cell = 1 Hence the formula is PQ. Coordination number of each = 8. |
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| 90177. |
A cubic solid is made of two elements A and B. Atoms of B are at the corners of the cube and of A, at the body centre. Determine the formula of the compound. |
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| 90178. |
A cubic solid A^(+)B^(-) has the B^(-) ions arranged as below. If the A^(+) ions occupy all the edge centres, the formula of solid is [O = B^(-)] |
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Answer» AB |
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| 90179. |
A cubic crystal possesses in all ........... planes of symmetry: |
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Answer» 9 |
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| 90180. |
A cube-shaped crystal of an alkali metal, 1.62 mm on an edge, was vapourized in a 500.0 mL evacuated flask.The pressure of the resulting vapour was 12.5 mm of Hg at 802^@C.The structure of the solid metal is known to be body-centered cubic.What is the atomic radius of the metal atom in picometers ? (R=0.082 It-atm/mol-K) (The radii of metals atoms as Li=152 pm, Na=186 pm,K=227 pm,Rb=248 pm) |
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Answer» Number of atoms`=nN_A=5.612xx10^19` So number of unit CELLS`=((nN_A)/2)=2.806xx10^19` So, if there are x unit cells along one edge of given cube.then Then total number of unit cells in the cubic crystal `=x^3=(nN_A)/2=2.806xx10^19 IMPLIES x=3.04xx10^6` Now if edge LENGTH of unit cell =a `implies ax=1.62 mm implies a=5.33xx10^(-9)mm implies a=533` pm `:. r=(sqrt3a)/4=230.3` pm |
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| 90181. |
A cubic crystal possesses in all ……. Elements of symmetry . |
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Answer» 9 |
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| 90182. |
A cubic crystal possesses : |
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Answer» 9 PLANE of SYMMETRY |
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| 90183. |
(A) Cu^(+) is highly stable in aqueous solution (R) E.C of Cu^(+) is [Ar]4s^1 3d^9 |
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Answer» Both (A) and (R) are TRUE and (R) is the CORRECT EXPLANATION of (A) |
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| 90184. |
(a) Cu^2+ ions are coloured but Zn^2+ ions are colourless . Give reason . (b) Write the formula to calculate spin only magnetic moment . |
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Answer» Solution : `Cu^2+` are coloured because of PRESENCE of UNPAIRED electrons . `Zn^2+` are colourless because of unpaired electrons . `MU`=`sqrt(N(n+2)` |
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| 90185. |
(A): Cu^(+) is diamagnetic (R): Ions with dl configuration are diamagnetic |
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Answer» Both A & R are true, R is the CORRECT explanation of A |
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| 90186. |
(A) Cu^+ ion is coloured in aqueous solutions (R) Four water molecules are coordinated to Cu^(+) ion |
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Answer» Both (A) and (R) are TRUE and (R) is the CORRECT EXPLANATION of (A) |
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| 90187. |
(A) Cu^(+) is diamagnetic (R) Ions With d^(10) configuration are diamagnetic |
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Answer» Both (A) and (R) are TRUE and (R) is the CORRECT EXPLANATION of (A) |
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| 90188. |
(A) : Cu^(+2)(aq) is more stable than Cu^(+)(aq) (R): Heat of hydration of Cu^(+2) is more than IP_(2) of Cu |
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Answer» Both (A) and (R) are TRUE and (R) is the CORRECT EXPLANATION of (A) |
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| 90189. |
(A) Cu has a unique behavior having a positive SRP. (R) Cu is unable to liberate H_2 from acids. |
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Answer» Both (A) and (R) are TRUE and (R) is the CORRECT EXPLANATION of (A) |
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| 90190. |
A+CS_(2)+HgCl_(2) give C_(2)H_(5)-N=C=S. Thus compound A is :- |
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Answer» `C_(2)H_(5)NH_(2)` |
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| 90191. |
(A) Crystals exhibiting Frenkel type defects do not show any change in density due to defect (R) In Frenkel defect the interstitial cations and cation vacancies are equal in number |
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Answer» Both (A) and (R) are true and (R) is the correct EXPLANATION of (A) |
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| 90192. |
(A): Crystalline solids are anisotropic (R): Crystalline solids are not as closely packed as amorphous solids. |
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Answer» Both (A) and (R) are true and (R) is the correct EXPLANATION of (A) |
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| 90193. |
(A) Crystalline solids are anisotropic in nature(R) Crystalline substances have short range order |
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Answer» Both (A) and (R) are TRUE and (R) is the CORRECT explanation of (A) |
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| 90194. |
A crystalline solid XY_(3) has ccp arrangement for its element Y. The element X occupies : |
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Answer» `66%` of TETRAHEDRAL holes `:.` X will occupy `1//3` RD of octahedral VOIDS `:.%` of octahedral voids occupied by X `=1/3xx100=33.3%` |
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| 90195. |
A crystalline solid XY_3 has ccp arrangement for its element Y. X occupies |
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Answer» 66% of tetrahedral VOIDS `:.` % of octahedral voids occupied by X `= 1/3 XX 100= 33.3%` |
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| 90196. |
A crystalline solid X reacts with dilute HCl to liberate a gas Y. The gas Y decolourises acidified KMnO_(4). When a gas Z is slowly passed into aqueous solution of Y, colloidal sulphur is obtained. X and Z could be respectively. |
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Answer» `Na_(2)S,SO_(3)` |
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| 90197. |
A crystalline solid 'X' reacts with dil. HCl to liberate a gas 'Y' . 'Y' decolourises acidified KMnO_4. When a gas'Z' is slowly passed into an aqueous solution of 'Y' , colloidal sulphur is obtained . 'x' and 'Z' could be , respectively. |
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Answer» `Na_2S, SO_3` |
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| 90198. |
A crystalline solid of pure substance has a face-centred cubic structure with a cell edge of 400 pm. If the density of the substance in the crystal is 8 g cm^(-3) then the number of atoms present in 256 g of the crystal is N xx 10^24. The value of N is............... |
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| 90199. |
A crystalline solid X reacts with dil HCI to liberate a gas Y. Y decolourises acidified KMnO4. Whena gas 'Z' is slowly passed into an aqueous solution of Y, colloidal sulphur is obtained. X and Z could be, respectively |
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Answer» `Na_2S, SO_3` ` H_2O + SO_2 to H_2SO_3` `H_2SO_3 + 2H_2S to 3S + 3H_2O` |
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| 90200. |
A crystalline solid dissolves in walur lo give colourless solution. On addition of aqueous silver nitrate, a yellow precipitate is formed. The solid is most likely to be |
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Answer» COPPER (II) oxide |
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