explain
Mass of solute = 18 g
Molar mass of benzene = 78.11 g/mol
P = 80% P0 = 0.8 P0
Relative lowering of vapour pressure = (P0 – P) / P0 = XB = Mole fraction of solute
(P0 – 0.8 P0) / P0 = nB / (nA + nB)
0.2 = nB / (nA + nB)
But number of solute particles is very less compare to solvent particles, hence
0.2 = nB / nA
Mole of solvent = 117 / 78.11 ≈ 1.50
Let molar mass of solute be ‘Y’
0.2 = (18 / Y) / 1.50
0.3 = 18/Y
Y = 60 g
Molar mass of solute is 60 g.
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