$\Delta$T_f = K_fm ......(1)

$\Delta$T_b = K_bm ......(2)

$\Rightarrow$ ${{\Delta {T_f}} \over {\Delta {T_b}}} = {{{K_f}} \over {{K_b}}}$ .....(3)

$\Delta$T_b = T₂ – T₁ = 100.18 – 100 = 0.18

k_f for water = 1.86 K kg mol^–1

k_b for water = 0.512 K kg mol^–1

${{\Delta {T_f}} \over {0.18}} = {{1.86} \over {0.512}}$

$\Rightarrow$ ${\Delta {T_f}}$ = 0.654

$\Delta$T_b = T₂ – T₁

$\Rightarrow$ 0.654 = 0 – T₂

$\Rightarrow$ T₂ = $-$ 0.654^oC