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What is freezing point of a liquid ? The...

What is freezing point of a liquid ? The freezing point of pure benzene is 278.4 K. Calculate the freezing point of the solution when 2.0 gram of a solute having molecular weight 100 gram is added to 100 gram of benzene. (`K_(f)` for benzene `=5.12 kg "mol"^(-1)`)

Text Solution

Verified by Experts

Freezing point of a liquid is a temperature at which the vapour pressure of solid is equal to the vapour pressure of the liquid.
Given,
`(T^(@))` Freezing point of benzene = 278.4 K
(T) Freezing point of solution = ?
`(W_(2))` Mass of solute `=2.0 g= 2.0xx10^(-3)kg`
`(M_(2))` Molar mass of solute `=100 g = 0.1 kg `
`K_(f)` of benzene `=5.12` kg `"mol"^(-1)`
`(W_(1))` Mass of solvent `=100 g = 0.1 kg`
`Delta T_(f)=(W_(2)xxK_(f)xx1000)/(M_(2)xxW_(1))`
`T^(@)-T=(2.0xx5.12xx10^(-3))/(0.100xx0.1)`
`278.4-T=1.024`
`278.4-T=1.024K`
`T=-1.024+278.4`
`=277.376` K
Freezing point of solution is 277.376 K.
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Knowledge Check

  • The decrease in freezing point of an aqueous solution of a substance is 1.395 K and that in the freezing point of a benzene solution of the same substance is 1.28 K. The substance :

    A
    dissociates in aqueous solution as well as in the benzene solution
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    dissociates in the aqueous solution and not in the benzene solution
  • The freezing point (.^(@)C) of a solution containing 0.1 g of K_(3)[Fe(CN)_(6)] (molecular weight 329) on 100 g of water (K_(f)=1.86 K kg mol^(-1))

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    `2.3 xx 10^(-2)`
    B
    `5.7 xx 10^(-2)`
    C
    `5.7 xx 10^(-3)`
    D
    `-1.20 xx 10^(-2)`
  • What happens to freezing point of benzene when naphthalene is added

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