(પાણી માટે $K_f=1.86\, K\, kg, mol^{-1}$ અને ઇથિલીન ગ્લાયકોલનું આણ્વિય દળ $= 62\, g\, mol^{-1}).$
$\Delta T_{f}=0-(-6)=6\,^oC$
As we know that
$\Delta T_{f}=K_{f} \times$ molality
$\Delta {T_f}\; = \;{k_f}\;\frac{w}{{mol.wt}}\; \times \;\frac{{1000}}{{wt.\;of\;solvent}}$
Substituting given values in formula
$6=\frac{1.86 \times 1000 \times\, w}{62 \times 4}$
$w=0.8\, \mathrm{kg}=800\, \mathrm{gm}$
$(R = 0.083 \,L\, bar \,mol^{-1}\, K{-1})$