Solved Sample Paper 1

Y

Reason — Potassium hydroxide (a base) reacts with nitric acid (an acid) to produce potassium nitrate (salt) and water.

HNO₃ + KOH ⟶ H₂O + KNO₃

single covalent bond

Reason — In a molecule of methane, one carbon atom forms four covalent bonds with four hydrogen atoms by sharing a pair of electrons between itself and each hydrogen (H) atom.

ammonia and oxygen

Reason —

$4\text{NH}_3 + 5\text{O}_2 \xrightarrow[700-800 \degree\text{C}]{\text{Pt}} 4\text{NO} + 6\text{H}_2\text{O} + 21.5 \text{K cals}$

increases

Reason — Melting point, boiling point, density etc. of the successive members of the homologous series increase with increase in molecular mass.

B

Reason — On heating sodium hydrogen carbonate, carbon dioxide gas is evolved which turns lime water milky.

Methanoic acid

Reason — The IUPAC name of formic acid [HCOOH] is methanoic acid.

2-butylene

Reason — Carbon to carbon double bond is found in 2-butylene.

Magnesium hydroxide

Reason — Magnesium hydroxide also known as milk of magnesia is a common component of antacids.

sodium zincate

Reason — Zn + 2NaOH ⟶ H₂ + Na₂ZnO₂ [sodium zincate]

C₃H₉

Reason — As, molecular formula = n x empirical formula

Hence, molecular formula = 3 x CH₃ = C₃H₉

Therefore, the molecular formula of the compound is C₃H₉

Only Q

Reason — SO₄^2- and OH^1- ions migrate to the anode but neither are discharged since the copper anode loses electrons more easily than SO₄^2- and OH^1-. Hence, copper anode shows a loss in mass.

Helium

Reason — Helium has highest ionization potential because ionization potential increases from left to right across a period and decreases down a group.

carbon cathode

Reason — At cathode :

2Al³⁺ + 6e^- ⟶ 2Al

At anode : 3O^2- - 6e^- ⟶ 3[O] ⟶ 3O₂

Al³⁺ are discharged in preference to Na¹⁺ and Ca²⁺ ions due to it's lower position in the electrochemical series.

reddish brown fumes formed at anode

Reason — Reddish brown fumes of bromine evolve at the anode and silvery gray metal lead is formed on the cathode.

Neither P nor Q

Reason — According to Avogadro's number (N_A) = 6.022 x 10²³ atoms are present in one mole of element.

In case of P,

Number of atoms in 14 g of N

= $\dfrac{\text{Given mass}}{\text{Atomic mass}}$ x N_A

= $\dfrac{14}{14}$ X 6.022 x 10²³

= 6.022 x 10²³

Number of atoms in 20 g of Ne

= $\dfrac{\text{Given mass}}{\text{Atomic mass}}$ x N_A

= $\dfrac{20}{20.18}$ X 6.022 x 10²³

= 0.99 x 6.022 x 10²³

= 5.96 x 10²³ atoms

As number of atoms in 14 g of nitrogen is almost equal to 20 g of Neon. Hence, statement P is incorrect.

In case of statement Q,

Number of atoms in 16 g of oxygen

= $\dfrac{\text{Given mass}}{\text{Atomic mass mass}}$ X N_A

= $\dfrac{16}{16}$ X 6.022 x 10²³

= 6.022 x 10²³

As number of atoms in 16 g of oxygen is equal to 14 g of nitrogen. Hence, statement Q is also incorrect.

(a) Methane [CH₄]

(b)

$\underset{\text{sodium acetate}}{\text{CH}_3\text{COONa}} + \underset{\text{sodalime}}{\text{NaOH}} \xrightarrow[\Delta]{\text{CaO}} \underset{\text{methane}}{\text{CH}_4} + \text{Na}_2\text{CO}_3$

The gas is collected by downward displacement of water.

(c) Decarboxylation reaction takes place. It happens due to elimination of a molecule of CO₂ from a carboxylic acid. In this reaction, CO₂ is removed as a carbonate.

(a) On moving down the group electron affinity decreases.

(b) The functional group present in acetic acid is —COOH.

