CBSE Class 12 Chemistry Sample Paper 2022 – 23 Solutions
CBSE Class 12 Chemistry Sample Paper 2022: Central Board of Secondary Education(CBSE) has released the CBSE Class 12 Chemistry Sample Paper 2022 – 23 on its official website on 16th September 2022.
SECTION A
(1) The major product of acid catalysed dehydration of 1-methylcyclohexanol is:
(a) 1-methylcyclohexane
(b) 1-methylcyclohexene
(c) 1-cyclohexylmethanol
(d) 1-methylenecyclohexane
Ans: 1-methylcyclohexene According to Saytzeff rule i.e highly substituted alkene is major product. Here dehydration reaction takes place, alkene is formed due to the removal of a water molecule.
(2) Which one of the following compounds is more reactive towards SN1 reaction?
(a) CH2=CHCH2Br
(b) C6H5CH2Br
(c)C6H5CH (C6H5)Br
(d) C6H5CH(CH3) Br
Ans: c C6H5CH (C6H5)Br C6H5CH (C6H5)+ carbocation formed is more stable
(3) KMnO4 is coloured due to:
(a) d-d transitions
(b) charge transfer from ligand to metal
(c) unpaired electrons in d orbital of Mn
(d) charge transfer from metal to ligand
Ans: (b) charge transfer from ligand to metal The Mn atom in KMnO4 has +7 oxidation state with electron configuration [Ar]3d 04s0 Since no unpaired electrons are present, d−d transitions are not possible. The molecule should, therefore, be colourless. Its intense purple due to L→M (ligand to metal) charge transfer 2p(L) of O to 3d(M) of Mn.
(4) Which radioactive isotope would have the longer half- life 15O or 19O? (Given rate constants for 15O and 19O are 5.63x 10-3 s -1 and k = 2.38 x 10-2s -1 respectively.)
(a) 15O
(b) 19O
(c) Both will have the same half-life
(d) None of the above, information given is insufficient
Ans: (a) 15O The rate constant for the decay of O-15 is less than that for O-19 . Therefore , the rate of decay of O-15 will be slower and will have a longer half life .
(5) The molar conductivity of CH3COOH at infinite dilution is 390 Scm2 /mol. Using the graph and given information, the molar conductivity of CH3COOK will be:
(a) 100 Scm2 /mol
(b) 115 Scm2 /mol
(c) 150 Scm2 /mol
(d) 125 Scm2 /mol
Ans: (b) 115 Scm2 /mol
ΛOCH3COOK = ΛO CH3COOH +ΛOKCl – ΛOHCl = 390 +150-425 = 115 Scm2 /mol
(6) For the reaction, A + 2B → Ab2 → the order w.r.t. reactant A is 2 and w.r.t. reactant B .What will be change in rate of reaction if the concentration of A is doubled and B is halved?
(a) increases four times
(b) decreases four times
(c) increases two times
(d) no change
Ans: (a) increases 4 times
Rate = [A]2 If [A] is doubled then Rate’ = [2A]2 = 4 [A]2 = 4 Rat
(7) Arrange the following in the increasing order of their boiling points:
A : Butanamine, B: N,N-Dimethylethanamine, C: N- Etthylethanaminamine
(a) a. C<B<A
(b) A<B<C
(c) A <C< B
(d) B<C<A
Ans: (d) B<C<A
In primary amine intermolecular association due to H-bonding is maximum while in tertiary it is minimum.
(8) The CFSE of [CoCl6] 3- is 18000 cm-1 the CFSE for [CoCl4] – will be:
(a) 18000 cm-1
(b) 8000cm-1
(c) 2000 cm-1
(d) 16000 cm-1
Ans: 8000 cm-1
∆t =(4/9 ) x 18000cm-1 =8000 cm-1
(9) What would be the major product of the following reaction?
Ans: C6 H5 – CH2– OC6h5 + HBr → A + B
(a) A= C6H5CH2OH , B= C6H6
(b) A=C6H5CH2OH ,B= C6H5Br
(c) A=C6H5CH3 ,B= C6H5Br
(d) A=C6H5CH2Br , B= C6H5OH
Ans: (d) A.=C6H5CH2Br , B = C6H5OH,
C6H5CH2OC6H5 H+ C6H5CH2OC6H5
(10) Which of the following statements is not correct for amines?
