Mass of ethylene glycol (antifreeze) to be added to 18.6 kg of water to protect the freezing point at $-24^{\circ} \mathrm{C}$ is ____ kg (Molar mass in $\mathrm{g}^{-1}$ for ${ }^{-1}$ for ethylene glycol $62, \mathrm{~K}_{\mathrm{f}}$ of water $=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ )

JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Solutions

Mass of ethylene glycol (antifreeze) to be added to 18.6 kg of water to protect the freezing point at

- 2 4^{\circ} C

is ____ kg (Molar mass in

g^{- 1}

for

^{- 1}

for ethylene glycol

62, K_{f}

of water

= 1.86 K kg mol^{- 1}

)

Answer: 15

Solution

Δ T_{f} = i K_{f} m

. Ethylene glycol is a non-electrolyte, so

i = 1

. With

Δ T_{f} = 24 K

K_{f} = 1.86 K kg mo l^{- 1}

, and 18.6 kg of water:

24 = 1 \times 1.86 \times \frac{W /62}{18.6}

W = \frac{24 \times 62 \times 18.6}{1.86} = 14880 g = 14.88 kg \approx 15 kg

JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Integer answer

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