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Narmada Gara
Asked: 2022-10-29T00:22:54+05:30 In:

`1.0 g` of a non-electrolyte solute( mol. Mass `250.0 g mol^(-1)`) was dissolved in `5.12 g` benzene. If the freezing point depression constant, `K_(f)` of benzene is `51.2 K kg mol^(-1)`, the freezing point of benzene will be lowered by:
A. `0.2 K`
B. `0.4 K`
C. `0.3 K`
D. `0.5 K`

`1.0 g` of a non-electrolyte solute( mol. Mass `250.0 g mol^(-1)`) was dissolved in `5.12 g` benzene. If the freezing point depression constant, `K_(f)` of benzene is `51.2 K kg mol^(-1)`, the freezing point of benzene will be lowered by:
A. `0.2 K`
B. `0.4 K`
C. `0.3 K`
D. `0.5 K`
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  1. Correct Answer – 2
    We need to calculate `DeltaT_(f)` (depression of freezing point of solvent) which is given as
    `DeltaT_(f)=iK_(f)m=iK_(f) W_(solute)/(mm_(solute)W_(solv))`
    we are given
    `i=1` as the solute is nonelectrolyte
    `K_(f)=5.12 K kg mol^(-1)`
    `W_(solute)=1.00 g`
    `W_(solvent)=51.2/1000kg`
    `mm_(solute)=250 g mol^(-1)`
    Substituting these results, we get
    `DeltaT_(f)=(1)(5.12 K kg mol^(-1))((1.00 g))/((250 g mol^(-1))(5.12/1000kg))`
    `=0.4 K`

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