(a) Most alkyl amines are more basic than ammonia solution.
(b) pKb value of ethylamine is lower than benzylamine
(c) CH3NH2 on reaction with nitrous acid releases NO2 gas.
(d) Hinsberg’s reagent reacts with secondary amines to form sulphonamides.
Ans: (c) CH3NH2 on reaction with nitrous acid releases NO2 gas Wrong statement . The evolution of nitrogen gas takes place.
(11) Which of the following tests/ reactions is given by aldehydes as well as ketones?
(a) Fehling’s test
(b) Tollen’s test
(c) 2,4 DNP test
(d) Cannizzaro reaction
Ans: (c) 2,4 DNP test
Fehling’s, Tollen’s and Cannizzao reaction is shown by alcohols only.
(12) Arrhenius equation can be represented graphically as follows:
The (i) intercept and (ii) slope of the graph are:
(a) (i) in A |
(ii) Ea/R |
(b) (i) A |
(ii) Ea |
(c) (i) In A |
(ii) – Ea/R |
(d) (i) A |
(ii) – Ea |
Ans: (c) (i)In A (ii) – Ea/R
(13) The number of ions formed on dissolving one molecule of FeSO4.(NH4)2SO4.6H2O in water is:
(a) 3
(b) 4
(c) 5
(d) 6
Ans: 1Fe2+, 2 SO4 2- and 2 NH4 + ions
(14) The oxidation of toluene to benzaldehyde by chromyl chloride is called
(a) Etard reaction
(b) Riemer-Tiemann reaction
(c) Stephen’s reaction
(d) Cannizzaro’s reaction
Ans: (a) Etard reaction
(15) Given below are two statements labelled as Assertion (A) and Reason (R)
Assertion (A): An ether is more volatile than an alcohol of comparable molecular mass.
Reason (R): Ethers are polar in nature.
Select the most appropriate answer from the options given below:
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Ans: (b) Both A and R are true but R is not the correct explanation of A. A and R are two different statements about ethers The correct reason is that hydrogen bonding does not exist amongst ether molecules.
(16) Given below are two statements labelled as Assertion (A) and Reason (R)
Assertion (A): Proteins are found to have two different types of secondary structures viz alpha-helix and beta-pleated sheet structure.
Reason (R): The secondary structure of proteins is stabilized by hydrogen bonding
Select the most appropriate answer from the options given below:
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Ans: (b) Both A and R are true but R is not the correct explanation of A.
(17) Given below are two statements labelled as Assertion (A) and Reason (R)
Assertion : Magnetic moment values of actinides are lesser than the theoretically predicted values.
Reason : Actinide elements are strongly paramagnetic.
Select the most appropriate answer from the options given below:
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Ans: (b) Both A and R are true but R is not the correct explanation of A. The magnetic moment is less as the 5f electrons of actinides are less effectively shielded which results in quenching of orbital contributions , they are strongly paramagnetic due to presence of unpaired electrons
(18) Given below are two statements labelled as Assertion (A) and Reason (R)
Assertion (A): Tertiary amines are more basic than corresponding secondary and primary amines in gaseous state
Reason (R): Tertiary amines have three alkyl groups which cause +I effect.
Select the most appropriate answer from the options given below:
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Ans: (a) Both A and R are true and R is the correct explanation of A.
SECTION B
(19) A first-order reaction takes 69.3 min for 50% completion. What is the time needed for 80% of the reaction to get completed? (Given: log 5 =0.6990, log 8 = 0.9030, log 2 = 0.3010)
Ans: .Half life t1/2 = 0.693 /k
k= 0.693/69.3 = 1/100 = 0.01 min-1
For first order reaction
= 2.303/ []/ []
= 2.303 /0.01 100/ 20
= 230.3 log 5 (log 5 =0.6990)
t= 160.9 min
(20) Account for the following:
(a) There are 5 OH groups in glucose
Ans: (a) Acetylation of glucose with acetic anhydride gives glucose pentaacetate which confirms the presence of five –OH groups. Since it exists as a stable compound, five –OH groups should be attached to different carbon atoms
(b) Glucose is a reducing sugar
Ans:
OR
What happens when D – glucose is treated with the following reagents
(a) Bromine water
Ans:
(b) HNO3
Ans:
(21) Give reason for the following:
(a) During the electrophilic substitution reaction of haloarenes, para substituted derivative is the major product.
