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For $B=5$ T, the same spin-only two-state probabilities apply, but the low-temperature argument is no longer small. At 300 K, $x=0.0111952$; at 2 K, $x=1.67928$. HenceThe 2 K sample is about $93.3\%$ of its saturation magnetization, with about $96.6\%$ of the moments in the aligned state. The two percentages describe different observables. A linear extrapolation would predict $M/M_s\simeq x=1.679$, exceeding saturation and therefore contradicting the bounded two-state population.

For completeness, at this 2 K, 5 T bias the dimensionless secant susceptibility is approximately ${\chi }_0\tanh x/x=0.01199$, while the differential susceptibility is ${\chi }_0{\mathrm{sech}}^{2}x=0.002805$ under the same prescribed-field approximation. Neither equals the zero-field Curie slope $0.02159$. Thus saturation reduces the incremental response even while the magnetization itself is large.

A $3d^9$ shell has one hole in the otherwise filled $d$ shell. Hund's rules give $S=1/2$, $L=2$, and, for a more-than-half-filled shell, $J=L+S=5/2$: the free-ion term is ${}^2{D}_{5/2}$. Its Landé factor isDistinguish the Curie effective moment from a saturated projection:The question then explicitly asks us to use orbital quenching in the crystal. Ligand fields split and mix orbital states, suppressing their freely orientable angular momentum. The spin-only approximation has $g\simeq 2$, $S=1/2$, so ${\mu }_{\mathrm{eff}}=\sqrt{3}{\mu }_B$ but the two field-axis projections are $\pm {\mu }_B$. The latter, not $\sqrt{3}{\mu }_B$, belong in the tanh argument and saturation magnetization.

The sulfate and water environment supplies mainly closed-shell, weakly magnetic surroundings, stabilizes localized Cu2+ sites and separates their moments so that exchange can be weak on the temperature scale considered. Paramagnetism here is an approximation of weakly interacting local spins, not a claim that dilution guarantees no magnetic ordering at any temperature. Elemental copper has a filled $3d$ band and itinerant conduction electrons rather than one localized Cu2+ spin per atom. Pauli spin response and orbital/core contributions then replace a local-spin Curie law; ordinary pure copper has a weak net diamagnetic response.^[ref]

There is one Cu2+ per formula unit, hence $n_s=N_A/V_m=5.5148\times 10^{27}$ m${}^{-3}$. The spin-only weak-field SI volume susceptibility isThese neglect exchange, anisotropic $g$ factors and background diamagnetism. At high field and low temperature, saturation requires the nonlinear response in a; a field-independent Curie value then describes only the zero-field slope.

Take two localized spin orientations with moments $\pm {\mu }_B$ along the applied magnetic field. This assumes an electron-like spin $g$ factor of two. Their energies are $\mp {\mu }_{B}B$, so the one-moment partition function is $Z_1=2\cosh x$, with $x={\mu }_{B}B/(k_{B}T)$. The thermal mean moment is $k_{B}T{\partial }_B\ln Z_1={\mu }_B\tanh x$, and the spin-one-half paramagnetic magnetization isFor $|x|\ll 1$, $M\simeq n{\mu }_B^{2}B/(k_{B}T)$. With the same SI convention $\chi =\partial M/\partial H$ and $B\simeq {\mu }_{0}H$,This Curie law is a zero-field or weak-field susceptibility. At arbitrary field the differential response is $(C/T){\mathrm{sech}}^{2}x$, and $M$ saturates at $n{\mu }_B$; extending the linear law to unlimited field would contradict that finite saturation magnetization.

A $J=1/2$ doublet has moment projections $\pm g_J{\mu }_B/2$. The requested normalization assumes $g_J=2$, so the projections are $\pm {\mu }_B$ and the two Zeeman energies are $\mp {\mu }_{B}B$. For independent localized ions, the one-ion partition function and thermal moment areTherefore the spin-one-half paramagnetic magnetization isIt saturates at $n{\mu }_B$ for a field strong compared with $k_{B}T/{\mu }_B$. For a different Landé g-factor, replace every moment ${\mu }_B$ by $g_J{\mu }_B/2$; $J=1/2$ alone does not fix that factor.