Water or hydride of oxygen Questions in English

(A) The solubility of an ionic compound like $NaCl$ depends on the dielectric constant of the solvent.
Higher dielectric constant leads to better solvation of ions,which increases solubility.
The dielectric constant of ordinary water $(H_2O)$ is approximately $81$,while that of heavy water $(D_2O)$ is approximately $80$.
Since the dielectric constant of ordinary water is higher,$NaCl$ is more soluble in ordinary water and less soluble in heavy water.
Therefore,both $(A)$ and $(R)$ are true and $(R)$ is the correct explanation of $(A)$.

(A) The $Calgon$ process is used for the removal of hardness of water. $Calgon$ is the trade name for sodium hexametaphosphate,$Na_2[Na_4(PO_3)_6]$,which removes $Ca^{2+}$ and $Mg^{2+}$ ions by forming soluble complexes.

(B) Heavy water,denoted as $D_2O$,has physical properties slightly different from ordinary water $(H_2O)$.
At $1 \ atm$ pressure,the freezing point of heavy water is $3.8 \ ^{\circ}C$.

(C) $SO_3 + D_2O \rightarrow D_2SO_4$ (Similar to $SO_3 + H_2O \rightarrow H_2SO_4$)
$CaC_2 + 2D_2O \rightarrow Ca(OD)_2 + C_2D_2$ (Similar to $CaC_2 + 2H_2O \rightarrow Ca(OH)_2 + C_2H_2$)
$Al_4C_3 + 12D_2O \rightarrow 4Al(OD)_3 + 3CD_4$ (Similar to $Al_4C_3 + 12H_2O \rightarrow 4Al(OH)_3 + 3CH_4$)
Thus,the correct match is $A-V, B-I, C-II$.

(C) moderator is a material used in a nuclear reactor to slow down the neutrons produced from fission.
Heavy water $(D_2O)$ is used as a moderator in nuclear reactors because it slows down neutrons effectively and also has a low probability of neutron absorption.

(A) Statement $1$ is correct because heavy water $(D_2O)$ is known to be injurious to the growth of both plants and animals.
Statement $2$ is incorrect because the reaction of aluminum carbide $(Al_4C_3)$ with heavy water $(D_2O)$ produces deuterated methane $(CD_4)$,not deuterated acetylene. The balanced chemical equation is: $Al_4C_3 + 12D_2O \rightarrow 4Al(OD)_3 + 3CD_4$.
Statement $3$ is correct because $BaCl_2 \cdot 2D_2O$ is a classic example of an interstitial deuterate (or deuterate hydrate).

(D) When heavy water $(D_2O)$ reacts with magnesium nitride $(Mg_3N_2)$,the deuterium atoms replace the hydrogen atoms in the products.
The balanced chemical equation for the reaction is:
$Mg_3N_2 + 6D_2O \longrightarrow 3Mg(OD)_2 + 2ND_3$
Thus,the products formed are magnesium deuteroxide $(Mg(OD)_2)$ and deuteroammonia $(ND_3)$.

(D) $(a). H_2O + H_2S \rightarrow H_3O^{+} + HS^{-}$
$(b). 3H_2O + N^{3-} \rightarrow NH_3 + 3OH^{-}$
$(c). 2H_2O + SiCl_4 \rightarrow SiO_2 + 4HCl$
$(d). 2F_2 + 2H_2O \rightarrow O_2 + 4HF$ (or $O_2 + 4F^{-} + 4H^{+}$)
Matching the products: $(a)-(i), (b)-(ii), (c)-(iv), (d)-(vi)$.

(C) Temporary hardness in water is caused by the presence of magnesium and calcium hydrogen carbonates.
When water containing $Mg(HCO_3)_2$ and $Ca(HCO_3)_2$ is boiled,these compounds decompose.
$Ca(HCO_3)_2$ decomposes to form calcium carbonate $(CaCO_3)$,which is a precipitate: $Ca(HCO_3)_2 \xrightarrow{\Delta} CaCO_3 \downarrow + H_2O + CO_2 \uparrow$.
$Mg(HCO_3)_2$ decomposes to form magnesium hydroxide $(Mg(OH)_2)$ because $Mg(OH)_2$ is less soluble than $MgCO_3$: $Mg(HCO_3)_2 \xrightarrow{\Delta} Mg(OH)_2 \downarrow + 2CO_2 \uparrow$.

(C) Viscosity of $H_2O$ at $25^{\circ} C \approx 0.89 \ cP$ (or $0.00089 \ Pa \cdot s$).
Viscosity of $D_2O$ at $25^{\circ} C \approx 1.10 \ cP$.
$\text{Ratio} = \frac{\text{Viscosity of } D_2O}{\text{Viscosity of } H_2O} = \frac{1.10}{0.89} \approx 1.24$.
Thus,the correct option is $C$.

(C) During the electrolysis of water,$H_{2}O$ molecules are decomposed into $H_{2}$ and $O_{2}$ gas more readily than $D_{2}O$ molecules due to the kinetic isotope effect.
As a result,the concentration of heavy water $(D_{2}O)$ in the remaining liquid increases over time.
Since the density of $D_{2}O$ $(1.107 \ g/cm^3)$ is higher than that of $H_{2}O$ $(1.00 \ g/cm^3)$,the overall density of the remaining liquid water increases.