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Published on: 04/11/2019
Download CBSE Class 12th Standard CBSE Chemistry question papers, sample papers, important questions, and previous year solved papers in PDF format. Get free study materials, NCERT solutions, and exam preparation resources for Class 12th Standard CBSE Chemistry
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1.
Compounds having general molecular formula \(AFe_{ 2 }O_{ 4 }\) are called ferrites and posses spinel type structures. Some common examples are \(MgFe_{ 2 }O_{ 4 }\) and \(ZnFe_{ 2 }O_{ 4 }\) They may be thought of being formed by replacing \(Fe^{ 2+ }\) ions present in \(Fe_{ 3 }O_{ 4 }\) by bivalent cations such as \(Mg^{ 2+ },Zn^{ 2+ }\) ions etc. Now answer the following questions :
(i) What types of materials are ferrites?
(ii) What are the main uses of ferrites?
2.
(i) For the reaction A➝B, the rate of reaction becomes twenty-seven times when the concentration of A is increased three times. What is the order of the reaction?
(ii) The activation energy of a reaction is 75.2 kJ mol-1 in the absence of a catalyst and it lowers to 50.14 kJ mol-1 with a catalyst. How many times will the rate of reaction grow in the presence of a catalyst if the reaction proceeds at 25°C?
3.
Mira's house had a fencing of iron rods. Her father suggested getting it painted. But she and her mother thought it was a waste of time and money as iron rods are strong:
(i) Whose opinion is acceptable according to you?
(ii) Give two other methods to prevent corrosion.
(iii) What is the chemical formula of rust?
(iv) Mention the values shown by Mira's father.
4.
(a) Define the following terms:
(i) Mole fraction,
(ii) Ideal solution.
(b) 15.0 g of an unknown molecular material is dissolved in 450 g of water. The resulting solution freezes at - 0.34 \(°\)C. What is the molar mass of the material? [Kf for water=1.86 K Kg mol-1]
5.
A compound A (C4H10O) is found to be soluble in concentrated sulphuric acid. (A) does not react with sodium metal or potassium permanganate. When (A) is heated with excess of HI, it gives a single alkyl halide. Deduce the structure of compound (A) and explain all the reactions involved.
6.
An organic compound (A) on treatment with ethyl alcohol gives a carboxylic acid (B) and compound (C). Hydrolysis of (C) under acidified conditions gives (B) and (D). Oxidation of (D) with KMnO4 also gives (B). (B) on heating with Ca(OH)2 gives (E) having moleuclar formula C3H6O. (E) does not give TOllens'test and does not reduce Fehiling's solution but forms 2, 4-dinitrophenyhydrazone. Identify (A),(B),(C),(D) and (E).
7.
Using the standard electrode potentials, predict if the reaction between the following is feasible:
(i) Fe3+(aq) and I-(aq)
(ii) Ag+(aq) and Cu(s)
(iii) Fe3+(aq) and Br-(aq)
(iv) Ag(s) and Fe3+(aq)
(v) Br2(aq) and Fe2+(aq)
8.
(a) Given below are the electrode potential values, Eo for the some of the first row of transition elements:
| Element | EoM2+/M (V) |
|
V(23) Cr(24) Mn(25) Fe(26) Co(27) Ni(28) Cu(29) |
-1.18 -0.91 -1.18 -0.44 -0.28 -0.25 +0.34 |
Explain the irregularities in these values on the basis of electronic structures of atoms.
(b) Complete the following reaction equations:
(i) Cr2O72-+Sn2++H+\(\longrightarrow \)
(ii) MnO4-+Fe2++H+\(\longrightarrow \)
1.
(i) Ferrites are magnetic material.
(ii) They are used in telephones and memory loops in computers.
2.
(i) \(r=k[R]^n\)
When concentration is increased three times, [R] = 3a
\(27r=k(3a)^n\)
\(\frac{27r}{r}=\frac{k(3a)^n}{ka^n}or27\)
\(=3^n \ or \ 3^3=3^n\)
\(\Rightarrow\ n=3\)
(ii) According to Arrhenius equation,
\(\log k=\log A-\frac{E_a}{2.303RT} \)
For uncatalysed reaction
\(\log k=\log A-\frac{E_a(2)}{2.303RT}\ \ \ \ \ \ .......(i) \)
For catalysed reaction
\(\log k_2=\log\ A \frac{E_a(2)}{2.303RT}\ \ \ \ \ ....(ii)\)
A is equal for both the reactions.
Subtracting equation (i) from equation (ii)
\(\log\frac{k_2}{k_1}=\frac{E_a(1)-E_a(2)}{2.303RT}\)
\(\Rightarrow\ \log\frac{k_2}{k_1}=\frac{(75.2-50.14)kJ\ mol^{-1}}{2.303\times8.314\ JK^{-1}\ mol^{-1}\times 298 K}\)
\(\Rightarrow\ \log\frac{k_2}{k_1}=4.39\)
\(\Rightarrow\ \frac{k_2}{k_1}=antilog(4.39)\)
\(=2.45\times 10^4\)
Rate of reaction increases by 2.45 x 104 times.
