Water or hydride of oxygen Questions in English

(D) The solubility of ionic compounds like $NaCl$ depends on the solvation energy of the ions.
In $D_2O$,the $O-D$ bond is stronger than the $O-H$ bond in $H_2O$ due to the higher mass of deuterium.
This leads to a lower dielectric constant and different solvation characteristics for $D_2O$ compared to $H_2O$.
Consequently,the solubility of $NaCl$ is lower in $D_2O$ than in $H_2O$.
Therefore,the statement that the solubility of $NaCl$ in $D_2O$ is more than in $H_2O$ is incorrect.

(B) Heavy water,denoted as $D_2O$,has a higher freezing point than ordinary water $(H_2O)$.
The freezing point of heavy water is $3.8$ $^oC$.

(B) Heavy water,denoted as $D_2O$,is an isotope of water where hydrogen is replaced by deuterium. It has a higher freezing point than ordinary water $(H_2O)$. The freezing point of heavy water is $3.8 \ ^oC$.

(A) . Heavy water $(D_2O)$ is used as a moderator in a nuclear reactor.
It slows down the speed of fast-moving neutrons to thermal energy levels.
It also acts as a coolant in some reactor designs.

(D) The correct answer is $(D)$.
Heavy water $(D_2O)$ is used as a moderator in nuclear reactors to slow down the speed of fast-moving neutrons.

(D) Heavy water is chemically known as deuterium oxide,represented by the formula $D_2O$.
It consists of the deuterium isotope of hydrogen ($D$ or $^2H$) instead of the common protium isotope ($H$ or $^1H$).

(B) $D_2O$ (heavy water) is used as a moderator in nuclear reactors to slow down fast-moving neutrons.

(D) Heavy water $(D_2O)$ is commonly used as a moderator in nuclear reactors to slow down fast neutrons. In some reactor designs,such as the $CANDU$ reactor,it also serves as a coolant to transfer heat from the fuel to the steam generators. Therefore,it acts as both a moderator and a coolant.

(A) The photolysis of water involves the splitting of water molecules by light energy.
The reaction is represented as: $H_2O \xrightarrow{h\nu} H^{+} + OH^{-}$.
Therefore,the correct option is $A$.

(B) In a molten ionic hydride (e.g.,$NaH$),the hydride ion $(H^-)$ is the anion.
During electrolysis,the anions migrate to the anode.
At the anode,the oxidation reaction occurs: $2H^- \rightarrow H_2(g) + 2e^-$.
Therefore,hydrogen gas is liberated at the anode.

(A) Surface tension is primarily determined by the strength of intermolecular forces,specifically hydrogen bonding.
Water $(H_2O)$ molecules exhibit strong hydrogen bonding due to their high polarity and the presence of two hydrogen atoms capable of forming bonds with oxygen atoms of adjacent molecules.
This extensive network of hydrogen bonds results in a higher cohesive force compared to organic solvents like benzene $(C_6H_6)$,methanol $(CH_3OH)$,or ethanol $(C_2H_5OH)$.
Therefore,water has the maximum surface tension among the given options.

(D) The white solid dissolves to form a blue solution.
Anhydrous $CuSO_4$ is a white powder. When it is added to dilute $H_2SO_4$,it absorbs water from the solution to form hydrated copper$(II)$ sulfate,which is $CuSO_4 \cdot 5H_2O$ (blue vitriol).
The reaction is: $CuSO_4(s) + 5H_2O(l) \to CuSO_4 \cdot 5H_2O(aq)$.
This hydrated salt dissolves in the aqueous medium to produce a characteristic blue-coloured solution.

(C) Plumbosolvency is the property of lead to dissolve in water that contains dissolved oxygen and carbon dioxide.
Ordinary water (soft water) often contains these dissolved gases,which react with lead to form lead hydroxide or lead carbonate,causing the lead to dissolve.
This is why lead pipes are not suitable for carrying drinking water.

(A) The best test to detect the presence of water in a liquid is using anhydrous copper$(II)$ sulfate $(CuSO_4)$.
When water is added to white anhydrous $CuSO_4$,it reacts to form hydrated copper$(II)$ sulfate $(CuSO_4 \cdot 5H_2O)$,which is blue in color.
The color change from white to blue confirms the presence of water.

