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What will be the degree of drociation of HA if im for \mathrm{NaCl}, \mathrm{HCl} and \mathrm{NaA} are 124.4, 425.9 and 100.5 respectively with conductivity of 0.002 HA is 6 \times 10^{-5} \mathrm{scm}^{-1}.

Option: 1

0.075


Option: 2

0.080


Option: 3

0.024


Option: 4

0.124


Answers (1)

best_answer

\Lambda_m^0(\mathrm{HCl})+\Lambda^{0}_{\mathrm{m}}(\mathrm{NaA})-\Lambda^{0}_\mathrm{~m}(\mathrm{Nacl}) \\\Lambda^0(\mathrm{HA})=429.5-126.4+100.5

\Lambda^0(\mathrm{HA})=400 \mathrm{scm}^2 \mathrm{~mol}^{-1}

\mathrm{K(\mathrm{HA})=86 \times 10^{-5} \mathrm{scm}^{-1}}

\mathrm{ \Lambda_m(H A)=\frac{K(H A) \times 1000}{\text { Molality of } H A}=\frac{6 \times 10^{-5} \times 1000}{0.002} }
\mathrm{ \Lambda m(H A)=30}

\mathrm{\alpha=\frac{\Lambda m(H A)}{\Lambda^{0}_m(H A)}=\frac{30}{400}=0.075}
 

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