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# Given,$\wedge _m^o$ ​ for $NaCl$, $HCl$ and $NaA$ are$126.4$,$\;425.9$ and $100.5{\text{ }}$$S.c{m^2}mo{l^{ - 1}}$ , respectively. If the conductivity of $0.001{\text{ }}M$ $HA$ is $5 \times {10^{ - 5}}S.c{m^{ - 1}}$ , degree of dissociation of $HA$ is:A) $0.75$B) $0.125$C) $0.25$D) $0.50$

Last updated date: 20th Jun 2024
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Hint: Reactions, for example, the one in the past model include the dissociation of an atom. Such reactions can be effectively depicted as far as the division of reactant particles that really separate to accomplish equilibrium in an example. This portion is known as the degree of dissociation.

The dissociation degree is the portion of unique solute atoms that have separated. It is typically demonstrated by the Greek image $\alpha$. All the more precisely, degree of dissociation alludes to the measure of solute separated into particles or revolutionaries per mole. If there should be an occurrence of solid acids and bases, the degree of dissociation will be near $1$. Less amazing acids and bases will have lesser degree of dissociation. There is a basic connection between this boundary and the van 't Hoff factor $i$. On the off chance that the solute substance separates into $n$ particles, at that point
$i = 1 + \alpha (n - 1)$
For example, for the accompanying dissociation
As $n = 2$, we would have that $i = 1 + \alpha$.
Thermodynamics of the response –
$\Delta G < 0$
The response is unconstrained and continues the forward way.
Law of Chemical equilibrium -
At a given temperature, the result of centralization of the response items raised to the particular stoichiometric coefficient in the reasonable synthetic condition isolated by the result of convergence of the reactants raised to their individual stoichiometric coefficients has a consistent worth.

${K_c} = \frac{{{{[C]}^{c\:{{[D]}^d}}}}}{{{{[A]}^a}\:{{[B]}^b}}}$
$[A],\:[B],\:[C]\:[D]$
are equilibrium concentration,
As we have learned in disassociation
we know
$\wedge _M^0(HA) = \wedge _M^0(HCL) + V(NaA) - \wedge _M^0(NaCl)$
$= 425.9 + 100.5 - 126.4$
$= 4005c{m^2}mo{l^{ - 1}}$
${ \wedge _M} = \frac{{1000K}}{M} = \frac{{1000 \times 5 \times {{10}^{ - 5}}}}{{{{10}^{ - 3}}}} = 505c{m^2}mo{l^{ - 1}}$
$\alpha = \frac{{{ \wedge _M}}}{{ \wedge _M^0}} = \frac{{50}}{{400}} = 0.125$
Hence, the correct option is (B).

Note:
Dissociation in science and natural chemistry is an overall cycle where particles (or ionic mixes, for example, salts, or edifices) independent or split into more modest particles, for example, atoms, particles, or revolutionaries, ordinarily in a reversible way. For example, when a corrosive disintegrates in water, a covalent connection between an electronegative particle and a hydrogen molecule is broken by heterolytic splitting, which gives a proton (${H^ + }$) and a negative particle. Dissociation is something contrary to affiliation or recombination.