(A) The molar mass of $MgSO_4$ is $120 \, g/mol$.
Since $120 \, g \, MgSO_4$ is equivalent to $100 \, g \, CaCO_3$,
the $30 \, ppm \, MgSO_4$ is equivalent to $\frac{100}{120} \times 30 = 25 \, ppm \, CaCO_3$.
The molar mass of $MgCl_2$ is $95 \, g/mol$.
Since $95 \, g \, MgCl_2$ is equivalent to $100 \, g \, CaCO_3$,
the $19 \, ppm \, MgCl_2$ is equivalent to $\frac{100}{95} \times 19 = 20 \, ppm \, CaCO_3$.
Total hardness of water = $25 + 20 = 45 \, ppm \, CaCO_3$ equivalent.

(A) $Calgon$ is the commercial name for sodium hexametaphosphate,which has the chemical formula $Na_2[Na_4(PO_3)_6]$.
It is used to soften hard water by sequestering calcium and magnesium ions.

(B) The hardness of water is expressed in terms of $CaCO_3$ equivalents in parts per million $(ppm)$.
This is done because $CaCO_3$ has a molecular weight of $100 \ g/mol$,which makes calculations convenient.
It represents the total concentration of calcium and magnesium salts present in the water,expressed as if they were all $CaCO_3$.

(C) In Clark's method for softening water,a calculated amount of lime,$Ca(OH)_2$,is used to remove temporary hardness by precipitating calcium carbonate and magnesium hydroxide.

(B) Temporary hardness is removed by adding $CaO$ (Clark's process).
$CaO + H_2O \to Ca(OH)_2$
$M(HCO_3)_2 + Ca(OH)_2 \to MCO_3 \downarrow + CaCO_3 \downarrow + 2H_2O$ (where $M = Ca, Mg$)
Permanent hardness is removed by adding $Na_2CO_3$ (Washing soda process).
$MSO_4 + Na_2CO_3 \to MCO_3 \downarrow + Na_2SO_4$ (where $M = Ca, Mg$)
Thus,$CaO$ removes temporary hardness and $Na_2CO_3$ removes permanent hardness.

(C) Water molecules exist as associated molecules due to $H$-bonding,which leads to a high boiling point.

(B) Statement $(A)$ is correct: Heavy water $(D_2O)$ is widely used as a moderator in nuclear reactors to slow down fast-moving neutrons.
Statement $(B)$ is correct: Due to stronger hydrogen bonding (deuterium bonding) in $D_2O$ compared to $H_2O$,the power of association is higher in $D_2O$.
Statement $(C)$ is incorrect: Heavy water is generally a poorer solvent than ordinary water because of the stronger intermolecular forces and higher viscosity.
Therefore,statements $(A)$ and $(B)$ are correct.

(D) Anhydrous copper sulfate $(CuSO_4)$ is white in color.
When it reacts with water,it forms hydrated copper sulfate $(CuSO_4 \cdot 5H_2O)$,which is blue in color.
$CuSO_4 \text{ (white)} + 5H_2O \rightarrow CuSO_4 \cdot 5H_2O \text{ (blue)}$
Therefore,the change in color from white to blue is a reliable scientific test to detect the presence of water in a liquid.

(A) $X$-ray studies of ice show that each oxygen atom is surrounded tetrahedrally by four other oxygen atoms at a distance of $276 \ pm$ by hydrogen bonds.
In the crystal lattice of ice, each oxygen atom is covalently bonded to two hydrogen atoms and hydrogen-bonded to two other hydrogen atoms from neighboring water molecules, resulting in a tetrahedral arrangement of four hydrogen atoms around each oxygen atom.

(A) The given reaction is the hydrolysis of magnesium chloride.
$MgCl_2 + 2H_2O \to Mg(OH)_2 + 2HCl$.
However,in the presence of heat,$Mg(OH)_2$ decomposes to $MgO$ and $H_2O$.
Thus,the reaction is $MgCl_2 + 2H_2O \to MgO + 2HCl + H_2O$.
Therefore,$X$ is $MgO$.

(C) Anhydrous $CaCl_2$ is commonly used as a drying agent in the laboratory to remove moisture from gases like natural gas.

