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Calculate the freezing point (in degrees C) of a solution made by dissolving 3.79 g of...

Calculate the freezing point (in degrees C) of a solution made by dissolving 3.79 g of anthracene{C14H10} in 81.1 g of benzene. The Kfp of the solvent is 5.12 K/m and the normal freezing point is 5.5 degrees C

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Answer #1

According to the queston , Here we have to use the formula delta Tf = Kfp m , here m is molality ,

Now , molar mass of anthracene is = 178.23

hence molality =[ 3.79 / 178.23 * 81.1 ] * 1000 = 0.2622 m

hecen delta Tf = 5.12 * 0.2622 = 1.342 0C

Hence the Freezing point is = 5.5 + 1.342 = 6.842 0C

Hence it is all about the given question thank you :)

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