Quiz about GRAPHICAL-INTERPRETATION IN GENERAL AWARENESS

Question 1

Consider the first order reaction `A rarr 2B`. Which of the following figure correctly describes the rate of disappearance of `A(r_(1))` and rate of appearance of `B(r_(2))` with time ?A. B. C. D.

Accepted Answer

NA

Question 2

Consider the following equilibrium `A(g) hArr 2B(g)` and the following graph Select the correct statements. A. Effect (I) represent the insertion of inert gas at constant volumeB. Effect (II) represent the insertion of inert gas at constant pressureC. Effect (III) represent the decrease in volume of containerD. Effect (IV) represent the addition of moles of gas B

Accepted Answer

NA

Question 3

Which graph represents the correct relationship between various velocities for an idela gas ?A. B. C. D.

Accepted Answer

NA

Question 4

The following graph is experimentally obtained for the reaction `A rarr 2B` at `25^(@)C`. The correct statement (s) for the reaction isare (Given ln 2 = 0.7) A. Time for `87.5%` reaction of A is 21 minB. The initial concentration of A was eMC. The time at which the concentration of A and B becomes equal is 7 minD. The rate of appearance of B is `(d[B])(Dt) = (0.2 min^(-1)) [A]`

Accepted Answer

NA

Question 5

Which of the following graph is correct w.r.t half life for a zero order reaction at various stages of reaction ?A. B. C. D.

Accepted Answer

NA

Question 6

A student analyzed tha data from a zero order reaction and obtained the graph shown. What labels should be attached to the X and Y axes respectively ? A. Time concentrationB. Time `1`concentrationC. Time ln (concentration)D. `1time` concentration

Accepted Answer

NA

Question 7

For an ideal gas four processes are marked as 123 and 4 on P-V diagram as shown in figure. The amount of heat supplied to the gas in the process 123 and 4 are `Q_(1)Q_(2)Q_(3)` and `Q_(4)` respectively then correct order of heat supplied to the gas is A. `Q_(1) gt Q_(2) gt Q_(3) gt Q_(4)`B. `Q_(1) lt Q_(2) lt Q_(3) lt Q_(4)`C. `Q_(1) gt Q_(2) gt Q_(4) gt Q_(3)`D. `Q_(1) gt Q_(4) gt Q_(2) gt Q_(3)`

Accepted Answer

NA

Question 8

Consider the cyclic process `R rarr S rarr R` as shown in the fig. You are told that one of the path is adiabatic and the other one isothermal. Which one of the following is (are) true ? A. Process `R rarr S` is isothermalB. Process `S rarr R` is adiabaticC. Process `R rarr S` is adiabaicD. Such a graph is not possible

Accepted Answer

NA

Question 9

In given cyclic process for an ideal gas  Path `B rarr C` is isoentropic. Then select the correct optionsA. `DeltaS_(A_B) = DeltaS_(C rarr A)`B. `DeltaS_(A rarr B) = DeltaS_(A rarr C)`C. `DeltaS_(Cyclic ne 0 )`D. `DeltaS_(C rarr A rarr B ne 0)`

Accepted Answer

NA

Question 10

Titration are one of the methods we can use to discover the precise concentration of solutions. A typical titration involves adding a solution from a burette to another solution in a flask. The end point of the titration is found by watching a colour change taking place.However a problem arises when a suitable indicator cannot be found or when the colour change involved are unclear. In these cases redox potential may sometimes come to the rescue.  A perticularly well known example (Fig.) is a method of discovering then concentration of iron (II) ions in a solution by titrating them with a solution of cerium (IV). The redox potential that are of interest here are`E_(Fe^(3+)|Fe^(2+))^(@)=0.77 V andE_(Ce^(4+|Ce^(3+)))^(@)=1.61V.` These tell us that cerium (IV) ions are the oxidising agents. They should react according to the equation `Fe^(2+) (aq) + Ce^(4+) (aq) to Fe^(3+) (aq) + Ce^(3+) (aq).` Now imagine that we know the concentration of the cerium (IV) ions solution in the burette. We want to measure the concentration of the cerium (IV) solution from the burette some of the iron (II) ions are will be oxidised. As a consequence the beaker would now contain large number of unreacted iron (II) ions but also some iron (III) ions as well. All of the cerium (III). The solution in the beaker now represents an iron `(III)iron(II)` half cell although not at standard conditions.Thus the emf of the cell will be near but not equal to `E_(Fe^(3+|Fe^(2+.)))^(@)`  If we continue to add cerium (IV) solution the number of iron (II) ions is gradually reduced and eventually only a very few are left (Table).At this stage the next few drops of cerium (IV) solution convert all the remaining irom (II) ions into iron (III) and some of the cerium (IV) ions are left unreacted. Once this happens we no longer have an iron `(III)iron(II)` half-cell. Instead we have a solution in which there is a large number of cerium (III) ions and a smaller number of cerium (IV) ions. The solution in the beaker now behaves as a cerium `(IV)cerium(III)`half-cell (although not a standard one.) Just before all the iron (II) ions are converted into iron (III) we have a cell with an EMF of around `+0.77V.` After all the iron (II) ions are oxidised we have a cell with an emf of about `+1.61 V. ` This rapid rise in emf occurs with the addition of just one drop of cerium (IV) solution. You should be able to understand why a graph of cell emf against volume of cerium (IV) solution of the iron (II) solution calculated in the usual way. When an ion is converted into a complex ion the redox potential change. you can see this in the combination of standard emf of the iron `(III)iron(II)` system `(+0.36 V).` The cyanide ion is said to stabilize the oxidation state of the iron. If you were to make up a cell `Pt(s)|underset(1mol dm^(-1)) (Fe(CN)_(6)^(3-))(aq)Fe(CN)_(6)^(4-)||underset(1mol dm^(-3))(Fe^(3+)(aq))Fe^(2+)(aq)|Pt(s)` what would be the emf and what would be the cell reaction?A. `(0.41V Fe(CN)_(6)^(4-)(aq)+Fe^(3+)(aq))(rarrFe(CN)_(6)^(3-)(aq)+Fe^(2+)(aq))`B. `(0.13V Fe(CN)_(6)^(4-)(aq)+Fe^(3+)(aq))(rarrFe(CN)_(6)^(3-)(aq)+Fe^(2+)(aq))`C. `(0.41V Fe(CN)_(6)^(3-)(aq)+Fe^(2+)(aq))(rarrFe(CN)_(6)^(4-)(aq)+Fe^(3+)(aq))`D. `(0.13V Fe(CN)_(6)^(3-)(aq)+Fe^(2+)(aq))(rarrFe(CN)_(6)^(4-)(aq)+Fe^(3+)(aq))`