(C) The boiling point of heavy water $(D_2O)$ is $101.4 ^o C$.
It is higher than that of ordinary water ($H_2O$,$100 ^o C$) due to stronger intermolecular forces (hydrogen bonding) in $D_2O$ compared to $H_2O$.

(B) As the atomic size increases,the van der Waals forces of attraction increase,leading to higher boiling points. $H_2O$ exhibits strong intermolecular $H$-bonding,which gives it an anomalously high boiling point. Among the remaining hydrides $(H_2S, H_2Se, H_2Te)$,$H_2S$ has the smallest molecular size and the weakest van der Waals forces. Therefore,$H_2S$ has the lowest boiling point.

(D) When $ZnCl_2 \cdot 2H_2O$ is heated strongly,it undergoes hydrolysis due to the presence of water of crystallization.
The reaction is as follows:
$ZnCl_2 \cdot 2H_2O \xrightarrow{\Delta} Zn(OH)Cl + HCl + H_2O$
Further heating of $Zn(OH)Cl$ leads to the formation of zinc oxide $(ZnO)$:
$2Zn(OH)Cl \xrightarrow{\Delta} ZnO + ZnCl_2 + H_2O$
However,in the context of standard chemistry problems regarding the dehydration of hydrated metal chlorides,the primary product formed upon strong heating is $ZnO$.

(B) Water $(H_2O)$ is known as the universal solvent because it has the ability to dissolve a wide variety of substances due to its polar nature and its ability to form hydrogen bonds. Therefore,the correct option is $B$.

(C) The addition of lime $(Ca(OH)_2)$ or calcium chloride $(CaCl_2)$ is a common method used to remove fluoride ions $(F^-)$ from water.
This process involves the precipitation of fluoride as calcium fluoride $(CaF_2)$,which is insoluble in water.
The chemical reaction is: $Ca^{2+} + 2F^- \rightarrow CaF_2(s)$.

(A) In ice,each water molecule forms $4$ hydrogen bonds.
Each oxygen atom is surrounded by $4$ hydrogen atoms,where $2$ are covalently bonded and $2$ are hydrogen bonded,resulting in a tetrahedral arrangement.

(A) The temperature of maximum density of $H_2O$ is $277.15 \, K$ $(4^{\circ}C)$.
The temperature of maximum density of $D_2O$ is $284.75 \, K$ $(11.6^{\circ}C)$.
Therefore,the correct option is $A$.

(C) Brackish water is water that has more salinity than freshwater,but not as much as seawater. It mostly contains $NaCl$ (sodium chloride) as the primary dissolved salt.

(B) The solubility of an ionic compound depends on the dielectric constant of the solvent.
$H_2O$ has a higher dielectric constant compared to $D_2O$.
Since a higher dielectric constant facilitates better dissociation of ionic compounds,the solubility of an ionic compound is higher in simple water $(H_2O)$ and lower in heavy water $(D_2O)$.

(A) The $Calgon$ process (using sodium hexametaphosphate,$Na_6P_6O_{18}$) is a well-known method used for the removal of permanent hardness of water by sequestering calcium and magnesium ions.
Therefore,the correct option is $(A)$.

(D) The exhausted permutite (hydrated sodium aluminium silicate,$Na_2Al_2Si_2O_8 \cdot xH_2O$) contains calcium and magnesium ions that have replaced the sodium ions during the water softening process.
To regenerate or revive the exhausted permutite,a concentrated solution of $10\% \,NaCl$ (brine) is passed through it.
The high concentration of $Na^+$ ions in the $NaCl$ solution displaces the $Ca^{2+}$ and $Mg^{2+}$ ions from the permutite,restoring it to its original sodium form.

(A) Heavy water $(D_2O)$ is used as a moderator in nuclear reactors to slow down fast-moving neutrons.
Heavy water has a higher boiling point $(374.42 \ K)$ compared to ordinary water $(373 \ K)$,which indicates that it is more associated due to stronger intermolecular forces.
The dielectric constant of ordinary water is higher than that of heavy water,making ordinary water a better solvent than heavy water.
Therefore,statements $(i)$ and $(ii)$ are correct.

(A) The statement regarding intramolecular hydrogen bonding is $FALSE$. In the condensed phase (liquid or solid),water molecules exhibit extensive $intermolecular$ hydrogen bonding,not $intramolecular$ hydrogen bonding.

