Q: Calculate the temporary and permanent hardness of water sample having the following the following constituents per litre:\$ Ca(HCO_{3})_{2} = 162\, mg, MgCl_{2} = 95 =\, mg, \$\$ NaCl = 585\, mg, Mg(HCO_{3})_{2} = 73\, mg, \$\$ CaSO_{4} = 136\, mg \$

mole of \$ Ca(HCO_{3})_{2} = \dfrac{162\times 10^{-3}mg}{162\,g/mole} = 1\times 10^{-3}\,moles \$Mole of \$ Ca(SO_{4}) = \dfrac{136\times 10^{-3}g}{136\,g/mole} = 1\times 10^{3}\,mole \$Total mole of \$ Ca = 2\times 10^{-3}\,mole \$mass of \$ CaCO_{3} = 2\times 10^{-3}\times 100 = 0.2\,g \$\$ \therefore \$ ppm (permanent hardness) \$ = \dfrac{6.2}{1000}\times 10^{6} = 200\,ppm \$Mole of \$ MgCl_{2} = \dfrac{95\times 10^{-3}}{95} = 1\times 10^{-3}\,mole \$ Mole Mg \$ (HCg)_{2} = \dfrac{73\times 10^{-3}}{146} = 5\times 10^{-4}\,mole \$mole Mg \$ = 1.5\times 10^{-4}\,mole \$Mole g \$ CaCO_{3} \$ (In terms of mg) \$ = 1.5\times 10^{-3} \$mass \$ = 1.5\times 10^{-3} = 0.150\,g \$ppm (temporary hardness) \$ = \dfrac{0.150}{100}\times 106 = 150\,ppm \$Hence, the correct option is \$\text{C}\$

Q: The drain cleaner, Drainex contains small bits of aluminum which react with caustic soda to produce dihydrogen. What volume of dihydrogen at \$20^o C\$ and one bar will be released when 0.15 g of aluminum reacts?

The reaction can be written as:-\$2Al_{(s)}+2NaOH_{(aq)}+2H_2O_{(l)}\longrightarrow {2NaAlO_2}_{(aq)}+{3H_2}_{(g)}\$Here, \$2\$ moles of \$Al\$ gives \$3\$ moles of \$H_2\$ gas.Moles of aluminium= \$\cfrac {0.15g}{27}=5.56\times 10^{-3}\$ moles\$\therefore\$ Moles of \$H_2\$ produced= \$\cfrac {3}{2}\times 5.56\times 10^{-3}\$=\$8.33\times 10^{-3}\$ moles\$\therefore\$ Volume of \$H_2\$=\$\cfrac {nRT}{P}\$=\$\cfrac {8.33\times 10^{-3}\times 0.0831\times 293}{1}\$=\$0.203\$ litres\$\therefore\$ Volume of dihydrogen released is \$0.203\$ litres

Q: Which of the following hydrides is electron-precise hydride?

Methane \$CH_4\$ is electron-precise hydride.Diborane \$B_2H_6\$ is electron-deficient hydrideAmmonia \$NH_3\$ and water \$H_2O\$ are electron-rich hydrides.Elements of group 14 form electron-precise (having required number of electrons to write the Lewis structure ) form precise hydrides.

Q: What causes the temporary and permanent hardness of water ?

The presence of calcium and magnesium bicarbonates \$Ca(HCO_3)_2\$ and \$Mg(HCO_3)_2\$ causes temporary hardness in water. The presence of soluble salts of calcium and magnesium, i.e., sulphates and chlorides of calcium and magnesium cause permanent hardness in water.

Q: Elements of which of the following groups do not form hydrides?

The region of the periodic table from the group \$7\$ to \$9\$ is referred to as the hydride gap as they do not form hydrides. Examples of such elements are Mn, Fe, Co, Ru etc. These elements do not form hydrides on account of low affinity for hydrogen in their normal oxidation states. Hence, option C is correct.