Accepted Answer

NA

Question 11

Titration are one of the methods we can use to discover the precise concentration of solutions. A typical titration involves adding a solution from a burette to another solution in a flask. The end point of the titration is found by watching a colour change taking place.However a problem arises when a suitable indicator cannot be found or when the colour change involved are unclear. In these cases redox potential may sometimes come to the rescue.  A perticularly well known example (Fig.) is a method of discovering then concentration of iron (II) ions in a solution by titrating them with a solution of cerium (IV). The redox potential that are of interest here are`E_(Fe^(3+)|Fe^(2+))^(@)=0.77 V andE_(Ce^(4+|Ce^(3+)))^(@)=1.61V.` These tell us that cerium (IV) ions are the oxidising agents. They should react according to the equation `Fe^(2+) (aq) + Ce^(4+) (aq) to Fe^(3+) (aq) + Ce^(3+) (aq).` Now imagine that we know the concentration of the cerium (IV) ions solution in the burette. We want to measure the concentration of the cerium (IV) solution from the burette some of the iron (II) ions are will be oxidised. As a consequence the beaker would now contain large number of unreacted iron (II) ions but also some iron (III) ions as well. All of the cerium (III). The solution in the beaker now represents an iron `(III)iron(II)` half cell although not at standard conditions.Thus the emf of the cell will be near but not equal to `E_(Fe^(3+|Fe^(2+.)))^(@)`  If we continue to add cerium (IV) solution the number of iron (II) ions is gradually reduced and eventually only a very few are left (Table).At this stage the next few drops of cerium (IV) solution convert all the remaining irom (II) ions into iron (III) and some of the cerium (IV) ions are left unreacted. Once this happens we no longer have an iron `(III)iron(II)` half-cell. Instead we have a solution in which there is a large number of cerium (III) ions and a smaller number of cerium (IV) ions. The solution in the beaker now behaves as a cerium `(IV)cerium(III)`half-cell (although not a standard one.) Just before all the iron (II) ions are converted into iron (III) we have a cell with an EMF of around `+0.77V.` After all the iron (II) ions are oxidised we have a cell with an emf of about `+1.61 V. ` This rapid rise in emf occurs with the addition of just one drop of cerium (IV) solution. You should be able to understand why a graph of cell emf against volume of cerium (IV) solution of the iron (II) solution calculated in the usual way. Imagine you were given a solution of potassium dichromate (VI) in a beaker and a solution of iron (II) sulphate in a burette. You do not know the concentration of dichromate (VI) ions but the concentration of the iron (II) solution is known. Your task it to carry out a redox titration using the two solutions in order to determine the concentration of dichromate (VI) ions. Sketch a graph showing how the emf changes in the course of the above titration. `E_(Cr_(2)O_(7)^(2-)Cr^(3+)) = 1.33 V E_(Fe^(3+)Fe^(2+)) = 0.77 V`A. B. C. D.

