CBSE Class 12 Chemistry Solved Question Paper: 2016

Download CBSE 12th Chemistry 2016 board exam solved paper in PDF format. Download link is available at the end of this article. This paper was held on 9th March 2016.

This solved CBSE Class 12 Chemistry question paper 2016 is very important for the preparation of coming Chemistry board exam. With this paper, students can easily understand the level of questions which can be asked in CBSE Class 12th Chemistry board exam.

Detailed explanations of each and every question are also available in the PDF of the solved paper and with these solutions, students can easily understand what concepts are involved in the given problems.

Some sample questions from the CBSE 12^th Chemistry 2016 board exam solved paper is given below:

Question:

Write the main reason for the stability for colloidal sols.

Solution:

Electrostatic stabilization and salvation.

Colloidal sols posses a definite type of electrical charge on the colloidal particles and the dispersion medium has an equal but opposite charge. Thus, the charge on colloidal particles is balanced with the equal and opposite charge of dispersion medium which makes the colloidal solution electrically neutral and stable.

CBSE Class 12 Chemistry Syllabus 2017 – 2018

Question:

Give an example of each of molecular solid and ionic solid.

Solution:

Molecular solid: sucrose/table sugar; Ionic solid: sodium chloride (NaCl).

Question:

Pb(NO₃)₂ on heating gives a brown gas which undergoes dimerization on cooling. Identify the gas.

Solution:

2 Pb(NO₃)₂  → 2 PbO + 4 NO₂ ↑+ O₂↑

2 NO₂→ N₂O₄

So, Gas is NO₂ (nitrogen dioxide)_.

Question:

(i) Gas (A) is more soluble in water than gas (B) at the same temperature. Which one of these two gases have higher value of K_H (Henry’s constant) and why?

(ii) In a non ideal solution, what type of deviation shows the formation of maximum boiling azeotropes?

Solution:

(i) According to Henry’s law; C = kP

Here C= concentration of gas in solution

K_H =Henry constant

P= Pressure

Hence as value of k increases, C; solubility of gas in solution increases. Therefore K_H will be higher for A

(ii) Negative deviation from Raoult’s Law.

Question:

The elements of 3d transition series are given as

Sc Ti V Cr Mn Fe Co Ni Cu Zn

Question:

Answer the following:

(i) Write the element which shows maximum number of oxidation states. Give reason.

(ii) Which element has the highest melting point?

(iii) Which element shows only +3 oxidation state?

(iv) Which element is a strong oxidizing agent in +3 oxidation state and why?

Solution:

(i) Mn shows +2 to +7 oxidation states due to presence of maximum number of unpaired electrons. The valence shell electronic configuration of 3d⁵4s² allows it to lose all seven valence electrons.

(ii) Due to presence of maximum unpaired electrons Cr forms strong metallic bonds and has the highest m.p.

(iii) Scandium Sc, (Z=21, 1s²2s²2p⁶3s²3p⁶3d¹4s²) exhibits only +3 oxidation state as its valence shell electronic configuration is 3d¹4s²

(iv)  Mn⁺³ is strongest oxidizing agent based on their value of electrode potential and the 3d⁵ valence shell configuration of +3 oxidation state is highly stable.

Question:

(a) The conductivity of 0.001 mol L^-1 solution of CH₃COOH is 3.095 × 10^-5 S cm^-1. Calculate its molar conductivity and degree of dissociation (α).

(b) Define electrochemical cell. What happens if external potential applied becomes greater than E^o_cell of electrochemical cell?

Solution:

(a) Molar conductance = Measured conductivity × 1000/ Electrolyte concentration= 39.05 s cm² mol^-1

Degree of dissociation = molar conductance / molar conductance at infinite dilution = 39.05 / 39.05 = 0.1

(b) A device which can generate electrical energy from chemical reactions is called as electrochemical cell. If the external potential is higher compare to E_cell, the flow of current is reversed and recharging of cell takes place.

Question:

Due to hectic and busy schedule, Mr Singh stared taking junk food in the lunch break and slowly became habitual of eating food irregularly to excel in his field. One day during meeting he felt severe chest pain and fell down. Mr Khanna, a close friend of Mr Singh, took him to doctor immediately. The doctor diagnosed that Mr Singh was suffering from acidity and prescribed some medicines.  Mr Khanna advised him to eat homemade food and change his lifestyle by doing yoga, meditation and some physical exercise. Mr Singh followed his friend’s advice and after few days he started feeling better.

