Model Paper 2

Correct the following statements and rewrite them :

(i) Sodium reacts with chlorine to form sodium chloride.

(ii) The electrolysis of lead bromide liberates lead and bromine.

(iii) Copper sulphate crystals are dehydrated by sulphuric acid.

(iv) Calcium nitrate reacts with sodium sulphate to form calcium sulphate.

(v) Metals of IA group are called alkaline earth metals.

Answer

(i) Sodium atom reacts with chlorine atom to form sodium chloride molecule.

(ii) The electrolysis of lead bromide liberates lead at cathode and bromine at anode.

(iii) Copper sulphate crystals are dehydrated by conc. sulphuric acid.

(iv) Calcium nitrate reacts with sodium hydroxide to form calcium hydroxide.

(v) Metals of II A group are called alkaline earth metals.

Write ionic reaction at the electrodes during the electrolysis of :

(i) copper II sulphate solution with copper electrodes,

(ii) acidulated water.

Answer

Dissociation of aq. copper sulphate

CuSO₄ ⇌ Cu²⁺ + SO₄^2- [ions present — Cu²⁺, H¹⁺, SO₄^2-, OH^1-]

H₂O ⇌ H¹⁺ + OH^1-

Reaction at cathode

Cu²⁺ + 2e^- ⟶ Cu [product copper metal]

Reaction at anode

Cu - 2e^- ⟶ Cu²⁺ [product nil - Cu²⁺ ions]

(ii) Dissociation of acidified water

H₂SO₄ ⇌ 2H¹⁺ + SO₄^2-

H₂O ⇌ H¹⁺ + OH^1- [ions H¹⁺, SO₄^2-, OH^1-]

Reaction at cathode

H¹⁺ + 1e^- ⟶ H x 4

2H + 2H ⟶ 2H₂ [product Hydrogen gas]

Reaction at anode

OH^1- - 1e^- ⟶ OH x 4

4OH ⟶ 2H₂O + O₂ [product oxygen gas]

(i) What is the atomicity of chlorine.

(ii) Name the compound formed when the above mentioned gas reacts with iron. Give equation.

(iii) Zinc sulphate reacts with sodium hydroxide first a little then in excess. Write a balanced equation for the two reactions with different conditions.

Answer

(i) Diatomic

(ii) Iron (III) chloride

2Fe + 3Cl₂ ⟶ 2FeCl₃

(iii) ZnSO₄ + 2NaOH ⟶ Na₂SO₄ + Zn(OH)₂ ↓

[Zn(OH)₂ + 2NaOH [in excess] ⟶ 2H₂O + Na₂ZnO₂]

If a crop of wheat replaces 30 kg of nitrogen per hectare of soil, what mass of the fertilizer calcium nitrate would be required to replace the nitrogen of a 5-hectare field ?

(at wt. N = 14, O = 16, Ca = 40)

Answer

Molecular weight of Ca(NO₃)₂
= 40 + 2[14 + 3(16)]
= 40 + 2[14 + 48]
= 40 + 2(62)
= 40 + 124 = 164 g

Total nitrogen that has to be replaced in 5 hectares of land = 30 x 5 = 150 kg = 150 x 10³ g

28 g of N is present in 164 g of calcium nitrate

∴ 150 x 10³ g of N is present in = $\dfrac{164}{28}$ x 150 x 10³ = 878.57 x 10³ g = 878.57 kg

Hence, 878.57 kg of fertilizer will be required

(i) Oxygen oxidises ethyne to produce 8.4 litres of carbon dioxide at STP. Calculate the volume of ethyne used in this reaction.

2C₂H₂ + 5O₂ ⟶ 4CO₂ + 2H₂O

(ii) A₂B is the empirical formula of a compound which has vapour density 25. If atomic weight of A and B are 10 and 5 respectively. Find it's molecular formula.

Answer

(i) [By Lussac's law]

$\begin{matrix} 2\text{C}_2\text{H}_2 & + & 5\text{O}_2 &\longrightarrow & 4\text{CO}_2 & + & 2\text{H}_2\text{O} & \\ 2 \text{ vol.} & : & 5 \text{ vol.} & \longrightarrow & 4\text{ vol.} & : & 2\text{ vol.} \end{matrix}$

To calculate the volume of ethyne gas

$\begin{matrix}\text{CO}_2 & : & \text{C}_2\text{H}_2 \\ 4 & : & 2 \\ 8.4 & : & x \end{matrix}$

$\therefore x = \dfrac{2}{4} \times 8.4 = 4.2 \text{ L}$

Hence, volume of ethyne gas required = 4.2 L.