Accepted Answer

NA

Question 12

Titration are one of the methods we can use to discover the precise concentration of solutions. A typical titration involves adding a solution from a burette to another solution in a flask. The end point of the titration is found by watching a colour change taking place.However a problem arises when a suitable indicator cannot be found or when the colour change involved are unclear. In these cases redox potential may sometimes come to the rescue.  A perticularly well known example (Fig.) is a method of discovering then concentration of iron (II) ions in a solution by titrating them with a solution of cerium (IV). The redox potential that are of interest here are`E_(Fe^(3+)|Fe^(2+))^(@)=0.77 V andE_(Ce^(4+|Ce^(3+)))^(@)=1.61V.` These tell us that cerium (IV) ions are the oxidising agents. They should react according to the equation `Fe^(2+) (aq) + Ce^(4+) (aq) to Fe^(3+) (aq) + Ce^(3+) (aq)`. Now imagine that we know the concentration of the cerium (IV) ions solution in the burette. We want to measure the concentration of the cerium (IV) solution from the burette some of the iron (II) ions are will be oxidised. As a consequence the beaker would now contain large number of unreacted iron (II) ions but also some iron (III) ions as well. All of the cerium (III). The solution in the beaker now represents an iron `(III)iron(II)` half cell although not at standard conditions.Thus the emf of the cell will be near but not equal to `E_(Fe^(3+|Fe^(2+.)))^(@)`  If we continue to add cerium (IV) solution the number of iron (II) ions is gradually reduced and eventually only a very few are left (Table).At this stage the next few drops of cerium (IV) solution convert all the remaining irom (II) ions into iron (III) and some of the cerium (IV) ions are left unreacted. Once this happens we no longer have an iron `(III)iron(II)` half-cell. Instead we have a solution in which there is a large number of cerium (III) ions and a smaller number of cerium (IV) ions. The solution in the beaker now behaves as a cerium `(IV)cerium(III)`half-cell (although not a standard one.) Just before all the iron (II) ions are converted into iron (III) we have a cell with an EMF of around `+0.77V.` After all the iron (II) ions are oxidised we have a cell with an emf of about `+1.61 V. ` This rapid rise in emf occurs with the addition of just one drop of cerium (IV) solution. You should be able to understand why a graph of cell emf against volume of cerium (IV) solution of the iron (II) solution calculated in the usual way. The cell shown below was set up `Pt|underset(1mol dm^(-3))(Fe^(3+)(aq)Fe^(2+)(aq))||underset(1mol dm^(-3)) (Br^(-)(aq)|Br_(2)(l))|Pt(s)` what would be the cell emf? If potassium cyanide solution were added to the left hand half cell (with due care!) what would you expect to happen to the emf of the cell ? `E_(Br_(2)Br^(-))^(@)= 1.07 V` and use data of previous question if required.A. `0.30 V emf will increase from 0.30 V to 0.41 V `B. `1.84 V emf will decrease from 1.84 V to 1.43 V `C. `0.30 V emf will increase from 0.30 V to 0.71 V `D. `0.30 V emf will increase from 0.30 V to 0.43 V `

Accepted Answer

NA

Question 13

Which one is the correct graph (fig.) for the corresponding acid base titration?A. B. C. D.

Accepted Answer

NA

Question 14

The curve represents the titration of a weak monoprotic acid. Ovar what `pH` range (s) will the acid being titrated serve as a buffer when mixed with its salt ?  (P) `pH 4 -6` (Q) `pH 9 -9` (R) `pH 12 -13`A. P onlyB. Q onlyC. P and R onlyD. PQ and R

Accepted Answer

NA

Question 15

A sample of 100 mL of a solution of a weak monoprotic acid of unknown concentration is titrated with `0.500 M NaOH` to give the titration curve shown. All of the statements are correct except A. Phenolphthalein would be a suitable indicator for this titrationB. A buffer solution is formed when 15 mL of `NaOH` is addedC. The `pK_(a)` of the acid is 4.0D. The initial concentration of the acid is `0.10M`

Accepted Answer

NA

Question 16

When weak base solution (50 ml of 0.1 `NH_(4)OH`) is titrated with strong acid 0.1 N HCl the pH of solution initially decreases fast and then decreases slowly till near equivalence point (as shown in the fig.) which of the following is true? A. The slow decrease of pH is due to formation of an acidic buffer solution after additio on some HCl.B. The slope of shown pH graph will be minimum when 25 ml of 0.1 N HCl is added.C. the slow decrease of pH is due to formation of a basic buffer solution.D. The initial fast decreament in pH is due to fast consumption of free `OH^(-) ` ions by HCl

Accepted Answer

NA

Question 17

1 mole of real gas changes its state from state -A ( bar 3L 100K) to state-B (2bar 5L 200K) at constant pressure and finally to state-C (3 bar 10 L 300K). If `DeltaU_(BC) = 110 J` and `C_(P_(m))` of gas `= 3R = 3x 8.3 JK^(-1) mol^(-1)` then choose the correct option (s).A. `W_(AB) = 830 J`B. `DeltaH_(AC) = 4600J`C. `DeltaU_(AC) = 2200 J`D. `DeltaU_(AC) = 1770J`

Accepted Answer

NA

Question 18

Two process are shown such that an ideal gas is taken from state 1 to state 3. Compare the following and choose the correct option(s). A. `W_(A) gt W_(B)`B. `DeltaU_(A) = DeltaU_(B)`C. `DeltaS_(A) = DeltaS_(B)`D. `q_(A) = q_(B)`

Accepted Answer

NA

Question 19

Which of the following is false regarding reversible adiabatic expansion of an ideal gas ?A. Plot of T us V is a straight line with slope equal to `gamma`B. Plot of ln T us ln V is a striaght line with slope equal to `gamma`C. Plot of ln T us ln V is a straight line with slope equal to `-gamma`D. Plot of ln T us ln V is a straight line with slope equal to `1 - gamma`

Accepted Answer

NA

Question 20

Which of the following graph is correct reversible adiabatic process for an ideal gasA. B. C. D.

