# The equation that is incorrect is:

Question:

The equation that is incorrect is:

1. $\left(\Lambda_{\mathrm{m}}^{0}\right)_{\mathrm{NaBr}}-\left(\Lambda_{\mathrm{m}}^{0}\right)_{\mathrm{Na}^{2} \mathrm{Cl}}=\left(\Lambda_{\mathrm{m}}^{0}\right)_{\mathrm{KBr}}-\left(\Lambda_{\mathrm{m}}^{0}\right)_{\mathrm{KCl}}$

2. $\left(\Lambda_{\mathrm{m}}^{0}\right)_{\mathrm{KCl}}-\left(\Lambda_{\mathrm{m}}^{0}\right)_{\mathrm{NaCl}}=\left(\Lambda_{\mathrm{m}}^{0}\right)_{\mathrm{KBr}}-\left(\Lambda_{\mathrm{m}}^{0}\right)_{\mathrm{NaBr}}$

3. $\left(\Lambda_{\mathrm{m}}^{0}\right)_{\mathrm{H}_{2} \mathrm{O}}=\left(\Lambda_{\mathrm{m}}^{0}\right)_{\mathrm{HCl}}+\left(\Lambda_{\mathrm{m}}^{0}\right)_{\mathrm{NaOH}}-\left(\Lambda_{\mathrm{m}}^{0}\right)_{\mathrm{NaCl}}$

4. $\left(\Lambda_{\mathrm{m}}^{0}\right)_{\mathrm{NaBr}}-\left(\Lambda_{\mathrm{m}}^{0}\right)_{\mathrm{Nal}}=\left(\Lambda_{\mathrm{m}}^{0}\right)_{\mathrm{KBr}}-\left(\Lambda_{\mathrm{m}}^{0}\right)_{\mathrm{NaBr}}$

Correct Option: , 4

Solution:

$\left(\Lambda_{\mathrm{m}}^{0}\right)_{\mathrm{NaBr}}-\left(\Lambda_{\mathrm{m}}^{0}\right)_{\mathrm{NaI}}$

$=\Lambda_{\mathrm{m}}^{0} \mathrm{Na}^{+}+\Lambda_{\mathrm{m}}^{0} \mathrm{Br}-\left(\Lambda_{\mathrm{m}}^{0} \mathrm{Na}^{+}+\Lambda_{\mathrm{m}}^{0} \mathrm{I}\right)$

$=\Lambda_{\mathrm{m}}^{0} \mathrm{Na}^{+}+\Lambda_{\mathrm{m}}^{0} \mathrm{Br}^{-\bar{r}}-\Lambda_{\mathrm{m}}^{0} \mathrm{Na}^{+}-\Lambda_{\mathrm{m}}^{0} \mathrm{I}$

$=\Lambda_{\mathrm{m}}^{0} \mathrm{Br}^{-}+\Lambda_{\mathrm{m}}^{0} \mathrm{I}^{-}$

$\left(\Lambda_{\mathrm{m}}^{0}\right)_{\mathrm{KBr}}-\left(\Lambda_{\mathrm{m}}^{0}\right)_{\mathrm{NaBr}}$

$=\Lambda_{\mathrm{m}}^{0} \mathrm{~K}^{+}+\Lambda_{\mathrm{m}}^{0} \mathrm{Br}^{-}-\left(\Lambda_{\mathrm{m}}^{0} \mathrm{Na}^{+}+\Lambda_{\mathrm{m}}^{0} \mathrm{Br}^{-}\right)$

$=\Lambda_{\mathrm{m}}^{0} \mathrm{~K}^{+}+\Lambda_{\mathrm{m}}^{0} \mathrm{~B} \overline{\mathrm{r}}-\Lambda_{\mathrm{m}}^{0} \mathrm{Na}^{+}-\Lambda_{\mathrm{m}}^{0} \mathrm{Br}^{-}$

$=\Lambda_{\mathrm{m}}^{0} \mathrm{~K}^{+}+\Lambda_{\mathrm{m}}^{0} \mathrm{Na}^{+}$

$\therefore\left(\Lambda_{\mathrm{m}}^{0}\right)_{\mathrm{NaBr}}-\left(\Lambda_{\mathrm{m}}^{0}\right)_{\mathrm{NaI}} \neq\left(\Lambda_{\mathrm{m}}^{0}\right)_{\mathrm{KBr}}-\left(\Lambda_{\mathrm{m}}^{0}\right)_{\mathrm{NaBr}}$