Chemistry, asked by himanshu199874, 11 days ago

The boiling point of a solution of 5 g of sulphur in 100 g of carbon disulphide is 0.474°C above that of pure
solvent. Determine the molecular formula of sulphur in this solvent. The boiling point of pure carbon disulphide
is 47°C and its heat of vaporisation is 84 calories per gram.​


Answered by kobenhavn

Answer: 246 g


Formula used for Elevation in boiling point :

\Delta T_b=k_b\times m


\Delta T_b=k_b\times \frac{\text{ Moles of solute}\times 1000}{\text{ Mass of solvent in Kg}}


\Delta T_b = change in boiling point

k_b = boiling point constant

m = molality

T_{solution}-T_{solvent}=2.34\times \frac{\text{ given mass}\times 1000}{\text {Mass of solvent in g}\times {\text {Molar mass of solute}}}

0.474=2.34\times \frac{5g\times 1000}{100g\times {\text {Molar mass of sulphur}}}

{\text {Molar mass of sulphur}}=246g

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