Accepted Answer

NA

Question 21

A `25.0 mL` sample of waste water is obtained to analyze for `Pb^(2+)` ions. This sample is evaporated to dryness and redissolved in `2.0 mL` of `H_(2)O` mixed with `2.0 mL` of a buffer solution and `2.0 mL` of a solution of dithizone then diluted to `10.0 mL`. The absorbance of the coloured `Pb^(2+)` dithizone complex is compared with the Beer-Lambert plot below. The absorbance of a proton of the final solution is `0.13`. What is the concentration of `Pb^(2+)` ions in the waste water in ppm ? A. `2.9`B. `7.2`C. 18D. 36

Accepted Answer

NA

Question 22

From the phase diagram of water and an aqueous solution containing non-volatile solute identify the incorrect option. A. At temperature `T_(0)` vapour pressure of solid and vapour pressure of liquid will be same .B. Order of vapour pressure `(P_(0)P_(1)P_(2))` are `P_(0) gt P_(1) gt P_(2)`.C. `P_(0) = P_(2) e(DeltaH_(fusion)[T_(0)-T_(1)])(T_(0)T_(1))`D. `P_(1) = P_(2)e (DeltaH_(vap)[T_(0)-T_(1)])(T_(0)T_(1))`

Accepted Answer

NA

Question 23

For a second order reaction plots are made for `(1)([A])` us time for the reaction `2A rarr` Product. Pick up the correct sentences.A. The graph will show straight line with slope `K(2 xx` rate constant.)B. The graph will show straight line with intercept `[A]_(0)`C. The graph will show straight line with slope `[A]_(0)`D. The graph will show straight line with intercept `(1)([A]_(0))`

Accepted Answer

NA

Question 24

Which of the following statements isare correct with respect to surface phenomenon ?A. Potassium ferrocyanide can cause greater coagulation in a basic dye as compared to `Na_(2)HPO_(3)`B. A starch aqua-sol can act as protective colloid for `Fe(OH)_(3)` solC. The slope of the Freundlich isotherm (log `(x)(m)` us log P) keeps on changing for a long range of pressure and is constant over a limited range of pressure.D. On mixing `AgNO_(3)` with large amount of KI and subjecting the colloidal state to electrophoresis coagulation is obtained at cathode.

Accepted Answer

NA

Question 25

In an experiment to determine the enthalpy of neutralisation of sodium hydroxide with sulphuric acid `50 cm^(3)` of `0.4M` sodium hydroxide were titreated thermometrically with `0.25M` sulphuric acid. Which of the following plots gives the correct representation ?A. B. C. D.

Accepted Answer

NA

Question 26

The plots of `(1)(x_(A))` (on y-axis) `(1)(y_(A))` (on x-axis) is linear with slope and intercept respectively. `p^(@)A =` vapour pressure of pure liquid A `p^(@)B =` vapour pressure of pure liquid BA. `(p_(A)^(@))(p_(B)^(@))` and `((p_(A)^(@)-p_(B)^(@)))(p_(B)^(@))`B. `(P_(a)^(@))(P_(b)^(@))` and `((P_(B)^(@)-P_(A)^(@)))(p_(B)^(@))`C. `(p_(B)^(@))(p_(A)^(@))` and `((p_(A)^(@)-p_(B)^(@)))(p_(B)^(@))`D. `(p_(B)^(@))(p_(A)^(@))` and `((p_(B)^(@)-p_(A)^(@)))(p_(B)^(@))`

Accepted Answer

NA

Question 27

Conductance measurements and be used to detect the end point of acid-base titrations. Which of the following plots correctly represents the end point of the titration of strong acid and a strong base?A. B. C. D.

Accepted Answer

NA

Question 28

Which of the following plots of radical probability function `4pi r^(2)Psi_(r)^(2)` is incorrectly labelled ?A. B. C. D.

Accepted Answer

NA

Explore topic-wise InterviewSolutions in .

Consider the first order reaction: `A rarr 2B`. Which of the following figure correctly describes the rate of disappearance of `A(r_(1))` and rate of appearance of `B(r_(2))` with time ?A. B. C. D.

Which graph represents the correct relationship between various velocities for an idela gas ?A. B. C. D.

Which of the following graph is correct w.r.t half life for a zero order reaction at various stages of reaction ?A. B. C. D.

A student analyzed tha data from a zero order reaction and obtained the graph shown. What labels should be attached to the X and Y axes, respectively ? A. Time, concentrationB. Time, `1//`concentrationC. Time, ln (concentration)D. `1//"time"`, concentration

In given cyclic process for an ideal gas Path `B rarr C` is isoentropic. Then select the correct options:A. `DeltaS_(A_B) = DeltaS_(C rarr A)`B. `DeltaS_(A rarr B) = DeltaS_(A rarr C)`C. `DeltaS_("Cyclic" ne 0 )`D. `DeltaS_(C rarr A rarr B ne 0)`

Which one is the correct graph (fig.) for the corresponding acid base titration?A. B. C. D.

