Question

0.48 g of an organic compound dissolved in 10.6 g of
benzene lowered the freezing point by 1.8°C. Find out
the molecular mass of the compound [molecular
freezing point constant for benzene = 50).
​

Answer 1

Answer:

The freezing depression of a solvent is dependent on the amount of solute in solution and the amount of solvent. You also need to know the freezing point of the pure solvent and its freezing point depression constant. These can be found in tables from a variety of sources

The freezing point of pure benzene is 5.5 deg C and its freezing point depression constant is -5.12 deg C/m where m is the molality of the solution (defined as the number of moles of solute divided by the kilograms of solvent).

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Answer 2

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Please find the answer to your question

Give Wt. of benzene (solvent),

W = Volume * density = 50 * 0.789 = 43.95 g

Wt. of compound (solute), w = 0.643 g

Mol. wt of benzene, M = 78

Mol. wt. of solute, m = ?

Depression in freezing point, ∆Tf = 5.51 – 5.03 = 0.48

Molal freezing constant, Kf = 5.12

Now we know that,

M – 1000 *Kf *w/W *∆Tf = 1000 *5.12 *0.643/43.95 *0.48 = 156.056

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