IIT JEE Syllabus - KEEEL

Mole fraction of benzene = 1 ( X2 = 1 ( 0.185 = 0.815. Prob 2. 1.2 gm of a non-volatile substance was dissolved in 100 gm of acetone at 20(C. The vapour pressure of the solution was found to be 182.5 torr. Calculate the molar mass of the substance (vapour pressure of acetone at 20(C is 185.0 torr) Sol. From equation . or M1 = 42.92 gm/mol. Prob 3. ................
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