The curve represents the titration of a weak monoprotic acid. Ovar what `pH` range (s) will the acid being titrated, serve as a buffer when mixed with its salt ? (P) `pH 4 -6` (Q) `pH 9 -9` (R) `pH 12 -13`A. P onlyB. Q onlyC. P and R onlyD. P,Q and R

Two process are shown such that an ideal gas is taken from state 1 to state 3. Compare the following and choose the correct option(s). A. `W_(A) gt W_(B)`B. `DeltaU_(A) = DeltaU_(B)`C. `DeltaS_(A) = DeltaS_(B)`D. `q_(A) = q_(B)`

Which of the following graph is correct reversible adiabatic process for an ideal gas:A. B. C. D.

In an experiment to determine the enthalpy of neutralisation of sodium hydroxide with sulphuric acid, `50 cm^(3)` of `0.4M` sodium hydroxide were titreated thermometrically with `0.25M` sulphuric acid. Which of the following plots gives the correct representation ?A. B. C. D.

Conductance measurements and be used to detect the end point of acid-base titrations. Which of the following plots correctly represents the end point of the titration of strong acid and a strong base?A. B. C. D.

Which of the following plots of radical probability function `4pi r^(2)Psi_(r)^(2)` is incorrectly labelled ?A. B. C. D.

151.	Consider the first order reaction: `A rarr 2B`. Which of the following figure correctly describes the rate of disappearance of `A(r_(1))` and rate of appearance of `B(r_(2))` with time ?A. B. C. D.
Answer» Correct Answer - C

152.	Consider the following equilibrium `A(g) hArr 2B(g)` and the following graph : Select the correct statements. A. Effect (I) represent the insertion of inert gas at constant volumeB. Effect (II) represent the insertion of inert gas at constant pressureC. Effect (III) represent the decrease in volume of containerD. Effect (IV) represent the addition of moles of gas B
Answer» Correct Answer - A::B::C::D

153.	Which graph represents the correct relationship between various velocities for an idela gas ?A. B. C. D.
Answer» Correct Answer - D

154.	The following graph is experimentally obtained for the reaction : `A rarr 2B`, at `25^(@)C`. The correct statement (s) for the reaction is/are (Given : ln 2 = 0.7) A. Time for `87.5%` reaction of A is 21 minB. The initial concentration of A was eMC. The time at which the concentration of A and B becomes equal is 7 minD. The rate of appearance of B is `(d[B])/(Dt) = (0.2 min^(-1)) [A]`
Answer» Correct Answer - A::B::D

155.	Which of the following graph is correct w.r.t half life for a zero order reaction at various stages of reaction ?A. B. C. D.
Answer» Correct Answer - C

156.	A student analyzed tha data from a zero order reaction and obtained the graph shown. What labels should be attached to the X and Y axes, respectively ? A. Time, concentrationB. Time, `1//`concentrationC. Time, ln (concentration)D. `1//"time"`, concentration
Answer» Correct Answer - A

157.	For an ideal gas four processes are marked as 1,2,3 and 4 on P-V diagram as shown in figure. The amount of heat supplied to the gas in the process 1,2,3 and 4 are `Q_(1),Q_(2),Q_(3)` and `Q_(4)` respectively, then correct order of heat supplied to the gas is: A. `Q_(1) gt Q_(2) gt Q_(3) gt Q_(4)`B. `Q_(1) lt Q_(2) lt Q_(3) lt Q_(4)`C. `Q_(1) gt Q_(2) gt Q_(4) gt Q_(3)`D. `Q_(1) gt Q_(4) gt Q_(2) gt Q_(3)`
Answer» Correct Answer - A

158.	Consider the cyclic process `R rarr S rarr R` as shown in the fig. You are told that one of the path is adiabatic and the other one isothermal. Which one of the following is (are) true ? A. Process `R rarr S` is isothermalB. Process `S rarr R` is adiabaticC. Process `R rarr S` is adiabaicD. Such a graph is not possible
Answer» Correct Answer - D

159.	In given cyclic process for an ideal gas Path `B rarr C` is isoentropic. Then select the correct options:A. `DeltaS_(A_B) = DeltaS_(C rarr A)`B. `DeltaS_(A rarr B) = DeltaS_(A rarr C)`C. `DeltaS_("Cyclic" ne 0 )`D. `DeltaS_(C rarr A rarr B ne 0)`
Answer» Correct Answer - B

