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he boiling point of benzene is 253.23 k. When 1.80 g of a non-volatile solute was dissolved in 90g of benzene, the boiling point is raised to 354.11k. Calculate the molar mass of the solute. Kb for benzene is 2.53 k Kg mol-1
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ans is 0.5016
∆Tb=KbWeight of solute /Mx weight of solvent)
∆Tb= boiling point of solution - boiling point of pure solvent
= 354.11-253.23=100.88
kb= 5.12
weight of solute = 1.8 g
weight of solvent = 90x10^-3 kg
molar mass = M
now using above formula solve the problem
∆Tb=KbWeight of solute /Mx weight of solvent)
∆Tb= boiling point of solution - boiling point of pure solvent
= 354.11-253.23=100.88
kb= 5.12
weight of solute = 1.8 g
weight of solvent = 90x10^-3 kg
molar mass = M
now using above formula solve the problem
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