Ans: At the ortho position, higher steric hindrance is there, hence para isomer is usually predominate and is obtained in the major amount.
(b) The product formed during SN 1 reaction is a racemic mixture.
Ans: During the SN 1 mechanism, intermediate carbocation formed is sp2 hybridized and planar in nature. This allows the attack of nucleophile from either side of the plane resulting in a racemic mixture.
OR
(a) Name the suitable alcohol and reagent, from which 2-Chloro-2-methyl propane can be prepared.
Ans: Tert butyl alcohol or 2-methyl propan-2-ol using Lucas reagent , mixture of concHCl and ZnCl2 the reaction will follow the SN 1 pathway
(b) Out of the Chloromethane and Fluoromethane , which one is has higher dipole moment and why?
Ans: Chloromethane is having higher dipole moment . Due to smaller size of fluorine the dipole moment of flouromethane is comparatively lesser.
(22) The formula Co(NH3)5CO3Cl could represent a carbonate or a chloride. Write the structures and names of possible isomers.
Ans: [Co(NH3)5CO3]Cl and [Co(NH3)5Cl]CO3
Pentaaminecarbonatocobalt(III)chloride
Pentaaminechloridocobalt(III)carbonate
(23) Corrosion is an electrochemical phenomenon. The oxygen in moist air reacts as follows:
O2(g) + 2H2O(l) + 4e– → 4OH– (aq).
Write down the possible reactions for corrosion of zinc occurring at anode, cathode, and overall reaction to form a white layer of zinc hydroxide.
Ans: Anode: Zn (s) → Zn 2+ (aq) + 2 e-
Cathode: O2(g) + 2H2O(l) + 4e– → 4OH– (aq).
2 Zn2+ (aq) + 4OH–àOverall: 2 Zn (s) 2+ + O2(g) + 2H2O(l) (aq)
2 Zn (s) + O2(g0 + 2H2O(I) → 2 Zn(OH)2 (ppt)
(24) .Explain how and why will the rate of reaction for a given reaction be affected when
(a) a catalyst is added
Ans: The rate of reaction will increase. The catalyst decreases the activation energy of the reaction therefore the reaction becomes faster.
(b) the temperature at which the reaction was taking place is decreased
Ans: The rate of reaction will decrease. At lower temperatures the kinetic energy of molecules decreases thereby the collisions decrease resulting in a lowering of rate of reaction.
(25) Write the reaction and IUPAC name of the product formed when 2-Methylpropanal (isobutyraldehyde) is treated with ethyl magnesium bromide followed by hydrolysis.
Ans: (CH3)2CHCHO + C2H5MgBr dry ether (CH3)2CHCH(C2H5)(OMgBr)
(CH3)2CHCH(C2H5)(OMgBr) H+ /H2O (CH3)2CHCH(C2H5)(OH)
2-Methylpentan-3-ol
SECTION C
(26) Write the equations for the following reaction:
(a) Salicylic acid is treated with acetic anhydride in the presence of conc. H2SO4
Ans:
(b) Tert butyl chloride is treated with sodium ethoxide.
Ans: (CH3)3CCl sodium ethoxide (CH3)2C=CH2 2methylpropene
(c) Phenol is treated with chloroform in the presence of NaOH
Ans: o-hydroxybezaldehyde will be formed
(27) Using Valence bond theory, explain the following in relation to the paramagnetic complex [Mn(CN)6] 3-
(a) type of hybridization
(b) magnetic moment value
(c) type of complex – inner, outer orbital complex
Ans:
(28) Answer the following questions:
(a) State Henry’s law and explain why are the tanks used by scuba divers filled with air diluted with helium (11.7% helium, 56.2% nitrogen and 32.1% oxygen)?
Ans: Henry’s law: the partial pressure of the gas in vapour phase (p) is proportional to the mole fraction of the gas (x) in the solution.
The pressure underwater is high, so the solubility of gases in blood increases. When the diver comes to surface the pressure decreases so does the solubility causing bubbles of nitrogen in blood, to avoid this situation and maintain the same partial pressure of nitrogen underwater too, the dilution is done.
