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(B) $(i)$ $D_2O$ ($2 	imes 1 + 8 = 10$ electrons) and $H_2O$ ($2 	imes 1 + 8 = 10$ electrons). So,$D_2O = H_2O$ is $CORRECT$.$(ii)$ The enthalpy of

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(B) $(i)$ $D_2O$ ($2 	imes 1 + 8 = 10$ electrons) and $H_2O$ ($2 	imes 1 + 8 = 10$ 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 $CORRECT$.

Find the number of $CORRECT$ order against mentioned property :-
$(i)$ $D_2O = H_2O$ (number of electrons)
$(ii)$ $D_2O > H_2O$ $(|\Delta H_f|)$
$(iii)$ $D_2O > H_2O$ $(|\Delta H_{vaporization}|)$
$(iv)$ $D_2O > H_2O$ (density)
$(v)$ $D_2O > H_2O$ (Molecular mass)

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Find the number of $CORRECT$ order against mentioned property :-$(i)$ $D_2O = H_2O$ (number of electrons)$(ii)$ $D_2O > H_2O$ $(|\Delta H_f|)$$(iii)$ $D_2O > H_2O$ $(|\Delta H_{vaporization}|)$$(iv)$ $D_2O > H_2O$ (density)$(v)$ $D_2O > H_2O$ (Molecular mass)