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K(f) for water is 1.86 K kg "mol"^(-1). ...

`K_(f)` for water is 1.86 K kg `"mol"^(-1)`. If your automobile radiator holds 1.0 kg of water, how many grams of ethylene glycol `(C_(2)H_(6)O_(6))` must you add to get the freezing point of the solution lowered to - `2.8^(@)C` ?

A

27 g

B

72 g

C

93 g

D

39 g

Text Solution

Verified by Experts

The correct Answer is:
C
`DeltaT_(f)=(K_(f)xx w_(2)xx1000)/(w_(1)xxM_(2))`
`DeltaT_(f)=0-(-2.8)=2.8^(@)`,
`K_(f)=1.86"K kg mol"^(-1)`
`w_(1)=1.0kg=1000g`
`2.8=(1.86xxw_(2)xx1000)/(1000xx62)`
`w_(2)=(2.8xx1000xx62)/(1000xx1.86)=93.3g`
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