The depression in freezing point of water containing ethylene glycol is $7.4^0\mathrm{C}$. The mass of ethylene glycol added in 500 g water is g (nearest integer)

$\left({\mathrm{K}}_{\mathrm{f}}=1.86 \mathrm{K} \mathrm{kg}{ \mathrm{mol}}^{-1}\right)$

Reduced freezing point = 7.4${}^{\circ }\mathrm{C}$
Ethylene glycol mass equals?
volume of water $=500 \mathrm{g}=0.5 \mathrm{kg}(1 \mathrm{kg}=1000 \mathrm{g})$
Formula used:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=i\times K_f\times \frac{\text{ Mass of ethylene glycol }}{\text{ Molar mass ofethylene glycol }\times \text{ Mass of water in }\mathrm{kg}}$
where,
$\Delta T_f$= change in freezing point =$7.4°\mathrm{C}$
I reprpesents the Van't Hoff factor whose value is 1 (for non electrolyte)
K_f= freezing point constant =$1.86°\mathrm{C}/\mathrm{m}$
m = molality
When you enter all the values provided in this formula, you get
$7.4^0\mathrm{C}=1\times \left(1.{86}^0\mathrm{C}/\mathrm{m}\right)\times \frac{xg}{62g/\mathrm{mol}\times 0.5 \mathrm{kg}}$$

x=123g

So, 500 g of water and 123 g of ethylene glycol were added.