NaCl is added to 1 litre water to such an extent that ${\mathrm{\Delta T}}_{\mathrm{f}}/{\mathrm{K}}_{\mathrm{f}}$ becomes to $\frac{1}{500}$, the weight of NaCl added is 

For NaCl, i = 2

$\text{∴\hspace{0.17em}}\frac{{\mathrm{\Delta T}}_{\mathrm{f}}}{{\mathrm{K}}_{\mathrm{f}}}=2\times {\mathrm{C}}_{\mathrm{m}}\mathrm{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\therefore \mathrm{\hspace{0.17em}}{\mathrm{\Delta T}}_{\mathrm{f}}={\mathrm{K}}_{\mathrm{f}}\times \mathrm{i}\times {\mathrm{C}}_{\mathrm{m}}$

$\begin{array}{l}\mathrm{or}, \frac{1}{500}=2\times \mathrm{molality}\\ \mathrm{or} \mathrm{molality} = \frac{1}{1000}\\ \mathrm{i}.\mathrm{e}.,\frac{1}{1000}\mathrm{moles} {\mathrm{Kg}}^{-1} \mathrm{water}\\ \mathrm{i}.\mathrm{e}., \frac{1}{1000} \mathrm{moles} \mathrm{per} \mathrm{litre} \mathrm{water}\\ \mathrm{=}\frac{1}{1000}\times 58.5 \mathrm{g} \mathrm{NaCl} = 0.0585 \mathrm{g} \mathrm{NaCl}\end{array}$