Ethylene glycol is used as coolant in car radiators, in order to prevent the solution from freezing at -0.3 $^{o}$ C. The amount of ethylene glycol to be added to 5 kg of water is (for water, $K_{f} = 1.86 K m^{- 1})$ :

20 g

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50 g

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40 g

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30 g

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Solution

The correct option is B 50 g
Let W g of $e t h y l e n e g l y c o l$ is added.
The molar mass of $e t h y l e n e g l y c o l = 62$ g/mol.
The depression in the freezing point $Δ T_{f} = K_{f} m$
$0.3 = 1.86 \times \frac{W}{62 g / m o l \times 5 k g}$
$W = 50$ g

Option B is correct.

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Similar questions

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Ethylene glycol is used as coolant in car radiators, in order to prevent the solution from freezing at -0.3oC. The amount of ethylene glycol to be added to 5 kg of water is (for water, Kf=1.86 Km−1):

The correct option is B 50 gLet W g of ethyleneglycol is added.The molar mass of ethyleneglycol=62 g/mol.The depression in the freezing point ΔTf=Kfm0.3=1.86×W62g/mol×5kgW=50 gOption B is correct.

Ethylene glycol is used as coolant in car radiators, in order to prevent the solution from freezing at -0.3 $^{o}$ C. The amount of ethylene glycol to be added to 5 kg of water is (for water, $K_{f} = 1.86 K m^{- 1})$ :

The correct option is B 50 g
Let W g of $e t h y l e n e g l y c o l$ is added.
The molar mass of $e t h y l e n e g l y c o l = 62$ g/mol.
The depression in the freezing point $Δ T_{f} = K_{f} m$
$0.3 = 1.86 \times \frac{W}{62 g / m o l \times 5 k g}$
$W = 50$ g

Option B is correct.