Q: Half litre each of three samples of \$H_{2}O_{2}\$ labelled \$10\$ volume, \$15\$ volume and \$20\$ volume are mixed and then diluted with equal volume of water. The relative strength of resultant \$H_{2}O_{2}\$ solution is:

Volume strength of \$H_{ 2 }{ O }_{ 2 } = 5.6\times \text{N}\$ \$(\text{N} =\$ Normality\$)\$Normality of \$10\$ vol of \$H_{ 2 }{ O }_{ 2 } = \dfrac{10}{5.6}\text{N}\$Normality of \$15\$ vol of \$H_{ 2 }{ O }_{ 2 } = \dfrac{15}{5.6} \text{N}\$Normality of \$20\$ vol of \$H_{ 2 }{ O }_{ 2 } = \dfrac{20}{5.6} \text{N}\$Let \$500\$ ml of each is mixed, then the total volume of mix \$= 1500\$ ml.Also, this is diluted to double its volume, i.e., \$= 3000\$ ml\$\text{N}\times 3000\$ = \$\dfrac { 10 }{ 5.6 } \times 500+\dfrac { 15 }{ 5.6 } \times 500+\dfrac { 20 }{ 5.6 } \times 500\$Normality of \$H_{ 2 }{ O }_{ 2 }\$ mix \$= 1.339\:\text{N}\$Vol. strength of \$H_{ 2 }{ O }_{ 2 } = 1.339 \times 5.6 = 7.5\$.

Question 1

Calculate the temporary and permanent hardness of water sample having the following the following constituents per litre:$ Ca(HCO_{3})_{2} = 162\, mg, MgCl_{2} = 95 =\, mg, $$ NaCl = 585\, mg, Mg(HCO_{3})_{2} = 73\, mg, $$ CaSO_{4} = 136\, mg $

Accepted Answer

mole of $ Ca(HCO_{3})_{2} = \dfrac{162\times 10^{-3}mg}{162\,g/mole} = 1\times 10^{-3}\,moles $Mole of $ Ca(SO_{4}) = \dfrac{136\times 10^{-3}g}{136\,g/mole} = 1\times 10^{3}\,mole $Total mole of $ Ca = 2\times 10^{-3}\,mole $mass of $ CaCO_{3} = 2\times 10^{-3}\times 100 = 0.2\,g $$ \therefore $ ppm (permanent hardness) $ = \dfrac{6.2}{1000}\times 10^{6} = 200\,ppm $Mole of $ MgCl_{2} = \dfrac{95\times 10^{-3}}{95} = 1\times 10^{-3}\,mole $ Mole Mg $ (HCg)_{2} = \dfrac{73\times 10^{-3}}{146} = 5\times 10^{-4}\,mole $mole Mg $ = 1.5\times 10^{-4}\,mole $Mole g $ CaCO_{3} $ (In terms of mg) $ = 1.5\times 10^{-3} $mass $ = 1.5\times 10^{-3} = 0.150\,g $ppm (temporary hardness) $ = \dfrac{0.150}{100}\times 106 = 150\,ppm $Hence, the correct option is $\text{C}$

Question 2

The drain cleaner, Drainex contains small bits of aluminum which react with caustic soda to produce dihydrogen. What volume of dihydrogen at $20^o C$ and one bar will be released when 0.15 g of aluminum reacts?

Accepted Answer

The reaction can be written as:-$2Al_{(s)}+2NaOH_{(aq)}+2H_2O_{(l)}\longrightarrow {2NaAlO_2}_{(aq)}+{3H_2}_{(g)}$Here, $2$ moles of $Al$ gives $3$ moles of $H_2$ gas.Moles of aluminium= $\cfrac {0.15g}{27}=5.56\times 10^{-3}$ moles$\therefore$ Moles of $H_2$ produced= $\cfrac {3}{2}\times 5.56\times 10^{-3}$=$8.33\times 10^{-3}$ moles$\therefore$ Volume of $H_2$=$\cfrac {nRT}{P}$=$\cfrac {8.33\times 10^{-3}\times 0.0831\times 293}{1}$=$0.203$ litres$\therefore$ Volume of dihydrogen released is $0.203$ litres

Question 3

Which of the following hydrides is electron-precise hydride?

Accepted Answer

Methane $CH_4$ is electron-precise hydride.Diborane $B_2H_6$ is electron-deficient hydrideAmmonia $NH_3$ and water $H_2O$ are electron-rich hydrides.Elements of group 14 form electron-precise (having required number of electrons to write the Lewis structure ) form precise hydrides.

Question 4

What causes the temporary and permanent hardness of water ?

Accepted Answer

The presence of calcium and magnesium bicarbonates $Ca(HCO_3)_2$ and $Mg(HCO_3)_2$ causes temporary hardness in water. The presence of soluble salts of calcium and magnesium, i.e., sulphates and chlorides of calcium and magnesium cause permanent hardness in water.

