A very small amount of non volatile solute (that does not dissociate) is dissolved in 56.8 cc of benzene (density is 0.889 g/cc). At room temperature the vapour pressure of the solution is 98.88 mm Hg while that of benzene is 100 mm Hg. Find the molality of the solution. If the freezing temperature of this solution is 0.73° lower than that of benzene, what is the value of molal freezing point depression constant of benzene.no spamming else I will report yr all questions and answers...
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According to Roult’s law
Po – p/po = w/m * M/W ;∆Tf = Kf * m
Substituting the given values;
1000 – 98.88/98.88 = w *78 *1000/m *W * 1000
w/m * 1000/W = 1.12 *1000/78.98.88 = 0.1452
∴ Molality = 0.1452 (∵ w/m * 1000/W = Molality)
Further ∆T = Kf molality
0.73 =Kf * 0.1452
Kf =5.027 K molality-1
Step-by-step explanation:
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