(ii) Empirical formula is A₂B

Empirical formula weight = 2(10) + 5 = 25

Vapour density (V.D.) = 25

Molecular weight = 2 x V.D. = 2 x 25

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{2 \times 25}{25} = 2$

∴ Molecular formula = n[E.F.] = 2[X₂Y] = A₄B₂

(i) Name the metalloid(s) of period 3.

(ii) Define atomic size.

(iii) What would be seen on mixing a solution of calcium chloride with a solution of sodium carbonate? Write an equation also.

(iv) Write an equation for the preparation of a hydroxide of a metal which is

1. Green

2. Light blue

Answer

(i) Silicon [Si]

(ii) Atomic size is the distance between the centre of the nucleus of an atom and it's outermost shell.

(iii) White precipitate of calcium carbonate appears.

CaCl₂ + Na₂CO₃ ⟶ 2NaCl + CaCO₃

(iv)

1. Green : Iron [II] hydroxide
   FeSO₄ + 2NaOH ⟶ Na₂SO₄ + Fe(OH)₂ ↓

2. Light blue : Copper [II] hydroxide
   CuSO₄ + 2NaOH ⟶ Na₂SO₄ + Cu(OH)₂ ↓

Give a chemical test in each case to distinguish between the following pairs of chemicals.

(i) Sodium chloride solution and sodium nitrate solution.

(ii) Sodium sulphate solution and sodium chloride solution.

(iii) Calcium nitrate solution and zinc nitrate solution.

Answer

(i) Add silver nitrate soln. to the given solns., sodium chloride reacts to form a white ppt. which is soluble in NH₄OH and insoluble in dil. HNO₃. The other soln. is sodium nitrate.

NaCl + AgNO₃ ⟶ AgCl + NaNO₃

NaNO₃ + AgNO₃ ⟶ no white ppt.

(ii) Sodium sulphate solution reacts with barium chloride to form white ppt. of barium sulphate and sodium chloride, whereas, no reaction takes place in case of sodium chloride solution because both of them have the same anion.

Na₂SO₄ + BaCl₂ ⟶ BaSO₄ ↓ [white ppt.] + 2NaCl

NaCl + BaCl₂ ⟶ no reaction

(iii) When NaOH is added to the given soln., Zn(NO₃)₂ reacts to form a gelatinous white ppt. which dissolves in excess of NaOH whereas, Ca(NO₃)₂ forms a milky white ppt. which is insoluble in excess of NaOH. Hence, the two can be distinguished.

Calculate

(i) The volume of 2.5 moles of a gas X at stp.

(ii) The weight of 2.8 dm³ of a basic gas you have learnt, at STP.

(iii) The molecular weight of a gas Y whose 5.6 litres of volume at STP weighs 3.5 gms.

(at wt of C = 12, X = 14, H = 1, O = 16)

Answer

(i) Vol. occupied by 1 mole = 22.4 lit

∴ Vol. occupied by 2.5 moles = 22.4 x 2.5 = 56 lit.

Hence, vol. occupied is 56 lit.

(ii) Let the basic gas be NH₃

Molecular mass of NH₃ = 14 +3(1) = 17 g

At STP, 1 mole of ammonia weighs 17 g and occupy 22.4 dm³ volume.

∴ 1 dm³ volume has weight = $\dfrac{17}{22.4}$ = 0.76 g

∴ 2.8 dm³ of NH₃ has weight = 2.8 x 0.76 = 2.128 g

Hence, weight of 2.8 dm³ of ammonia = 2.128 g

(iii) 5.6 L of volume at STP weighs 3.5 gms.

∴ 22.4 L of volume weighs = $\dfrac{3.5}{5.6}$ x 22.4 = 14 g

Hence, molecular weight = 14g

Concentrated nitric acid oxidises phosphorus to phosphoric acid as

P + HNO₃ ⟶H₃PO₄ + H₂O + NO₂

Balance the equation and answer the following questions based on it.

(i) What mass of phosphoric acid can be prepared from 6.2 g of phosphorus?

(ii) What mass of nitric acid will be consumed at the same time ?

(iii) What would be the volume of steam produced at the same time if measured at 760 mm. Hg pressure and 273°C.

(Atomic wt of H = 1, N = 14, O = 16, P = 31) ?

Answer

Balanced equation is :

P + 5HNO₃ ⟶ H₃PO₄ + H₂O + 5NO₂

(i) Gram molecular mass of phosphorus = 31 g

gram molecular mass of H₃PO₄ = 3(1) + 31 + 4(16) = 3 + 31 + 64 = 98 g

31 g of P makes 98 g of phosphoric acid

∴ 6.2 g of P will make $\dfrac{98}{31}$ x 6.2 = 19.6 g

Hence, mass of phosphoric acid prepared = 19.6 g

(ii) Gram molecular mass of nitric acid (HNO₃) = 1 + 14 + 3(16) = 63 g

31 g of P needs 63 g of nitric acid

∴ 6.2 g of P will need $\dfrac{63}{31}$ x 6.2 = 12.6 g of nitric acid

Hence, mass of nitric acid needed = 12.6 g

(iii) 31 g of P gives 22.4 lit of steam

∴ 6.2 g of P will give $\dfrac{22.4}{31}$ x 6.2 = 4.48 L

Hence, volume of steam produced = 4.48 L

What is an 'alloy'. Give composition of

(i) duralumin

(ii) brass

(iii) fuse metal.