3.
(i) Mira's father is right as it prevents the iron rods from corrosion.
(ii) Galvanization and cathodic protection.
(iii) Fe2O3.x H2O
(iv) Scientific knowledge and care for his property.
4.
(a) (i) The ratio of number of moles of one component to the total number of moles of solution is known as mole fraction.
(ii) The solution which follows Raoult's law over entire range of concentrations at specific temperature is called ideal solution.
(b) WB = 15 g, WA = 450 g, \(\Delta\)Tf = 0.34 K, MB = ?
\({ M }_{ B }=\frac { 1000\times { K }_{ f }\times { W }_{ B } }{ \Delta { T }_{ f }\times { W }_{ A } } \)
\(=\frac { 1000\times 1.86 \ K \ kg \ { mol }^{ -1 }\times 15 \ g }{ 0.34K\times 450 \ g } \)
= 182.35 g/mol
5.
(i) Since compound A (C4H10O) does not react with Na metal or KMnO4, it cannot be an alcohol.
(ii) Since compound A dissolves in cone, H2SO4, it may be an ether.
(iii) Since ether A on heating with excess of HI gives a single alkyl halide, therefore, ether (A) must be symmetrical. Now the only symmetrical ether having M.F. C4HIOO is diethyl ether (CH3CH2OCH2CH3).
6.
(i) Since compound (E) with molecular formula, C3H60 does not reduce Tollens' reagent and Fehling's solution but forms 2, 4-dinitrophenylhydrazone, it must be a ketone. But the only possible ketone having the molecular formula, C3H6O is acetone or propanone. Thus, compound (E) is acetone or (propanone) CH3COCH3·
(ii) Since acetone (E) is obtained by heating compound (8) with Ca(OH)2 therefore, (B) must be acetic acid (ethanoic acid), CH3COOH.
(iii) Since (D) on oxidation with KMn04 gives acetic acid (8), therefore, (D) must be ethyl alcohol (ethanol), CH3CH2OH.
(iv) Since acetic acid (8) and ethyl alcohol (D) are obtained by hydrolysis of (C) under acidic conditions, therefore, (C) must be ethyl acetate (ethyl ethanoate), CH3COOC2H5
(v) Since ethyl acetate (C) and acetic acid (8) are obtained by treatment of compound (A) with ethyl alcohol, therefore, compound (A) must be acetic anhydride (ethanoic anhydride), (CH3COO)2O.
(vi) All the reactions involved in this problem can now be explained as follows
7.
(i) \(2Fe^{3+}+2I^-\longrightarrow2Fe^{2+}+I_2\)
\(E^0_{cell}=E^0_{I_2/ I^-}-E^0_{Fe^{3+}/Few^{2+}}=0.54 V - (0.77 V) = 0.54 V - 0.11 v = - 0.23 V\)
It is not feasible therefore E0cell is -ve ΔGo = +ve
(ii) 2Ag+(aq) + Cu(s) ⟶ Cu2+(aq) + Ag(s)
\(E^0_{cell} = E^0_{Ag^+/Ag} -E^0_{Cu^{2+}/Cu} = +0.80 V - 0.34 V = 0.46 V\)
It is feasible because E0cell is +ve
∴ ΔG0 = - ve
(iii) 2Fe3+(aq) + 2Br-(aq) ⟶ 2Fe2+(aq) + Br2'
\(E^0_{cell} = E^0_{Fe^{3+}/Fe^{2+}}- E^0_{Br_2/Br^-} = 0.77 V - 1.09 V = - 0.32 V\)
Since E0cell is -ve, ΔGo = +ve therefore, reaction is non spontaneous.
(iv) Ag(s) + Fe3+(aq) ⟶ Fe2+(aq) + Ag+(aq)
\(E^0_{cell} = E^0 _{Fe^{3+}/Fe^{2+}}-E^0_{Ag^+/Ag} = 0.77 V - 0.80 V = - 0.03 V\)
since E0cell is -ve, ΔGo is +ve therefore, reaction is non spontaneous.
(v) Br2 + 2Fe2+(aq) ⟶ 2Br-(aq) + 2Fe3+(aq)
\(E^0_{cell} = E^0_{Br^2/Br^-} E^0_{Fe^{3+}/Fe^{2+}}=1.09 V - 0.77 V = 0.32 V\)
Since E0cell is +ve, ΔGo = -ve, therefore, reaction will be spontaneous.
8.
(a) It is due to irregular variations in sum of first and second ionisation energies and sublimation energies. It is also due to stability of electronic configuration.
(b) (i) Cr2O72- + 3Sn2+ + 14H+➝ 2Cr3+ + 3Sn4+ + 7H2O
(ii) MnO4- + 5Fe2+ + 8H+ ➝ SFe3+ + Mn2+ + 4H2O
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