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  • The molar conductance at infinite dilution of <img alt="\mathrm{BaCl}_2, \mathrm{NaCl}$ and $\mathrm{NaOH}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BBaCl%7D_2%2C%20%5Cmathrm

The molar conductance at infinite dilution of \mathrm{BaCl}_2, \mathrm{NaCl}$ and $\mathrm{NaOH}  are 285 \times 10^{-4}, 122.5 \times 10^{-4}$ and $245 \times 10^{-4} \mathrm{sm}^2$ mol $^{-1} respectively.

The molar conductance at incite dilution for \mathrm{Ba}(\mathrm{OH})_2 is.

Option: 1

422 \times 10^{-3} \mathrm{sm}^2 \mathrm{~mol}^{-1}


Option: 2

530 \times 10^{-4} \mathrm{sm}^2 \mathrm{~mol}^{-1}


Option: 3

224 \times 10^{-3} \mathrm{sm}^2 \mathrm{~mol}^{-1}


Option: 4

1.28 \times 10^3 \mathrm{~mol}^{-1}


Answers (1)

best_answer

\mu_{\mathrm{BaCl}_2}^{\infty}=\mu_{\mathrm{Ba}^{2+}}^{\infty}+2 \mu_{\mathrm{Cl}^{-}}^{\infty} \\   .....(1)

\mu_{\mathrm{NaCl}}^{\infty}=\mu_{\mathrm{Na}^{+}}^{\infty}+ \mu_{\mathrm{Cl}^{-}}^{\infty} \\       .....(2)

\mu_{\mathrm{NaOH}}^{\infty}=\mu_{\mathrm{Na}^{+}}^{\infty}+ \mu_{\mathrm{HO}^{-}}^{\infty} \\   .....(3)

1+2(3)-2(2) \\

\mu_{(\mathrm{Ba}(\mathrm{OH})_2)}=\left(285 \times 10^{-4}\right)+2\left(245 \times 10^{-4}\right)-2\left(122.5 \times 10^{-4}\right)

=530 \times 10^{-4} \mathrm{sm}^2 \mathrm{~mol}^{-1} \\

Posted by

Ritika Harsh

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