Answer

Alloy — An alloy is a homogeneous mixture of two or more metals or of one or more metals with certain non-metallic elements.

(i) Duralumin — Al [95%], Mg [0.5%], Mn [0.5%], Cu [4%]

(ii) Brass — Cu [60-80%], Zn [40-20%]

(iii) Fuse metal — Pb [50%], Sn [50%]

Define

(i) Acid

(ii) molar volume

(iii) neutralisation.

Answer

(i) Acid — An acid is a compound which when dissolved in water yields hydronium ions [H₃O⁺] as the only positively charged ion.

(ii) Molar volume — It is the volume occupied by 1 mole of a gas at STP.

(iii) Neutralization — It is the process due to which [H⁺] ions of an acid react completely or combine with [OH^-] ions of a base to give salt and water only.

(i) Define pH. If the hydrogen ion concentration of a solution is 10^-12 moles per litre, calculate its pH value.

(ii) What fluid is injected when yellow wasps or white ants bite? How can the pain be relieved?

(iii) A solution of Iron (III) chloride has a pH less than 7. Is the solution acidic or alkaline. Write the equation of its hydrolysis.

(iv) If iron is reacted with dilute sulphuric acid, what will be the product formed ?

Answer

(i) pH is defined as the negative logarithm [to the base 10] of the hydrogen ion concentration expressed in moles/litre.

Thus, pH = -log₁₀H⁺. It represents the strength of acids and alkalis, expressed in terms of hydrogen ion concentration [H⁺ aq.]

pH = -log₁₀H⁺ = -log₁₀[10^-12] = 12

Hence, pH value = 12

(ii) When yellow wasps or white ants bite, they inject formic acid into the skin. So by applying base it neutralises the acid and gives relief.

(iii) Solution will be acidic as pH is less than 7.

Hydrolysis of Iron (III) chloride:

FeCl₃ ⟶ Fe³⁺ + 3Cl^-

H₂O ⟶ H⁺ + OH^-

Further, the ferric ions partially combine with hydroxide ions to form ferric hydroxide.

Fe³⁺ + 3OH^- ⟶ Fe(OH)₃

This precipitation removes the hydroxyl ions [OH^-] from the solution leaving a relative excess of hydrogen ions and this excess of hydrogen ions makes the solution acidic.

(iv) Iron sulphate will be formed

Fe + H₂SO₄ ⟶ FeSO₄ + H₂ ↑

What do you observe when —

(i) Ammonia is passed over heated copper oxide.

(ii) Conc. H₂SO₄ is added to blue vitriol.

(iii) Acetylene is passed into bromine water.

(iv) NaCl is added to silver nitrate solution.

(v) Conc. HCl acid is exposed to atmosphere.

Answer

(i) Ammonia gas reduces black copper [II] oxide to brown Copper.

2NH₃ + 3CuO ⟶ 3Cu + 3H₂O + N₂ ↑

(ii) Blue hydrated copper sulphate becomes white anhydrous copper sulphate

$\text{CuSO}_4.5\text{H}_2\text{O} \xrightarrow{\text{Conc. H}_2 \text{SO}_4} \text{CuSO}_4 + 5\text{H}_2\text{O}$

(iii) Reddish brown bromine dissolved in water turns colourless when acetylene (ethyne) is passed through it.

C₂H₂ + Br₂ ⟶ C₂H₂Br₂ + Br₂ ⟶ C₂H₂Br₄

(iv) NaCl reacts with silver nitrate solution to form a white ppt.

NaCl + AgNO₃ ⟶ AgCl ↓ [white ppt.] + NaNO₃

(v) Due to strong affinity for water, concentrated hydrochloric acid pulls moisture from the air towards itself. This moisture forms droplets of water and hence, a cloud can be seen.

(i) Give 2 tests to distinguish between HCl solution and Ammonia solution.

(ii) Distinguish between MnO₂ and CuO

Answer

(i) Two tests to distinguish between HCl solution and Ammonia solution:

1. HCl turns moist blue litmus paper red, whereas, ammonia gas turns moist red litmus blue.
2. With HCl phenolphthalein solution remains colourless whereas, ammonia gas turns it pink.

