ઠારબિંદુ $= 0^o$ સે $ - $ $\Delta T_f$
$-0.186 =$ $\Delta T_f$
$\Delta T_f$ $= 0.186$
$\Delta T_f$ = $K_f$ $\times$ $m $
$0.186 = 1.86 $ $\times m$ $m = 0.1$
$\Delta T_b = K_b \times m$
$\Delta T_b$ $= 0.512$ $\times$ $0.1$ $\Delta T_b$ $= 0.0512^o$ સે
$(K_f =-1.86\,^o\, C/m)$
(નજીકનાં પૂર્ણાંકમાં રાઉન્ડ ઑફ) $\left[ K _{ b }=0.52 \,K \,kg \,mol ^{-1}\right]$