After reading the above passage, answer the following:

(i) What are the values (at least two) displayed by Mr Khanna?

(ii) What are antacids? Give one example.

(iii) Would it be advisable to take antacids for a long period of time? Give reason.

Solution:

(i) Concern for his friend and knowledge

(ii) Antacids are type of medicines which mainly consist of some mild basic compounds. These alkaline compounds neutralize the effect of stomach acid and reduce the symptoms of heartburn, relieve pain. For example aluminium hydroxide, magnesium carbonate, magnesium trisilicate.

(iii) No, we should not take antacids in excess as it may cause reduction in stomach pH resulting in digestion related problems.

Question:

(a) The conductivity of 0.001 mol L^-1 solution of CH₃COOH is 3.095 × 10^-5 S cm^-1. Calculate its molar conductivity and degree of dissociation (α).

(b) Define electrochemical cell. What happens if external potential applied becomes greater than E^o_cell of electrochemical cell?

Solution:

(a) Molar conductance = Measured conductivity × 1000/ Electrolyte concentration= 39.05 s cm² mol^-1

Degree of dissociation = molar conductance / molar conductance at infinite dilution = 39.05 / 39.05 = 0.1

(b) A device which can generate electrical energy from chemical reactions is called as electrochemical cell. If the external potential is higher compare to E_cell, the flow of current is reversed and recharging of cell takes place.

Question: Calculate the boiling point of solution when 4 g of MgSO₄ (M=120 g mol^-1) was dissolved in 100 g of water, assuming MgSO₄undergoes complete ionization.

(K_b for water = 0.52 K kg mol^-1)

Solution:

According to the question:

K_b  =0.52 K kg/mol,    Molar mass of solute MgSO₄ = 120.366 g/mol, Mass of MgSO₄ = 4g, mass of water (solvent)= 100g

Hence, Moles of MgSO₄ = 4 / 120.4 = 0.033 moles

And Molarity of  MgSO₄= 0.033 × 1000/100 = 0.33 mol/L

Since MgSO₄ undergoes complete ionization, one mole of MgSO₄ obtains 2 mols of constituent ions; hence i=2

∆T­ =iK_b m

 2 × 0.52 × 0.33 = 0.343 K

Hence boiling point would be = 0.343 K

Question:

Define the following terms:

(i) Lycophilic colloid

(ii) Zeta potential

(iii) Associated colloids

Solution:

(i) Lyophilic colloids: They are liquid loving colloids therefore when they are mixed with the suitable liquid; there will be very strong force of attraction exists between colloidal particles and liquid.

(ii) Zeta potential: It is measurement of the magnitude of the electrostatic between charged particles such as in colloidal solutions. Therefore it is one of the fundamental parameters known to affect stability.

(iii) Associated colloids: They are also called as micelle. They are strong electrolytes at low concentration and show colloidal properties at high concentration. That is because of aggregation of particles. These aggregated particles are called as associated colloids or micelles such as Sodium stearate (C₁₇H₃₅COONa).

Question:

Give reasons for the following:

(i) Aniline does not undergo Friedel-Crafts reaction.

(ii) (CH₃)₂NH is more basic than (CH₃)₃N in an aqueous solution.

(iii) Primary amines have higher boiling point than tertiary amines.

Solution:

(i) Due to basic nature of aniline, it forms anelinium ion in acidic medium of Friedal-craft reaction. Therefore, the benzene ring of aniline can’t undergo electrophilic substation reaction.

(ii) (NH₃)₃N cannot release its lone pair due to steric hindrance of 3 methyl groups around nitrogen atom.

(iii) Primary amines form intermolecular hydrogen bond which requires more energy to boil the solution. Tertiary amines do not have any hydrogen atoms for formation of hydrogen bonds.

CBSE Class 12 Chemistry Sample Paper: 2018

Question:

For the first order thermal decomposition reaction, the following data were obtained:

C₂H₅Cl _(g) à C₂H_{4 (g)} + HCl _(g)

Time/second    Total pressure/atm

0                                0.30

300                          0.50

Calculate the log constant.