(c) Dilute sulphuric acid is a weak acid and a good electrolyte.

(d) Compounds which contains one —COOH group are known as monocarboxylic acid.

(e) Quicklime is not used to dry HCl gas because CaO is alkaline in nature.

(a) Neutralization reaction

(b) Sodium (Na)

(c) Alkali

(d) Ammonia

(e) Nitric oxide

(a) Structural formulae are shown below:

(1) But-1-ene

(2) Butanoic acid

(3) Propanol

(b) IUPAC names of the organic compounds are:

(1) Methanoic acid

(2) 3-methyl-hexane

The compound Q is sodalime [NaOH]

$\underset{\text{sodium ethanoate}}{\text{CH}_3\text{COONa}} + \underset{\text{sodalime}}{\text{NaOH}} \xrightarrow[\Delta]{\text{CaO}} \underset{\text{methane}}{\text{CH}_4} + \text{Na}_2\text{CO}_3$

(a) Hydrogen gas

(b) Sulphur dioxide gas

(a) Pb(NO₃)₂ + 2HCl ⟶ PbCl₂ ↓ + 2HNO₃

(b) ZnO + 2KOH ⟶ K₂ZnO₂ + H₂O

(c) AgNO₃ + NaCN ⟶ AgCN ↓ + NaNO₃

(a) (1) False
Corrected Statement — According to Mendeleev, periodic properties of elements are periodic functions of their atomic weight.

(a) (2) False.
Corrected Statement — Metals form oxide that are basic in nature.

(b) Pb(NO₃)₂ + 2NH₄OH ⟶ Pb(OH)₂ ↓ + 2NH₄NO₃

(a) Covalent bond — The chemical bond that is formed between two combining atoms by mutual sharing of one or more pair of electrons is called a covalent bond. Depending on the number of electron pairs shared : bond is single [-], double [=], or triple [≡] covalent.

(b) Co-ordinate bond — The bond formed between two atoms by sharing a pair of electrons, provided entirely by one of the combining atoms but shared by both is called a coordinate bond.

Simplest ratio of whole numbers = C : H : Cl = 1 : 1 : 2

Hence, empirical formula is CHCl₂

(a) Conditions required for the conversion of potassium nitrate to potassium bisulphate are:

1. Temperature of the reaction should not exceed 200°C.
2. All glass apparatus should be used.
3. Reaction:
   KNO₃ + H₂SO₄ $\xrightarrow{\lt 200 \degree\text{C}}$ KHSO₄ + HNO₃

(b) Conditions required for the conversion of nitrogen and hydrogen to ammonia are:

1. Temperature should be between 450-500°C.
2. Pressure should be above 200 atm.
3. Catalyst used is finely divided iron.
4. Promoter used is traces of molybdenum or Al₂O₃.
5. Reaction:
   N₂ + 3H₂ $\xrightleftharpoons[450-500\text{°C},\text { above }200 \text{ atm}]{\text{Fe and Mo}}$ 2NH₃ + Δ

(c) Conditions required for the conversion of ammonia to nitric oxide are:

1. Temperature should be around 800°C.
2. Catalyst used is Platinum.
3. Reaction:
   4NH₃ + 5O₂ $\xrightarrow[800 \degree \text{C}]{\text{Pt}.}$ 4NO↑ + 6H₂O + Δ
   2NO + O₂ ⟶ 2NO₂ (brown gas)

1. Duralumin
2. Stainless steel
3. Solder

(a) Platinum gauze

(b) Oleum or pyrosulphuric acid (H₂S₂O₇)

(a) The products formed are Ammonium chloride and Water.
NH₄OH + HCl ⟶ NH₄Cl + H₂O

(b) The products formed are Zinc chloride and Hydrogen gas.
ZnS + 2HCl ⟶ ZnCl₂ + H₂S ↑

(a) Na < Mg < Al < Si < Cl

(b) Li > Be > C > O > F

(c) Si < Al < Mg < Na

(a) Set Y is known as noble gases.

(b) In set X oxygen has the highest electronegativity.

(c) The X set contains the non-metals.

(a) CaO is used as a drying agent. In order to get dry ammonia, the gas is passed through a drying tower containing lumps of quicklime [CaO].

