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The addition of 0.643 g of a compound to...

The addition of `0.643 g` of a compound to `50 mL` of benzene (density 0.879 g `mL^(-1)`) lowers the freezing point from `5.51` to `5.03^(@)C`. If `K_(f)` for benzene is `5.12`, calculate the molecular weight of the compound.

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Addition of 0.643g of a compound to 50mL of benzene (density: 0.879g mL^(-1)) lower the freezing point from 5.51^(@)C to 5.03^(@)C . If K_(f) for benzene is 5.12 K kg mol^(-1) , calculate the molar mass of the compound.

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.

Addition of 0.643 g of compound to 50 ml of benzene (density =0.879mol^(-1) ) lower the freezing point from 50.51^(@)C . If k_(f) for benzene is 5.12 the molecule of the compound is

A solution of 0.643 g of an organic compound in 50 mL of benzene (density 0.879 g//mL ) lowered its freezing point from 5.51^(@)C to 5.03^(@)C . Calculate the molecular weight of solid. (K_(f) for benzene is 5.12 K mol^(-1) kg )

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