[5 Mark Question Answer]

Question 14 Marks

The diagram represents fractional distillation for separation mixtures. Answer the following

$1.$ Can two immiscible liquids be separated by this process.
$2.$ Separation of liquids by this process is based on which physical property?
$3.$ If methyl alcohol $\&$ water are to be separated, which liquid would remain in flask ‘$X$’ after condensation.
$4.$ Give a reason for the above answer.
$5.$ State the purpose of the fractionating column in the apparatus.

Answer

$1.$ No, they can be separated by separating funnel.
$2.$ The physical property on which separation is based "Difference in their boiling points."
$3.$ Water having boiling point $100^{\circ} C$ will remain in flask ' $X$ '.
$4.$ Alcohol having lower $B.P. 78^{\circ} C$ will evaporate first and condense in the conical flask - receiver ' $Y^{\prime}$.
$5.$ The upper part of the 'Fractionating column is cooler, so as the hot vapours rise up in the column, they get cooled $($condense$)$ and trickle back into the distillation flask ' $X$ '.

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Question 24 Marks

Explain with diagrams the process used to $-$ separate the following substances from the given mixtures.
$(a)$ Ammonium chloride from a mixture of $-$ ammonium chloride $\&$ potassium chloride.
$(b)$ Iron from a mixture of $-$ iron $\&$ copper
$(c)$ Sulphur from a mixture of $-$ sulphur $\&$ copper.
$(d)$ Potassium nitrate from a mixture of $-$ potassium nitrate $\&$ potassium chlorate.
$(e)$ Lead carbonate [insoluble] from a mixture of $-$ lead carbonate $\&$ water.
$(f)$ Lead nitrate [soluble] from a mixture of $-$ lead nitrate $\&$ water Le. lead nitrate solution.
$(g)$ Carbon tetrachloride from a mixture of $-$ carbon tetrachloride $[$heavier component$]\ \&$ water.
$(h)$ Benzene from a mixture of $-$ benezene $\left[\right. b.p. \left.80^{\circ} C \right]$ toluene $\left[ b . p . \ 110^{\circ} CJ\right.]$.
$(i)$ Different dyes $-$ in their liquid constituent ink.

Answer

$(a)$ By sublimation : On heating the mixture in evaporating dish, ammonium chloride sublimes on the walls of funnel and potassium chloride remains in evaporating dish.

Here $A$ is Ammonium Chloride and $B$ is Potassium Chloride.
$(b)$ By magnetic separation : By bringing a magnet near the mixture iron pieces can be separated which will cling to the magnet.

$(c)$ By solvent extraction : Mixture of copper and sulphur is added to the beaker containing solvent carbon disulphide and stirred well. Sulphur dissolves. Put this mixture on filter paper in the funnel. Copper remains on filter paper and sulphur passes into the beaker as filtrate. Sulphur separates as carbon disulphide evaporates.

Here $A$ is for Copper and $B$ is Sulphur.
$(d)$ Potassium nitrate $\ce{KNO_3}$ is more soluble than potassium chlorate $\ce{KClO_3}$.
On heating to get saturated solution and on cooling the saturated solution less soluble $\ce{(KClO_3)}$ crystallise out. More soluble $\ce{KNO_3}$ is filtered out from hot saturated solution , and is recrystallised from hot water and dried.

$(e)$ Evaporation : Lead carbonate can be separated by evaporation. On evaporation, water evaporates leaving behind solid lead carbonate which has higher $\ce{M.P}$.

$(f)$ Lead nitrate is separated from soluble lead nitrate solution by crystallisation.

$(g)$ By separing funnel, heavier $\ce{CCl_4}$ carbon tetrachloride form the lower layer is separated when tap is opened and is collected in the flask. Water the lighter top layer remains in the funnel.

$(h)$ By fractional distillation, miscible low boiling point benzene $(B.P. 80^\circ C)$ evaporates on heating the mixture and condenses in and collects in flask $‘Y’$ where as higher boiling pt. Toluene $(B.P. 110^\circ C)$ remains in flask $‘X’$ after condensation.

$(i)$ By chromatography : Different dyes $[$solid constituents i. e. $\ce{A, B, C, D]}$ in ink which is the liquid constituent. By placing the ink spot containing different solid constituents $[$dyes$]$ on the filter paper. Filter paper is hung with it’s lower end completely dipped in the solvent.

The solvent flows over the ink spot and the solid constituents $[$dyes $\ce{‘A’, ‘B’, ‘C’, ‘D’]}$ separate out.

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Question 34 Marks

State any one method $–$ to separate the following mixtures$-$
$(a)$ Two solid mixtures one of which $–$ directly changes into vapour on heating.
$(b)$ Two solid mixtures one of which $–$ dissolves in a particular solvent and other does not
$(c)$ A solid-liquid mixture containing $–$ an insoluble solid in the liquid component
$(d)$ A solid-liquid mixture containing $–$ a soluble solid in the liquid component
$(e)$ A liquid-liquid mixture containing $–$ two immiscible liquids having different densities
$(f)$ A liquid-liquid mixture containing $–$ two miscible liquids having different boiling points.
$(g)$ A liquid-gas mixture containing $–$ a gas dissolved in a liquid component.
$(h)$ A gas-gas mixture containing $–$ two gases with different densities.
$(i)$ A mixture of different solid constituents $–$ in a liquid constituent.