(b) Mo acts as a promoter i.e., it increases the efficiency of the catalyst.

(a) No. of moles of oxygen atoms = $\dfrac{12.044 \times 10^{23}}{6.022 \times 10^{23}}$ = 2 moles

Mass of oxygen atoms = no. of moles x atomic mass

= 2 x 16 = 32 g.

∴ The mass of 12.044 x 10²³ oxygen atoms is 32 grams.

(b) Gram molecular mass of ethylene (C₂H₄) = 2(12) + 4(1) = 24 + 4 = 28 g

Vapour density = $\dfrac{\text{Molecular weight}}{2}$ = $\dfrac{28}{2}$ = 14

Hence, vapour density is 14

(a) Cryolite and fluorspar are components of the electrolyte.

(b) Addition of cryolite lowers the fusion temperature from 2050°C to 950°C and enhances conductivity. Both cryolite & fluorspar act as a solvent for the electrolytic mixture and increase its conductivity since pure alumina is almost a non-conductor of electricity.

(a) Calcium oxide [CaO]

(b) The walls appeared shiny on the 3rd day due to the reaction of calcium hydroxide with carbon dioxide present in the air to produce white-colored calcium carbonate.

CaO + H₂O ⟶ Ca(OH)₂ + heat
Ca(OH)₂ (aq) + CO₂ (g) ⟶ CaCO₃ (s) + H₂O (l)

(c) Precipitation reaction

(a) The following particles are present in:

1. Lead chloride — Pb²⁺ and Cl^1-
2. Potassium nitrate — K¹⁺ and NO₃^1-

(b) The layer of powdered coke is sprinkled over the electrolytic mixture as:

1. It prevents burning of carbon electrodes in air at the emergence point from the bath.
2. It minimizes or prevents heat loss by radiation.

Electronic configuration of ₈B is [2, 6]. It needs two electrons to complete its octet.

Electronic configuration of ₂₀A is [2, 8, 8, 2]. It wants to donate two electrons in order to obtain a stable octet configuration.

Hence, ionic bond is formed as ₂₀A loses two electrons and ₈B gains two electrons and both attain a stable octet.

₈B [2, 6] is present in group VIA and ₂₀A [2, 8, 8, 2] is present in group in IIA as the group of an elements is predicted from the number of electrons in the valence shell.

(a) Two uses of ammonia are:

1. Liquid ammonia is used as a refrigerant in ice plants.
2. Ammonia is used in the manufacture of nitrogenous fertilisers like ammonium sulphate, ammonium nitrate, urea, etc.

(b) Balanced equations for the conversions are given below:

1. AlCl₃ (aq) + 3NH₄OH (aq) ⟶ Al(OH)₃ ↓ + 3NH₄Cl (aq)

2. 4NH₃ + 5O₂ $\xrightarrow[800 \degree \text{C}]{\text{Pt}.}$ 4NO↑ + 6H₂O + Δ
   2NO + O₂ ⟶ 2NO₂ (brown gas)

(a) The functional group present is -COOH (carboxylic acid).

(b) The functional group present is -OH (alcohol).

Given,

empirical formula = CH₂O

Hence, empirical formula mass = 12 + 2(1) + 16 = 12 + 2 + 16 = 30 g

molecular formula = n x empirical formula

where,

n = $\dfrac{\text{molecular mass}}{\text{empirical formula mass}}$

∴ n = $\dfrac{180}{30}$ = 6

Hence, molecular formula = 6 x CH₂O = C₆H₁₂O₆

Therefore, the molecular formula of the compound is C₆H₁₂O₆

(a) Phosphorous [P]

(b) Sodium [Na]

(c) Helium [He]

(a) 2CH₃I + 2Na $\xrightarrow[\text{ether}]{\text{dry}}$ C₂H₆ ↑ + 2NaI

(b) CaC₂ + 2H₂O ⟶ C₂H₂ ↑ + Ca(OH)₂

(c) C₂H₅OH $\xrightarrow[170\degree\text{C}]{\text{Conc. H}_2\text{SO}_4}$ C₂H₄ ↑ + H₂O