Browse Questions

# Which of the following expressions is correct?

$\begin{array}{1 1}\Lambda_m^{\infty}(NH_4OH)=\Lambda_m^{\infty}(NH_4Cl)+\Lambda_m^{\infty}(NaOH)-\Lambda_m^{\infty}(NaCl)\\\Lambda_m^{\infty}(NH_4OH)=\Lambda_m^{\infty}(NH_4Cl)-\Lambda_m^{\infty}(NaOH)+\Lambda_m^{\infty}(NaCl)\\\Lambda_m^{\infty}(NH_4OH)=\Lambda_m^{\infty}(NH_4Cl)-\Lambda_m^{\infty}(NaOH)-\Lambda_m^{\infty}(NaCl)\\\Lambda_m^{\infty}(NH_4OH)=\Lambda_m^{\infty}(NH_4Cl)+\Lambda_m^{\infty}(NaOH)+\Lambda_m^{\infty}(NaCl)\end{array}$

Can you answer this question?

Answer : $\Lambda_m^{\infty}(NH_4OH)=\Lambda_m^{\infty}(NH_4Cl)+\Lambda_m^{\infty}(NaOH)-\Lambda_m^{\infty}(NaCl)$
$\Lambda_m^{\infty}(NH_4OH)=\lambda_m^{\infty}(NH_4^+)+\lambda_m^{\infty}(OH^-)$
$\Rightarrow \lambda_m^{\infty}(NH_4^+)+\lambda_m^{\infty}(Cl^-)+\lambda_m^{\infty}(Na^+)+\lambda_m^{\infty}(OH^-)-\lambda_m^{\infty}(Na^+)-\lambda_m^{\infty}(Cl^-)$
$\Rightarrow \Lambda_m^{\infty}(NH_4OH)=\Lambda_m^{\infty}(NH_4Cl)+\Lambda_m^{\infty}(NaOH)-\Lambda_m^{\infty}(NaCl)$
answered Jul 18, 2014