160.	Titration are one of the methods we can use to discover the precise concentration of solutions. A typical titration involves adding a solution from a burette to another solution in a flask. The end point of the titration is found by watching a colour change taking place.However, a problem arises when a suitable indicator cannot be found, or when the colour change involved are unclear. In these cases, redox potential may sometimes come to the rescue. A perticularly well known example (Fig.) is a method of discovering then concentration of iron (II) ions in a solution by titrating them with a solution of cerium (IV). The redox potential that are of interest here are`E_(Fe^(3+)\|Fe^(2+))^(@)=0.77 V "and"E_(Ce^(4+\|Ce^(3+)))^(@)=1.61V.` These tell us that cerium (IV) ions are the oxidising agents. They should react according to the equation `Fe^(2+) (aq) + Ce^(4+) (aq) to Fe^(3+) (aq) + Ce^(3+) (aq).` Now imagine that we know the concentration of the cerium (IV) ions solution in the burette. We want to measure the concentration of the cerium (IV) solution from the burette, some of the iron (II) ions are will be oxidised. As a consequence the beaker would now contain large number of unreacted iron (II) ions , but also some iron (III) ions as well. All of the cerium (III). The solution in the beaker now represents an iron `(III)//iron(II)` half cell although not at standard conditions.Thus , the emf of the cell will be near, but not equal to `E_(Fe^(3+\|Fe^(2+.)))^(@)` If we continue to add cerium (IV) solution , the number of iron (II) ions is gradually reduced and eventually only a very few are left (Table).At this stage the next few drops of cerium (IV) solution convert all the remaining irom (II) ions into iron (III), and some of the cerium (IV) ions are left unreacted. Once this happens we no longer have an iron `(III)//"iron"(II)` half-cell. Instead we have a solution in which there is a large number of cerium (III) ions and a smaller number of cerium (IV) ions. The solution in the beaker now behaves as a cerium `(IV)//"cerium"(III)`half-cell (although not a standard one.) Just before all the iron (II) ions are converted into iron (III) we have a cell with an EMF of around `+0.77V.` After all the iron (II) ions are oxidised we have a cell with an emf of about `+1.61 V. ` This rapid rise in emf occurs with the addition of just one drop of cerium (IV) solution. You should be able to understand why a graph of cell emf against volume of cerium (IV) solution of the iron (II) solution calculated in the usual way. When an ion is converted into a complex ion, the redox potential change. you can see this in the combination of standard emf of the iron `(III)//"iron"(II)` system `(+0.36 V).` The cyanide ion is said to stabilize the oxidation state of the iron. If you were to make up a cell `Pt(s)\|underset(1"mol dm"^(-1)) (Fe(CN)_(6)^(3-))(aq),Fe(CN)_(6)^(4-)\|\|underset(1"mol dm"^(-3))(Fe^(3+)(aq)),Fe^(2+)(aq)\|Pt(s)` what would be the emf and what would be the cell reaction?A. `{:(0.41V Fe(CN)_(6)^(4-)(aq)+,Fe^(3+)(aq)),(,rarrFe(CN)_(6)^(3-)(aq)+Fe^(2+)(aq)):}`B. `{:(0.13V Fe(CN)_(6)^(4-)(aq)+,Fe^(3+)(aq)),(,rarrFe(CN)_(6)^(3-)(aq)+Fe^(2+)(aq)):}`C. `{:(0.41V Fe(CN)_(6)^(3-)(aq)+,Fe^(2+)(aq)),(,rarrFe(CN)_(6)^(4-)(aq)+Fe^(3+)(aq)):}`D. `{:(0.13V Fe(CN)_(6)^(3-)(aq)+Fe^(2+)(aq)),(,rarrFe(CN)_(6)^(4-)(aq)+Fe^(3+)(aq)):}`
Answer» Correct Answer - A

161.	Titration are one of the methods we can use to discover the precise concentration of solutions. A typical titration involves adding a solution from a burette to another solution in a flask. The end point of the titration is found by watching a colour change taking place.However, a problem arises when a suitable indicator cannot be found, or when the colour change involved are unclear. In these cases, redox potential may sometimes come to the rescue. A perticularly well known example (Fig.) is a method of discovering then concentration of iron (II) ions in a solution by titrating them with a solution of cerium (IV). The redox potential that are of interest here are`E_(Fe^(3+)\|Fe^(2+))^(@)=0.77 V "and"E_(Ce^(4+\|Ce^(3+)))^(@)=1.61V.` These tell us that cerium (IV) ions are the oxidising agents. They should react according to the equation `Fe^(2+) (aq) + Ce^(4+) (aq) to Fe^(3+) (aq) + Ce^(3+) (aq).` Now imagine that we know the concentration of the cerium (IV) ions solution in the burette. We want to measure the concentration of the cerium (IV) solution from the burette, some of the iron (II) ions are will be oxidised. As a consequence the beaker would now contain large number of unreacted iron (II) ions , but also some iron (III) ions as well. All of the cerium (III). The solution in the beaker now represents an iron `(III)//iron(II)` half cell although not at standard conditions.Thus , the emf of the cell will be near, but not equal to `E_(Fe^(3+\|Fe^(2+.)))^(@)` If we continue to add cerium (IV) solution , the number of iron (II) ions is gradually reduced and eventually only a very few are left (Table).At this stage the next few drops of cerium (IV) solution convert all the remaining irom (II) ions into iron (III), and some of the cerium (IV) ions are left unreacted. Once this happens we no longer have an iron `(III)//"iron"(II)` half-cell. Instead we have a solution in which there is a large number of cerium (III) ions and a smaller number of cerium (IV) ions. The solution in the beaker now behaves as a cerium `(IV)//"cerium"(III)`half-cell (although not a standard one.) Just before all the iron (II) ions are converted into iron (III) we have a cell with an EMF of around `+0.77V.` After all the iron (II) ions are oxidised we have a cell with an emf of about `+1.61 V. ` This rapid rise in emf occurs with the addition of just one drop of cerium (IV) solution. You should be able to understand why a graph of cell emf against volume of cerium (IV) solution of the iron (II) solution calculated in the usual way. Imagine you were given a solution of potassium dichromate (VI) in a beaker and a solution of iron (II) sulphate in a burette. You do not know the concentration of dichromate (VI) ions , but the concentration of the iron (II) solution is known. Your task it to carry out a redox titration using the two solutions in order to determine the concentration of dichromate (VI) ions. Sketch a graph showing how the emf changes in the course of the above titration. `E_(Cr_(2)O_(7)^(2-)//Cr^(3+)) = 1.33 V, E_(Fe^(3+)//Fe^(2+)) = 0.77 V`A. B. C. D.
Answer» Correct Answer - B