(B) $(i)$ $D_2O$ ($2 \times 1 + 8 = 10$ electrons) and $H_2O$ ($2 \times 1 + 8 = 10$ electrons). So,$D_2O = H_2O$ is $CORRECT$.
$(ii)$ The enthalpy of formation $|\Delta H_f|$ for $D_2O$ is higher than $H_2O$ due to stronger $O-D$ bonds compared to $O-H$ bonds. So,$D_2O > H_2O$ is $CORRECT$.
$(iii)$ The enthalpy of vaporization $|\Delta H_{vaporization}|$ for $D_2O$ is higher than $H_2O$ due to stronger intermolecular hydrogen bonding in $D_2O$. So,$D_2O > H_2O$ is $CORRECT$.
$(iv)$ The density of $D_2O$ $(1.107 \ g/mL)$ is higher than $H_2O$ $(1.000 \ g/mL)$. So,$D_2O > H_2O$ is $CORRECT$.
$(v)$ Molecular mass of $D_2O$ $(20 \ u)$ is higher than $H_2O$ $(18 \ u)$. So,$D_2O > H_2O$ is $CORRECT$.
All $5$ statements are $CORRECT$.

(D) $1$. The catenation tendency of phosphorus $(P)$ is greater than that of nitrogen $(N)$ because $P-P$ single bonds are stronger than $N-N$ single bonds due to the absence of lone pair-lone pair repulsion in $P-P$ bonds.
$2$. $PCl_3$ is indeed produced by the reaction of white phosphorus $(P_4)$ with thionyl chloride $(SOCl_2)$: $P_4 + 8SOCl_2 \rightarrow 4PCl_3 + 4SO_2 + 2S_2Cl_2$.
$3$. According to the Bronsted-Lowry theory,water can act as both a proton donor $(H_2O \rightarrow H^+ + OH^-)$ and a proton acceptor $(H_2O + H^+ \rightarrow H_3O^+)$,making it amphoteric.
$4$. Permanent hardness of water is caused by the presence of dissolved chlorides and sulfates of calcium and magnesium. Boiling only removes temporary hardness (caused by bicarbonates). Therefore,the statement that permanent hardness is removed by boiling is incorrect.

(D) Clark's process is a method used for the removal of temporary hardness from water by adding a calculated amount of lime,$Ca(OH)_2$.
The chemical reaction involved is:
$Ca(HCO_3)_2 + Ca(OH)_2 \rightarrow 2CaCO_3 \downarrow + 2H_2O$
Thus,the chemical used is calcium hydroxide,$Ca(OH)_2$.

(D) Temporary hardness of water is caused by the presence of dissolved bicarbonates of calcium and magnesium,specifically $Ca(HCO_3)_2$ and $Mg(HCO_3)_2$.
These can be removed by boiling,which converts them into insoluble carbonates:
$Ca(HCO_3)_2 \rightarrow CaCO_3 \downarrow + H_2O + CO_2$
$Mg(HCO_3)_2 \rightarrow MgCO_3 \downarrow + H_2O + CO_2$

(D) Permanent hardness of water is caused by the presence of soluble salts of magnesium and calcium in the form of chlorides and sulphates in water.
$Na_6P_6O_{18}$ (Calgon) is used to remove permanent hardness by sequestering $Ca^{2+}$ and $Mg^{2+}$ ions.
$Na_2CO_3$ (Washing soda) removes permanent hardness by precipitating calcium and magnesium ions as carbonates.
Zeolite $(NaAlSiO_4)$ is used in the ion-exchange method to remove permanent hardness.
$Ca(OH)_2$ (Slaked lime) is used for the removal of temporary hardness of water (Clark's method),not permanent hardness.

(B) Temporary hardness is caused by dissolved calcium or magnesium hydrogencarbonate.
Clark's method is a chemical process used to remove temporary hardness by adding a calculated amount of slaked lime,which is $Ca(OH)_{2}$.
In this process,the lime reacts with the bicarbonates to form insoluble precipitates that can be filtered out.
The chemical reactions are:
$Ca(HCO_{3})_{2} + Ca(OH)_{2} \rightarrow 2CaCO_{3}(\downarrow) + 2H_{2}O$
$Mg(HCO_{3})_{2} + 2Ca(OH)_{2} \rightarrow 2CaCO_{3}(\downarrow) + Mg(OH)_{2}(\downarrow) + 2H_{2}O$
Therefore,$Ca(OH)_{2}$ is the compound used.