Question 5

Write chemical reactions to show the amphoteric nature of water.

Accepted Answer

Amphoteric nature of water is represented by following chemical reactions.(1) Water as a base:$\displaystyle H_2O(l)+H_2S(g) \rightleftharpoons H_3O^+(aq) + HS^-(aq) $(2) Water as an acid:$\displaystyle H_2O(l) + NH_3(aq) \rightleftharpoons OH^-+NH_4^+ $(3) Self ionization of water in which water simultaneous acts as acid and base.$\displaystyle 2H_2O \rightarrow H_3O^+ + OH^- $

Question 6

Elements of which of the following groups do not form hydrides?

Accepted Answer

The region of the periodic table from the group $7$ to $9$ is referred to as the hydride gap as they do not form hydrides. Examples of such elements are Mn, Fe, Co, Ru etc. These elements do not form hydrides on account of low affinity for hydrogen in their normal oxidation states. Hence, option C is correct.

Question 7

What do you understand by (i) electron-deficient, (ii) electron-precise, and (iii) electron-rich compounds of hydrogen?

Accepted Answer

(i) Electron deficient compounds of hydrogen do not have sufficient number &#10;of electrons to form normal covalent bonds. Examples include hydrides of&#10; group 13 such as $\displaystyle BH_3, AlH_3 $ etc. They exist in &#10;polymeric forms such as $\displaystyle B_2H_6, Al_2H_6 $. Thus, they &#10;make up their deficiency.(ii) Electron precise compounds of hydrogen contain sufficient valence &#10;electrons to form covalent bonds. Examples include group 14 hydrides &#10;$\displaystyle CH_4,SiH_4, GeH_4, SnH_4, PbH_4 $ etc. These compounds &#10;have tetrahedral shape.(iii) Electron rich compounds of hydrogen contain excess valence electrons to form covalent bonds.Examples include group 15 and group 17 hydrides such as $\displaystyle NH_3, PH_3, HCl, HBr $ etc. These compounds contains lone pairs of electrons.

Question 8

What happens when dil. hydrochloric acid is added to iron filings?

Accepted Answer

When dilute Hydrochloric acid is added to iron filings, iron chloride & hydrogen gas is produced.$Fe+2HCl\longrightarrow FeCl_2+H_2$The iron displaces hydrogen from hydrochloric acid to form iron (II) chloride & hydrogen gas. This is a single displacement reaction.Thus the answer is option (A) Hydrogen gas and iron chloride are produced.

Question 9

What do you understand by the term 'auto-protolysis' of water? What is its significance?

Accepted Answer

The reaction between two water molecules to produce hydronium ion and hydroxide ion is known as autoprotolysis of water. This is self ionization of water.$\displaystyle 2H_2O \rightarrow H_3O^++OH^- $ This reaction indicates the amphoteric nature of water. It can act as an acid as well as a base. One water molecule donates electron whereas other water molecule accepts electron.

Question 10

Half litre each of three samples of $H_{2}O_{2}$ labelled $10$ volume, $15$ volume and $20$ volume are mixed and then diluted with equal volume of water. The relative strength of resultant $H_{2}O_{2}$ solution is:

Accepted Answer

Volume strength of $H_{ 2 }{ O }_{ 2 } = 5.6\times \text{N}$     $(\text{N} =$ Normality$)$Normality of $10$ vol of $H_{ 2 }{ O }_{ 2 } = \dfrac{10}{5.6}\text{N}$Normality of $15$ vol of $H_{ 2 }{ O }_{ 2 } = \dfrac{15}{5.6} \text{N}$Normality of $20$ vol of $H_{ 2 }{ O }_{ 2 } = \dfrac{20}{5.6} \text{N}$Let $500$ ml of each is mixed, then the total volume of mix $= 1500$ ml.Also, this is diluted to double its volume, i.e., $= 3000$ ml$\text{N}\times 3000$ = $\dfrac { 10 }{ 5.6 } \times 500+\dfrac { 15 }{ 5.6 } \times 500+\dfrac { 20 }{ 5.6 } \times 500$Normality of $H_{ 2 }{ O }_{ 2 }$ mix $= 1.339\:\text{N}$Vol. strength of $H_{ 2 }{ O }_{ 2 } = 1.339 \times 5.6 = 7.5$.