162.	Titration are one of the methods we can use to discover the precise concentration of solutions. A typical titration involves adding a solution from a burette to another solution in a flask. The end point of the titration is found by watching a colour change taking place.However, a problem arises when a suitable indicator cannot be found, or when the colour change involved are unclear. In these cases, redox potential may sometimes come to the rescue. A perticularly well known example (Fig.) is a method of discovering then concentration of iron (II) ions in a solution by titrating them with a solution of cerium (IV). The redox potential that are of interest here are`E_(Fe^(3+)\|Fe^(2+))^(@)=0.77 V "and"E_(Ce^(4+\|Ce^(3+)))^(@)=1.61V.` These tell us that cerium (IV) ions are the oxidising agents. They should react according to the equation `Fe^(2+) (aq) + Ce^(4+) (aq) to Fe^(3+) (aq) + Ce^(3+) (aq)`. Now imagine that we know the concentration of the cerium (IV) ions solution in the burette. We want to measure the concentration of the cerium (IV) solution from the burette, some of the iron (II) ions are will be oxidised. As a consequence the beaker would now contain large number of unreacted iron (II) ions , but also some iron (III) ions as well. All of the cerium (III). The solution in the beaker now represents an iron `(III)//iron(II)` half cell although not at standard conditions.Thus , the emf of the cell will be near, but not equal to `E_(Fe^(3+\|Fe^(2+.)))^(@)` If we continue to add cerium (IV) solution , the number of iron (II) ions is gradually reduced and eventually only a very few are left (Table).At this stage the next few drops of cerium (IV) solution convert all the remaining irom (II) ions into iron (III), and some of the cerium (IV) ions are left unreacted. Once this happens we no longer have an iron `(III)//"iron"(II)` half-cell. Instead we have a solution in which there is a large number of cerium (III) ions and a smaller number of cerium (IV) ions. The solution in the beaker now behaves as a cerium `(IV)//"cerium"(III)`half-cell (although not a standard one.) Just before all the iron (II) ions are converted into iron (III) we have a cell with an EMF of around `+0.77V.` After all the iron (II) ions are oxidised we have a cell with an emf of about `+1.61 V. ` This rapid rise in emf occurs with the addition of just one drop of cerium (IV) solution. You should be able to understand why a graph of cell emf against volume of cerium (IV) solution of the iron (II) solution calculated in the usual way. The cell shown below was set up `Pt\|underset(1"mol dm"^(-3))(Fe^(3+)(aq),Fe^(2+)(aq)),\|\|underset(1"mol dm"^(-3)) (Br^(-)(aq)\|Br_(2)(l)),\|Pt(s)` what would be the cell emf? If potassium cyanide solution were added to the left hand half cell (with due care!) what would you expect to happen to the emf of the cell ? `E_(Br_(2)//Br^(-))^(@)= 1.07 V` and use data of previous question, if required.A. `0.30 V, "emf will increase from 0.30 V to" 0.41 V `B. `1.84 V, "emf will decrease from 1.84 V to" 1.43 V `C. `0.30 V, "emf will increase from 0.30 V to" 0.71 V `D. `0.30 V, "emf will increase from 0.30 V to" 0.43 V `
Answer» Correct Answer - C

163.	Which one is the correct graph (fig.) for the corresponding acid base titration?A. B. C. D.
Answer» Correct Answer - A::B::C::D

164.	The curve represents the titration of a weak monoprotic acid. Ovar what `pH` range (s) will the acid being titrated, serve as a buffer when mixed with its salt ? (P) `pH 4 -6` (Q) `pH 9 -9` (R) `pH 12 -13`A. P onlyB. Q onlyC. P and R onlyD. P,Q and R
Answer» Correct Answer - A

165.	A sample of 100 mL of a solution of a weak monoprotic acid of unknown concentration is titrated with `0.500 M NaOH` to give the titration curve shown. All of the statements are correct except: A. Phenolphthalein would be a suitable indicator for this titrationB. A buffer solution is formed when 15 mL of `NaOH` is addedC. The `pK_(a)` of the acid is 4.0D. The initial concentration of the acid is `0.10M`
Answer» Correct Answer - C

166.	When weak base solution (50 ml of 0.1 `NH_(4)OH`) is titrated with strong acid 0.1 N HCl, the pH of solution initially decreases fast and then decreases slowly till near equivalence point (as shown in the fig.) which of the following is true? A. The slow decrease of pH is due to formation of an acidic buffer solution after additio on some HCl.B. The slope of shown pH graph will be minimum when 25 ml of 0.1 N HCl is added.C. the slow decrease of pH is due to formation of a basic buffer solution.D. The initial fast decreament in pH is due to fast consumption of free `OH^(-) ` ions by HCl
Answer» Correct Answer - B::C::D

