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Addition of 0.40 g of a compound to 45.5...

Addition of `0.40 g` of a compound to `45.5 m L` of benzene (density `0.879 g mL^(-1)`) lowers the freezing point from `5.51^(@)C` to `4.998^(@)C`.If `K_(f)` for benzene is `5.12 K kg mol^(-1)`,calculate the molar mass of the compound.

Text Solution

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The given values are:
`W_("solute")=0.40 g`,`K_(f)=5.12 K m^(-1)`
`rho_("benzene")=0.879 g mL^(-1)`
`V_("benzene")=45.50 mL`
`DeltaT_(f)=T_(f)^(@)-T_(f)=5.51-5.03=0.212^(@)C`
Now weight of benzene =`V xx rho`
`=45.50xx0.879`
`=40.00 g`
Molar mass of solute,`M_("solute")` is calculated as
`Mw_("solute")=(K_(f)xxW_("solute")xx1000)/(W_("solvent")xxDeltaT_(f))`
`=(5.12xx0.40xx1000)/(40xx0.512)=400`
Thus, the molecular weight of solute is `400 g mol^(-1)`.
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