(C) To obtain de-mineralised water (also known as de-ionized water),all dissolved mineral salts must be removed.
This is achieved by passing water through an ion-exchange resin system.
First,the water passes through a cation exchange resin in the $H^{+}$ form,which replaces all metal cations (like $Ca^{2+}$,$Mg^{2+}$,$Na^{+}$) with $H^{+}$ ions.
Second,the water passes through an anion exchange resin in the $OH^{-}$ form,which replaces all anions (like $Cl^{-}$,$SO_4^{2-}$,$HCO_3^{-}$) with $OH^{-}$ ions.
The released $H^{+}$ and $OH^{-}$ ions combine to form water $(H^{+} + OH^{-} \rightarrow H_2O)$.
Therefore,the correct method uses both $H^{+}$ cation exchanger and $OH^{-}$ anion exchanger.

(D) $1$. Calgon's method uses sodium hexametaphosphate $(Na_6P_6O_{18})$,which is commercially called 'Calgon'. It keeps calcium and magnesium ions in solution as complex ions.
$2$. The permutit method uses hydrated sodium aluminium silicate $(Na_2Al_2Si_2O_8 \cdot xH_2O)$,also known as zeolite,to exchange $Na^{+}$ ions with $Ca^{2+}$ and $Mg^{2+}$ ions.
$3$. The synthetic resin method uses ion-exchange resins to remove all soluble mineral salts (both cations and anions) from water,resulting in demineralized or deionized water.
$4$. Since all the given statements are correct,the correct option is $D$.

(B) Clark's method is specifically used to remove the temporary hardness of water,which is caused by the presence of bicarbonates of calcium and magnesium. In this process,calculated amounts of slaked lime,$Ca(OH)_{2}$,are added to the water.
The chemical reactions are as follows:
$Ca(HCO_{3})_{2} + Ca(OH)_{2} \rightarrow 2 CaCO_{3} \downarrow + 2 H_{2}O$
$Mg(HCO_{3})_{2} + 2 Ca(OH)_{2} \rightarrow 2 CaCO_{3} \downarrow + Mg(OH)_{2} \downarrow + 2 H_{2}O$
Other methods like washing soda,Calgon's,and ion-exchange are used to remove both temporary and permanent hardness.
Hence,option $B$ is correct.

(A) The chemical reaction for the removal of temporary hardness using lime water is:
$Ca(HCO_3)_2 + Ca(OH)_2 \rightarrow 2CaCO_3 \downarrow + 2H_2O$
From the balanced chemical equation,$1 \ mole$ of $Ca(OH)_2$ reacts with $1 \ mole$ of $Ca(HCO_3)_2$ to precipitate calcium carbonate.
Therefore,the number of moles of $Ca(OH)_2$ required is $1$.

(B) The critical temperature $(T_c)$ of $H_2O$ $(647 \ K)$ is higher than that of $NH_3$ $(405 \ K)$ due to stronger hydrogen bonding in $H_2O$.
The standard boiling point of water is $100 \ ^oC$ $(373.15 \ K)$,which is a correct statement.
The critical volume $(V_c)$ of $H_2O$ $(56 \ cm^3/mol)$ is less than that of $NH_3$ $(72 \ cm^3/mol)$.
Therefore,statements $(ii)$ and $(iii)$ are correct.

(A) Clark's method is a process used to remove temporary hardness of water caused by calcium bicarbonate $(Ca(HCO_3)_2)$ and magnesium bicarbonate $(Mg(HCO_3)_2)$.
In this method,a calculated amount of lime $(Ca(OH)_2)$ is added to the hard water.
The lime reacts with the bicarbonates to form insoluble calcium carbonate $(CaCO_3)$ and magnesium hydroxide $(Mg(OH)_2)$,which can then be filtered out.
The chemical reactions are:
$Ca(HCO_3)_2 + Ca(OH)_2 \rightarrow 2CaCO_3 \downarrow + 2H_2O$
$Mg(HCO_3)_2 + 2Ca(OH)_2 \rightarrow 2CaCO_3 \downarrow + Mg(OH)_2 \downarrow + 2H_2O$