(b) Assume that argon exerts a partial pressure of 6 bar. Calculate the solubility of argon gas in water. (Given Henry’s law constant for argon dissolved in water, KH = 40kbar)
Ans: . p = KH x
(29) Give reasons for any 3 of the following observations:
(a) Aniline is acetylated before nitration reaction.
Ans: Aniline is acetylated, before nitration reaction in order to avoid formation of tarry oxidation products and protecting the amino group, so that p -nitro derivative can be obtained as major product.
(b) pKb of aniline is lower than the m-nitroaniline.
Ans: pKb of aniline is lower than the m-nitro aniline.The basic strength of aniline is more that m-nitroaniline . pkb value is inversely proportional to basic strength. Presence of Electron withdrawing group decrease basic strength.
(c) Primary amine on treatment with benzenesulphonyl chloride forms a product which is soluble in NaOH however secondary amine gives product which is insoluble in NaOH.
Ans: Due to the presence of acidic hydrogen in the N-alkylbenzenesulphonamide formed by the treatment of primary amines.
(d) Aniline does not react with methyl chloride in the presence of anhydrous AlCl3 catalyst.
Ans: Aniline does not react with methylchloride in the presence of AlCl3 catalyst , because aniline is a base and AlCl3 is Lewis acid which lead to formation of salt.
(30) (a) Identify the major product formed when 2-cyclohexylchloroethane undergoes a dehydrohalogenation reaction. Name the reagent which is used to carry out the reaction.
Ans: The major product formed when 2-cyclohexylchloroethane undergoes dehydrohalogenation reaction is 1- cyclohexylethene. The reagent which is used to carry out the reaction is ethanolic KOH.
(b) Why are haloalkanes more reactive towards nucleophilic substitution reactions than haloarenes and vinylic halides?
Ans: Haloalkanes are more reactive than haloarenes and vinylic halides because of the presence of partial double bond character C-X bond in haloarenes and vinylic halides. Hence they do not undergo nucleophilic reactions easily.
Or
(a) Name the possible alkenes which will yield 1-chloro-1-methylcyclohexane on their reaction with HCl. Write the reactions involved.
Ans: Methylenecyclohexane
(b) Allyl chloride is hydrolysed more readily than n-propyl chloride. Why?
Ans: Allyl chloride shows high reactivity as the carbocation formed in the first step is stabilised by resonance while no such stabilisation of carbocation exists in the case of n-propyl chloride.
SECTION D
(31) Strengthening the Foundation: Chargaff Formulates His “Rules”
Many people believe that James Watson and Francis Crick discovered DNA in the 1950s. In reality, this is not the case. Rather, DNA was first identified in the late 1860s by Swiss chemist Friedrich Miescher. Then, in the decades following Miescher’s discovery, other scientists–notably, Phoebus Levene and Erwin Chargaff–carried out a series of research efforts that revealed additional details about the DNA molecule, including its primary chemical components and the ways in which they joined with one another. Without the scientific foundation provided by these pioneers, Watson and Crick may never have reached their groundbreaking conclusion of 1953: that the DNA molecule exists in the form of a three-dimensional double helix
Chargaff, an Austrian biochemist, as his first step in this DNA research, set out to see whether there were any differences in DNA among different species. After developing a new paper chromatography method for separating and identifying small amounts of organic material, Chargaff reached two major conclusions: (i) the nucleotide composition of DNA varies among species.
(ii) Almost all DNA, no matter what organism or tissue type it comes from maintains certain properties, even as its composition varies. In particular, the amount of adenine (A) is similar to the amount of thymine (T), and the amount of guanine (G) approximates the amount of cytosine (C). In other words, the total amount of purines (A + G) and the total amount of pyrimidines (C + T) are usually nearly equal. This conclusion is now known as “Chargaff’s rule.” Chargaff’s rule is not obeyed in some viruses. These either have single- stranded DNA or RNA as their genetic material.
Answer the following questions:
(a) A segment of DNA has 100 adenine and 150 cytosine bases. What is the total number of nucleotides present in this segment of DNA?
