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

`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_(2))` must you add to get the freezing point of the solution lowered to `-2.8^(@)C` ?

A

72 g

B

93 g

C

39 g

D

27 g

Text Solution

Verified by Experts

The correct Answer is:
B
`Delta T_(f)= i xx k_(f)xx m`
`2.8=1xx1.86xx(x)/(62xx1)`
`x=(2.8xx62)/(1.86)=93 gm`
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K_(f) for waer 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_(2)) must you add to get the freezing point of the solution lowered to -2.8^(@)C ?

45 g of ethylene glycol (C_(2) H_(6)O_(2)) is mixed with 600 g of water. The freezing point of the solution is (K_(f) for water is 1.86 K kg mol^(-1) )

(i) Prove that depression in freezing point is a colligative property. (ii) 45 g of ethylene glycol (C_(2)H_(6)O_(2)) is mixed with 600g of water . Calculate the freezing point depression. ( K_(f) for water = 1.86 k kg mol^(-1) )

45 g of ethylene glycol C_(2)H_(6)O_(2) is mixed with 600 g of water. Calculate (a) the freezing point depression and (b) the freezing point of solution. Given K_(f)=1.86 K kg mol^(-1) .

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