(Given log 2= 0.301, log 3=0.4771, log 4=0.6021)

Solution:

K= (2.303/t) ×  log [a/a-x]

Here:

• K = Rate constant for first order reaction

• t= time

• a= Initial pressure

• (a‒x) = pressure after time t

K = (2.303 /300) log [0.30 / 0.30 -0.20]

K = (2.303 × 0.4771) / 300 or  K= 0.00366  sec^-1

Question:  

An element crystallizes in a b.c.c. lattice with cell edge of 500pm. The density of the element is 7.5g cm^-3. How many atoms are present in 300 g of the element?

Solution:

According to the question:

a = 500 pm or 500 x 10^-10 cm

m= 300g and z=2

We know that m=M/N_A; here, M= molar mass

Since, d (density) = zM/a³ N_A

Putting the given values on above equation to calculate M

M= 7.5 × (500)³ × 10^-30 × 6.02 × 10²³ / 2= 282.1 g/mol

As we know that Molar mass (M) = mass of compound× (N_A / Number of atoms)

Putting all the values to calculate number of atoms

282.1 = 300 × 6.02 × 10²³ / Number of atoms

Or Number of atoms = 300 × 6.02 × 10²³ /282.1 =  6.4 x 10²³

Question:

(i) SO₂  is reducing while TeO₂ is an oxidizing agent.

 (ii) Nitrogen does not form pentahalide.

 (iii) ICl is more reactive than I₂.

Solution:

(i) Because of inert pair effect.

(ii) Because of lack of d orbital, N cannot form 5 covalent bonds.

(iii) ICl is interhallogen compound in which I atom is partially electropositive and chlorine is electronegative. In other words, we can say that ICl consists of ionic characters whereas I₂ is pure covalent compound.

Question:

(I) Write the structural difference between starch and cellulose.

(II) What type of linkage is present in nucleic acids?

(III) Give an example of each fibrous protein and globular protein.

Solution.

(I) Starch is branched homopolysaccharide of D-glucose and contains two types of glucose polymer, amylase and amylopectin. The long, unbranched chains of D-glucose residues connected by alpha1, 4 linkages are termed as amylase. The branched polymer of glucose wherein alpha 1, 4 glycosidic linkages join successive glucose with branch points having alpha 1, 6 linkages. Cellulose is linear unbranched polymer of glucose residues linked together by beta 1, 4 linkage.

(II) Phosphodiester linkage in which the 5’-phosphate group of one nucleotide unit is joined to the 3’-hydroxyl group of the next nucleotide (a covalent bond).

(III) Fibrous protein: silk; globular protein: egg albumin.

Question:

What happens when?

(I) SO₂ gas is passed through an aqueous solution Fe³⁺ salt?

(II) XeF₄ reacts with SbF₅?

Solution:

(I) When SO₂ gas is passed through an aqueous solution Fe³⁺ salt, SO₂ serves as reducing agent and red color of Fe (III) solution is turned into green color Fe (II).

SO₂  + 2 Fe³⁺ + 2 H₂O à 2 Fe²⁺ + SO₄^2- + 4 H⁺

(II) The xenon fluorides react with strong Lewis acids to form complexes.

 XeF₄ + SbF₅ ¾¾® [XeF₃] ⁺ [SbF₆]^-    

Question:

(i) Gas (A) is more soluble in water than gas (B) at the same temperature. Which one of these two gases have higher value of K_H (Henry’s constant) and why?

(ii) In a non ideal solution, what type of deviation shows the formation of maximum boiling azeotropes?

Solution:

(i) According to Henry’s law; C = kP

Here C= concentration of gas in solution

K_H =Henry constant

P= Pressure

Hence as value of k increases, C; solubility of gas in solution increases. Therefore K_H will be higher for A

(ii) Negative deviation from Raoult’s Law.

Question: Pb(NO₃)₂ on heating gives a brown gas which undergoes dimerization on cooling. Identify the gas.

Solution.

2 Pb(NO₃)₂  → 2 PbO + 4 NO₂ ↑+ O₂↑

2 NO₂→ N₂O₄

So, Gas is NO₂ (nitrogen dioxide)_.

Question:

Write the main reason for the stability for colloidal sols.

Solution:

Electrostatic stabilization and salvation.

Colloidal sols posses a definite type of electrical charge on the colloidal particles and the dispersion medium has an equal but opposite charge. Thus, the charge on colloidal particles is balanced with the equal and opposite charge of dispersion medium which makes the colloidal solution electrically neutral and stable.

Download CBSE Class 12 Chemistry 2016 Board Exam Solved Paper