Ans: A = 100 so T = 100
C=150 so G = 150
Total nucleotides = 100+100+150+150 =500
(b) A sample of hair and blood was found at two sites. Scientists claim that the samples belong to same species. How did the scientists arrive at this conclusion?
Ans: They studied the nucleotide composition of DNA. It was the same so they concluded that the samples belong to same species.
(c) The sample of a virus was tested and it was found to contain 20% adenine, 20% thymine, 20 % guanine and the rest cytosine. Is the genetic material of this virus (a) DNA- double helix (b) DNA-single helix (c) RNA? What do you infer from this data?
Ans: A = T = 20%
But G is not equal to C so double helix is ruled out.
The bases pairs are ATGC and not AUGC so it is not RNA
The virus is a single helix DNA virus
OR
How can Chargaff’s rule be used to infer that the genetic material of an organism is double- helix or single- helix?
Ans: According to Charagraff rule, all double helix DNA will have the same amount of A and T as well as C will be same amount as G. If this is not the case then the helix is single stranded.
(32) Henna is investigating the melting point of different salt solutions. She makes a salt solution using 10 mL of water with a known mass of NaCl salt. She puts the salt solution into a freezer and leaves it to freeze. She takes the frozen salt solution out of the freezer and measures the temperature when the frozen salt solution melts. She repeats each experiment.
S.No | Mass of the salt used in g |
1 | 0.3 |
2 | 0.4 |
3 | 0.5 |
4 | 0.6 |
5 | 0.8 |
6 | 1.0 |
Melting point in 0C | |
Readings Set 1 | Reading Set 2 |
-1.9 | -1.9 |
2.5 | -2.6 |
-3.0 | -5.5 |
-3.8 | -3.8 |
-5.1 | -5.0 |
-6.4 | -6.3 |
Assuming the melting point of pure water as 0oC, answer the following questions:
(a) One temperature in the second set of results does not fit the pattern. Which temperature is that? Justify your answer.
Ans: 3rd reading for 0.5 g there has to be an increase in depression of freezing point and therefore decrease in freezing point so also decrease in melting point when amount of salt is increased but the trend is not followed on this case.
(b) Why did Henna collect two sets of results?
Ans: two sets of reading help to avoid error in data collection and give more objective data
(c) In place of NaCl, if Henna had used glucose, what would have been the melting point of the solution with 0.6 g glucose in it?
Ans: ΔTf (glucose) = 1 x Kf x 0.6 x 1000/ 180 x 10
ΔTf (NaCl) = 2 x Kf x 0.6 x 1000/ 58.5 x 10
3.8 = 2 x Kf x 0.6 x 1000 /58.5 x 10
Divide equation 1 by 2
ΔT(glucose0=/3.8 = 58.5/2 x 180
ΔTf (glucose = 0.62 Freezing point or Melting point = – 0.62 oC
Or
What is the predicted melting point if 1.2 g of salt is added to 10 mL of water? Justify your answer.
Ans: depression in freezing point is directly proportional to molality (mass of solute when the amount of solvent remains same)
0.3 g depression is 1.9 oC
0.6 g depression is 3.8 oC
1.2 g depression will be 3.8 x2 = 7.6 oC
SECTION E
(33) (a) Why does the cell voltage of a mercury cell remain constant during its 10 lifetime?
Ans: The cell potential remains constant during its life as the overall reaction does not involve any ion in solution whose concentration can change during its life time.
(b) Write the reaction occurring at anode and cathode and the products of electrolysis of aq KCl.
Ans: KCl (aq) à K+ (aq) + Cl- (aq)
cathode: H2O(l) + e- à ½ H2 (g) + OH- (aq)
anode: Cl- (aq) à ½ Cl2 (aq) + e-
net reaction:
KCl (aq) + H2O (l) à K+ (aq) +OH- (aq) + ½ H2 (g) + ½ Cl2 (g)
(c) What is the pH of HCl solution when the hydrogen gas electrode shows a potential of -0.59 V at standard temperature and pressure?