167.	1 mole of real gas changes its state from state -A ( bar, 3L, 100K) to state-B (2bar, 5L, 200K) at constant pressure and finally to state-C (3 bar, 10 L, 300K). If `DeltaU_(BC) = 110 J` and `C_(P_(m))` of gas `= 3R = 3x 8.3 JK^(-1) mol^(-1)` then choose the correct option (s).A. `W_(AB) = 830 J`B. `DeltaH_(AC) = 4600J`C. `DeltaU_(AC) = 2200 J`D. `DeltaU_(AC) = 1770J`
Answer» Correct Answer - B::C

168.	Two process are shown such that an ideal gas is taken from state 1 to state 3. Compare the following and choose the correct option(s). A. `W_(A) gt W_(B)`B. `DeltaU_(A) = DeltaU_(B)`C. `DeltaS_(A) = DeltaS_(B)`D. `q_(A) = q_(B)`
Answer» Correct Answer - B::C::D

169.	Which of the following is false regarding reversible adiabatic expansion of an ideal gas ?A. Plot of T us V is a straight line with slope equal to `gamma`B. Plot of ln T us ln V is a striaght line with slope equal to `gamma`C. Plot of ln T us ln V is a straight line with slope equal to `-gamma`D. Plot of ln T us ln V is a straight line with slope equal to `1 - gamma`
Answer» Correct Answer - A::B::C

170.	Which of the following graph is correct reversible adiabatic process for an ideal gas:A. B. C. D.
Answer» Correct Answer - B

171.	A `25.0 mL` sample of waste water is obtained to analyze for `Pb^(2+)` ions. This sample is evaporated to dryness and redissolved in `2.0 mL` of `H_(2)O`, mixed with `2.0 mL` of a buffer solution and `2.0 mL` of a solution of dithizone then diluted to `10.0 mL`. The absorbance of the coloured `Pb^(2+)` dithizone complex is compared with the Beer-Lambert plot below. The absorbance of a proton of the final solution is `0.13`. What is the concentration of `Pb^(2+)` ions in the waste water in ppm ? A. `2.9`B. `7.2`C. 18D. 36
Answer» Correct Answer - A

172.	From the phase diagram of water and an aqueous solution containing non-volatile solute, identify the incorrect option. A. At temperature `T_(0)`, vapour pressure of solid and vapour pressure of liquid will be same .B. Order of vapour pressure `(P_(0),P_(1),P_(2))` are `P_(0) gt P_(1) gt P_(2)`.C. `P_(0) = P_(2) e(DeltaH_("fusion")[T_(0)-T_(1)])/(T_(0)T_(1))`D. `P_(1) = P_(2)e (DeltaH_(vap)[T_(0)-T_(1)])/(T_(0)T_(1))`
Answer» Correct Answer - C

173.	For a second order reaction, plots are made for `(1)/([A])` us time for the reaction, `2A rarr` Product. Pick up the correct sentences.A. The graph will show straight line with slope `K(2 xx` rate constant.)B. The graph will show straight line with intercept `[A]_(0)`C. The graph will show straight line with slope `[A]_(0)`D. The graph will show straight line with intercept `(1)/([A]_(0))`
Answer» Correct Answer - A::D

174.	Which of the following statements is/are correct with respect to surface phenomenon ?A. Potassium ferrocyanide can cause greater coagulation in a basic dye as compared to `Na_(2)HPO_(3)`B. A starch aqua-sol can act as protective colloid for `Fe(OH)_(3)` solC. The slope of the Freundlich isotherm (log `(x)/(m)` us log P) keeps on changing for a long range of pressure and is constant over a limited range of pressure.D. On mixing `AgNO_(3)` with large amount of KI and subjecting the colloidal state to electrophoresis, coagulation is obtained at cathode.
Answer» Correct Answer - A::B::C

175.	In an experiment to determine the enthalpy of neutralisation of sodium hydroxide with sulphuric acid, `50 cm^(3)` of `0.4M` sodium hydroxide were titreated thermometrically with `0.25M` sulphuric acid. Which of the following plots gives the correct representation ?A. B. C. D.
Answer» Correct Answer - B

176.	The plots of `(1)/(x_(A))` (on y-axis) `(1)/(y_(A))` (on x-axis) is linear with slope and intercept respectively. `p^(@)A =` vapour pressure of pure liquid A `p^(@)B =` vapour pressure of pure liquid BA. `(p_(A)^(@))/(p_(B)^(@))` and `((p_(A)^(@)-p_(B)^(@)))/(p_(B)^(@))`B. `(P_(a)^(@))/(P_(b)^(@))` and `((P_(B)^(@)-P_(A)^(@)))/(p_(B)^(@))`C. `(p_(B)^(@))/(p_(A)^(@))` and `((p_(A)^(@)-p_(B)^(@)))/(p_(B)^(@))`D. `(p_(B)^(@))/(p_(A)^(@))` and `((p_(B)^(@)-p_(A)^(@)))/(p_(B)^(@))`
Answer» Correct Answer - B

177.	Conductance measurements and be used to detect the end point of acid-base titrations. Which of the following plots correctly represents the end point of the titration of strong acid and a strong base?A. B. C. D.
Answer» Correct Answer - A

178.	Which of the following plots of radical probability function `4pi r^(2)Psi_(r)^(2)` is incorrectly labelled ?A. B. C. D.
Answer» Correct Answer - A