(ii) Distinguish between MnO₂ and CuO:

(a) When each of the compound is heated with conc. hydrochloric acid, greenish yellow (chlorine) gas is evolved in case of manganese dioxide and filtrate is brownish in colour whereas, no chlorine gas is evolved in case of copper oxide and filtrate is bluish in colour.

MnO₂ + 4HCl ⟶ MnCl₂ + 2H₂O + Cl₂

CuO + 2HCl ⟶ CuCl₂ + H₂O

(b) On adding ammonium hydroxide to the above filtrate, no ppt. is formed in case of MnO₂, whereas, in case of copper oxide, pale blue ppt. is formed. The ppt. dissolves in excess of ammonium hydroxide forming azure blue colour.

Write an ionic reaction at the electrodes during electroplating of silver over brass spoon after giving the ionisation of the electrolyte.

Answer

Dissociation of sodium silver cyanide

Na[Ag(CN)₂] ⇌ Na¹⁺ + Ag¹⁺ + 2CN^1-

H₂O ⇌ H¹⁺ + OH^1-

Reaction at cathode [article to be plated]

Ag¹⁺ + 1e^1- ⟶ Ag [Ag deposited on article]

Reaction at anode [block of active-silver]

Ag - 1e^1- ⟶ Ag¹⁺ [product nil - Ag¹⁺ ions]

Draw an electron dot diagram of :

(i) a non-polar hydrocarbon molecule.

(ii) a polar covalent molecule.

Answer

Electron dot diagram of :

(i) a non-polar hydrocarbon molecule : Methane

(ii) a polar covalent molecule : HCl

State three harmful effects of acid rain.

Answer

Harmful effects of acid rain are:

1. Decolourises leaf pigments, retards growth of crops and reduces rate of photosynthesis.
2. Acid rain gets absorbed by plants, animals and directly or indirectly toxicity enters food chain affecting humans.
3. Corrodes metallic surfaces, disintegrates paper and leather, weakens building material such as statues, sculptures, marbles, limestone [CaCO3] etc.

State the elements of group 17. What are they called and why ?

Answer

The elements of group 17 are Fluorine, Chlorine, Bromine, Iodine, and Astatine.

Group 17 elements are called halogens. The name halogen (Greek, halo = sea or salt + gen = producing) means 'salt former'. Group 17 elements form salts, hence, they are called Halogens.

Define.

(i) Ionisation potential

(ii) Electron affinity

How do they vary in VII A group or IIIrd period.

Answer

(i) Ionization Potential is the amount of energy required to remove an electron from the outer most shell of an isolated gaseous atom.

(ii) Electron affinity is the amount of energy released when an atom in the gaseous state accepts an electron to form an anion.

Both ionization potential and electron affinity decreases down the group VII A.

Both ionization potential and electron affinity increases as one moves from left to right in IIIrd period.

Name the common refrigerant. How does it deplete ozone layer ?

Answer

The main refrigerants used are Freon chlorofluorocarbons (CFC). They deplete ozone layer and contribute to global warming. The chlorofluorocarbons are decomposed by ultraviolet rays to highly reactive chlorine which is produced in the atomic form.

$\text{CF}_2\text{Cl}_2 \xrightarrow{\text{Ultraviolet}} \text{CF}_2\text{Cl} + \text{Cl} [\text{free radical}]$

The free radical [Cl] reacts with ozone and chlorine monoxide is formed.

Cl + O₃ [ozone] ⟶ ClO + O₂

This causes depletion of ozone. Chlorine monoxide further reacts with atomic oxygen and produces more free radicals.

ClO + O ⟶ Cl + O₂

Again this free radical [Cl] destroys ozone, and the process continues thereby giving rise to ozone depletion.

Draw the structures of,

(i) an alcohol with 2 carbon atoms

(ii) an alkyne with 3 carbon atoms

Give their IUPAC names.

Answer

(i) Ethanol [C₂H₅OH]

(ii) Propyne [C₃H₄]

Write the equations for the following and name the products formed.

(i) Ethyl alcohol reacts with sulphuric acid at 170°C.

(ii) Acetic acid reacts with conc. sodium hydroxide solution.

(iii) Acetic acid reacts with potassium hydrogen carbonate.

Answer

(i) Products formed — ethene and water.

$\underset{\text{ ethyl alcohol}}{\text{C}_2\text{H}_5\text{OH}} \xrightarrow[170\degree\text{C}]{\text{Conc. H}_2\text{SO}_4\text{[excess]}} \underset{ \text{ethene}}{\text{C}_2\text{H}_4} ↑ + \text{H}_2\text{O}\$

(ii) Products formed — sodium acetate and water

CH₃COOH + NaOH ⟶ CH₃COONa + H₂O

(iii) Products formed are potassium acetate, water and carbon dioxide gas.

CH₃COOH + KHCO₃ ⟶ CH₃COOK + H₂O + CO₂ ↑