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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

`93 g`

B

`39 g`

C

`27 g`

D

`72 g`

Text Solution

Verified by Experts

The correct Answer is:
A
Given`K_(f) = 1.86 K kg mol^(-1)`
`Delta T_(f) = 0-(-2.8) = 2.8^(@)C`
Mass of solvent `= 1.0 kg`, Mass of solute `=?`
Molecular mass of solute `= 62`
`:.` molarity, `m = ("Weight of solute")/(("Molecular mass of solute")/(Mass of solvent (g)")) xx 100`
`m = (w//62)/(1000)xx1000 = (w)/(62)`
Now, `Delta T_(f) = K_(f) xx m`
`2.8 = 1.86 xx (w)/(62) rArr w = (62 xx 2.8)/(1.86) = 93 g`
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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) )