$\therefore \,\,\frac{{\Delta {T_{b\,(A)}}}}{{\Delta {T_{b\,(B)}}}}\, = \,\frac{{{K_{b(A)}}}}{{{K_{b(B)}}}}$ as ${m_A}\, = \,{m_B}$
$\therefore \,\,\frac{{\Delta {T_{b\,(A)}}}}{{\Delta {T_{b\,(B)}}}}\, = \frac{1}{5}$
[ આપેલ : $\mathrm{R}=0.083 \mathrm{~L} \mathrm{bar} \mathrm{mol}^{-1} \mathrm{~K}^{-1}$ ]
$\mathrm{NaCl}$ નું સંપૂર્ણ વિયોજન થાય છે તે ધારી લો.
[ઉપયોગ : ${R}=0.083\, {~L}\, bar \,{mol}^{-1} \,{~K}^{-1}$ ]