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Question

calculate the mass of the compound whose molar mass is 256 gram per mole to be dissolved in 75 gram of benzene to elevate its boiling point by 0.48 Kelvin where KB for benzene is 3.28 Kelvin kg per mole

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Solution

Dear Student

ΔTb = i*molality of solution * Kb
Kb =3.28 K kg mol-1
ΔTb =0.48K
Assuming the value of i to be 1 (since nothing is specified)
the molality of solution = ΔTb / Kb = 0.48 / 3.28 = 0.146 mol kg-1
molality = mass of solute/(molar mass of solute * mass of solvent (in kg))
0.146 = mass of solute / (256* 75*10-3)
Hence mass of solute = 0.146*256* 0.075 gram = 2.8 grams

Regards

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