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Question

When a non-volatile solute of 10 g is added to 150 g of carbon disulphide, the boiling point of the solution raised to 0.55C above that of pure solvent. The boiling point of pure carbon disulphide is 46.30C and its heat of vapourisation is 84 cal/g. Determine the molar mass of the solute.

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Solution

We know,
Boiling point elevation constant,
Kb=RT2b1000 Lv
where,
Tb is boiling point of pure solvent in K
R in cal/mol/K
Lv is latent heat of vaourisation of solvent in cal/g
Thus,
Kb=2×(46.30+273)21000×84=2.43
Boiling point elevation of a solution is given by,
Tb=Kb×m
m=TbKb=0.552.43
And also, m=moles of solutewt. of solvent in grams×1000
=10M×1000150=2003M
where, M is the molar weight of the non-volatile solute.
Thus, 2003M=0.552.43; M=294.5 g/mol
Thus, the molar mass of non-volatile solute is 294.5 g/mol

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