Answer

$(a)$ The method used is sublimation : Sublimable solid sublimes on heating i.e. changes into vapours and condenses on cooling is separated leaving behind non$-$sublimable solid.
$(b)$ The method is solvent extraction : Soluble solid dissolves in solvent leaving behind the other insoluble solid. The dissolved solid is recovered by evaporation.
$(c)$ The method is filtration : The insoluble solid is obtained on the filter paper as residue.
$(d)$ The method is evaporation or distillation : Liquid evaporates leaving behind solid.
$(e)$ The method is separating funnel: Heavier liquid collects in the flask below an opening tap, while lighter layer remains in the separating funnel.
$(f)$ Method is fractional distillation : The liquid with lower
boiling point collects in the receiver while the liquid with higher $B.P$. remains in distillation flask.
$(g)$ The method used is boiling the mixture : Solubility of gas decreases with increase in temperature, so gas escapes when mixture is boiled and collected separately.
$(h)$ The method is diffusion : The lighter gas diffuses more rapidly on passing the rough porous partition where as heavier gas diffuses less rapidly on passing through porous partition.
$(i)$ The method used is chromatography.

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Question 44 Marks

Explain the term ‘mixture’. Differentiate between homogenous $\&$ heterogenous mixtures. State why brass is considered as a homogenous mixture while a mixture of iron $\&$ sulphur – heterogenous. Give an example of two liquids which form $–$
$(a)$ homogenous
$(b)$ heterogenous $–$ mixtures.

Answer

Mixture : A mixture is made up of two or more substances elements or compounds or both mechanically mixed together in any proportion. A mixture retain the properties of its constituent elements or compounds.
Differenance between homogenous & heterogenous mixtures :
Homogenous mixture :
There constituents are uniformly mixed.
The properties and composition are same throughout the mixture Two solids are Brass $[Cu + Zn]$
Heterogeneous mixture :
There constituents are not uniformly mixed.
The properties and composition are very throughout the mixture Two solids are Iron $+$ sulphur
$\text{BRASS}$ is $\text{HOMOGENOUS}$ because mixture is just the same through out and its constituents cannot be distinguished from each other i.e. composition is same through out.
Whereas in Heterogenous mixture of iron and sulphur, the constituents. Can be separated easily by a magnet, the particles can be distinguished from each other and composition is not uniform.
$(a)$ Two liquids which form Homogenous mixture are Alcohol and water.
$(b)$ Two liquids which form Heterogenous mixture are oil and water.

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Question 54 Marks

State five different characteristics of compounds. Give three differences between elements $\&$ compounds with relevant examples.

Answer

Five characteristics of a compound :
$(a)$ Components in a compound are in a definite proportion.
$(b)$ Compound is always homogenous $[$i.e. identical composition$]$.
$(c)$ Particles in a compound are of one kind.
$(d)$ Compounds have a definite set of properties.
$(e)$ Elements in the compound do not retain their original properties.
$(f)$ Components in a compound can be separated by chemical means only.
Three differences between elements and compounds :
Element :
They contains one kind of atoms only e.g. element $(Na)$ Sodium and $[Cl]$ chlorine.
They elements have ther own set of properties e.g. $[H]$ Hydrogen is combustible.
They cannot be broken down into two or more simpler substances by physical chemical means.
Compound :
They contains two or more kinds of atoms, e.g. compound $-$ water $\ce{[H_2O]}$.
They Properties of compounds are entirly new i.e. water $\ce{[H_2O]}$ is a liquid and can extinguish fire.
They Cannot be broken down into two or more simpler substances by physical or substances by chemical means.

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Question 64 Marks

Elements are broadly classified into metals $\&$ non$-$metals. State six general differences in physical properties of metals $\&$ non$-$metals. State two metals & two non-metals which contradict with the general physical properties $–$ giving y between metalloids $\&$ noble gases.

Answer

Metallic elements :
$($a$)$ Have lustre $–$ i.e. shine.
$($b$)$ Are malleable $–$ i.e. can be beaten into sheets.
$($c$)$ Are ductile $–$ i.e. can be drawn into wires.
$($d$)$ Are good conductors $–$ of heat and electricity.
$($e$)$ have high $–$ melting and boiling points.
$($f$)$ Have high $–$ density.
$($g$)$ Contain $–$ one type of atoms $–$ monoatomic
Non-metallic Elements :
$($a$)$ Do no have lustre.
$($b$)$ Are non-malleable $–$ i.e. cannot be beaten into sheets.
$($c$)$ Are non-ductile $–$ i.e. cannot be drawn into wires.
$($d$)$ Are poor conductors $–$ of heat and electricity.
$($e$)$ Have low $–$ melting and boiling points.
$($f$)$ How low $–$ density.
$($g$)$ Contain $–$ monoatomic or diatomic atoms.
Two metals which contradict properties :
$($i$)$ Mercury $–$ is liquid at room temperature whereas metals are mostly solid.
$($ii$)$ Zinc $–$ is $\text{NON-MALLEABLE}$ and $\text{NON-DUCTILE}$ contradicts metal which are $\text{MALLEABLE}$ and $\text{DUCTILE.}$
Two Non$-$metals are :
$($i$)$ Iodine $–$ is lustrous contradicts non-metals have no lustre.
$($ii$)$ Graphite $–$ is good conductor of electricity where as non¬metals are poor conductor of electricity.

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Question 74 Marks

‘The modern periodic table consists of elements arranged according to their increasing atomic numbers’. With reference to elements with atomic numbers $1$ to $20$ only in the periodic table $–$ differentiate them into $–$ metallic elements, metalloids, non$-$metals $\&$ noble gases.

Answer

Names and symbols of metal, non$-$metals, metalloids and noble gases out of $1^{st} 20$ elements.

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