Chapter 15, Problem 15F.4AE

Interpretation Introduction

Interpretation:

The spin only molar susceptibility of $CuS{O}_4.5H_{2}O$ at ${25}^{o}C$ has to be calculated.

Concept Introduction:

The spin only molar susceptibility can be calculated using following equation.

${\chi }_m=\frac{N_A{g}_e^2{\mu }_0{\mu }_B^{2}S\left(S+1\right)}{3kT}$

Where,

  $\begin{array}{l}{\chi }_m=molarsusceptibility\\ N_A=6.022\times {10}^{23}mo{l}^{-1}\\ g_e^{}=2.0023\\ {\mu }_0=\left(4\pi \times {10}^{-7}{T}^2{J}^{-1}{m}^3\right)\\ {\mu }_B^{}=9.274\times {10}^{-24}J{T}^{-1}\\ k=1.381\times {10}^{-23}J{K}^{-1}\\ S=\frac{n}{2}\\ n=number\text{}ofunpairedelectrons\\ T=temperature\end{array}$