$\left(\mathrm{R}=0.083 \mathrm{~L}\right.$ bar mol-1 $\mathrm{K}^{-1}$ નો ઉપયોગ કરો)
\( \text { Slope }=R T \)
\( 25.73=0.083 \times T \)
\(T=\frac{25.73}{0.083}=309.47 \approx 310 \mathrm{~K}\)
\(\therefore \quad \text { Temperature in }{ }^{\circ} \mathrm{C} \)\( =310-273 \)
\( =37^{\circ} \mathrm{C}\)
$(M.wt.$ of $CuCl_2 =134.4 $ અને $K_b = 0.52\, , molal^{-1})$
$(R =0.083\, L\, bar \,K ^{-1} \,mol ^{-1})$ (નજીકનો પૂર્ણાંક)