Mass of ethylene glycol (antifreeze) to be added to 18 . 6kg of water to protect the freezing point at - 24 ° C is kg (Molar mass in gmol ^ - 1 for ethylene glycol 62, K _ f of water = 1 . 86Kkg mol ^ - 1 )

Correct answer: 0 Given, Δ T _ f = - 24 ^ 0 C Molar mass of Ethylene glycol= 62 Δ T _ f = i K _ f × molality 24 = ( 1 ) × 1 . 86 × (W)/(62 × 18 . 6) W = 14880gm = 14 . 880kg

JEE Main 20241 FebruaryEvening Shift

Exam: JEE Main
Year: 2024
Date: 1 February
Shift: Evening Shift
Subject: Chemistry
Chapter: Solutions

JEE Main 2024 · 1 February · Evening Shift · Chemistry · Solutions

Mass of ethylene glycol (antifreeze) to be added to 18 . 6kg of...

Mass of ethylene glycol (antifreeze) to be added to

18 . 6kg

of water to protect the freezing point at

- 24 ° C

kg

(Molar mass in

{gmol}^{- 1}

for ethylene glycol

62, {K}_{f}

of water

= 1 . 86Kkg{mol}^{- 1}

)

Correct answer

Step-by-step solution

Given, ${\Delta T}_{f} = - {24}^{0}C$

Molar mass of Ethylene glycol= $62$

${\Delta T}_{f} = i{K}_{f} \times$ molality

$24 = ( 1 ) \times 1 . 86 \times \dfrac{W}{62 \times 18 . 6}$

$W = 14880gm$

$= 14 . 880kg$

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