The boiling point elevation is 1.6617
The given data is
0.253 of naphthalene
190 g of liquid benzene
ΔTb = Kb M
molarity = moles of solute / kilograms of solvent
moles of naphthalene is 0.01248
molarity = 0.01248 / 0.190
= 0.06568 M
ΔTb = Kb M
= 2.53 × 0.6568
= 1.6617
Hence the boiling point elevation is 1.6617
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