Ans: . Given, potential of hydrogen gas electrode = −0.59 V Electrode reaction: H+ + e– → 0.5 H2
Applying Nernst equation,
E (H+ /H2) = Eo (H+ /H2) – 0.059 log [H2] 1/2 n [H+ ]
Eo (H+ /H2) = 0 V
E (H+ /H2) = -0.59 V
n = 1
[H2] =1 bar
−0.59 = 0 – 0.059 ( – log [H+ ] )
−0.59 = −0.059pH
∴ pH = 10
Or
(a) Molar conductivity of substance “A” is 5.9×103 S/m and “B” is 1 x 10-16 S/m. Which of the two is most likely to be copper metal and why?
Ans: “A” is copper, metals are conductors thus have high value of conductivity.
(b) What is the quantity of electricity in Coulombs required to produce 4.8 g of Mg from molten MgCl2? How much Ca will be produced if the same amount of electricity was passed through molten CaCl2? (Atomic mass of Mg = 24 u, atomic mass of Ca = 40 u).
Ans: Mg2+ + 2e- à Mg
1 mole of magnesium ions gains two moles of electrons or 2F to form 1 mole of Mg 24 g Mg requires 2 F electricity 4.8 g Mg requires 2 x4.8/24 = 0.4 F = 0.4 x96500 = 38600C
Ca2+ + 2e- Caà
2 F electricity is required to produce 1 mole =40 g Ca 0.4 F electricity will produce 8 g Ca
(c) What is the standard free energy change for the following reaction at room temperature? Is the reaction spontaneous?
Sn(s) + 2Cu2+ (aq) à Sn2+ (aq) + 2Cu+ (s)
Ans: F = 96500C, n=2,
Sn2+ (aq) + 2e– → Sn(s) –0.14V
Eocell = Eocathode – Eo anode
= 0.15 – (-0.14) = 0.29V
ΔGo = -nFEo cell
= -2 x96500x 0.29 = 55970 J/mol
(34) A hydrocarbon (A) with molecular formula C5H10 on ozonolysis gives two products (B) and ( C). Both (B) and (C) give a yellow precipitate when heated with iodine in presence of NaOH while only (B) give a silver mirror on reaction with Tollen’s reagent.
(a) Identify (A), (B) and (C).
Ans: A is an alkene
(b) Write the reaction of B with Tollen’s reagent
Ans: B is an aldehyde with –CH3 group
(c) Write the equation for iodoform test for C
Ans:
(d) Write the equation for iodoform test for C
OR
An organic compound (A) with molecular formula C2Cl3O2H is obtained when (B) reacts with Red P and Cl2. The organic compound (B) can be obtained on the reaction of methyl magnesium chloride with dry ice followed by acid hydrolysis.
(a) Identify A and B
Ans: CCl3COOH (B): CH3COOH
(b) Write down the reaction for the formation of A from B. What is this reaction called?
Ans: CH3COOH (i)Red P / Cl2 CCl3COOH, Hell Volhard Zelinsky reaction
(c) Give any one method by which organic compound B can be prepared from its corresponding acid chloride.
Ans: CH3COCl H2O CH3COOH
(d) Which will be the more acidic compound (A) or (B)? Why?
Ans: A will be more acidic due to presence of 3 Cl groups (electron withdrawing groups) which increase acidity of carboxylic acid.
(e) Write down the reaction to prepare methane from the compound (B)
Ans: CH3COOH (i)NaOH, CaO (ii) heat CH4 + Na2CO3
(35) Answer the following:
(a) Why are all copper halides known except that copper iodide?
Ans: Cu 2+ oxidizes iodide ion to iodine.
(b) Why is the Eo (V3+/V2+) value for vanadium comparatively low?
Ans: The low value for V is related to the stability of V2+ (half-filled t2g level)
(c) Why HCl should not be used for potassium permanganate titrations?
Ans: Permanganate titrations in presence of hydrochloric acid are unsatisfactory since hydrochloric acid is oxidised to chlorine.
(d) Explain the observation, at the end of each period, there is a slight increase in the atomic radius of d block elements.
Ans: The d orbital is full with ten electrons and shield the electrons present in the higher s-orbital to a greater extent resulting in increase in size.
(e) What is the effect of pH on dichromate ion solution?
Ans: The chromates and dichromates are interconvertible in aqueous solution depending upon pH of the solution. Increasing the pH (in basic solution)of dichromate ions a colour change from orange to yellow is observed as dichromate ions change to chromate ions.
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