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Elevation in Boiling Point

The b.p. of a...

Question

The b.p. of a solution of 5g of sulphur in 100g of carbon disulphide is ${0.476}^{o}$ above that of pure solvent. Determine the moles of sulphur in this solvent. The b.p. of pure carbon disulphide is ${46.30}^{o} C$ and its heat of vaporation is 84.1 calories per gram.

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Solution

$Δ T_{b} = 0.47 6^{0} C K_{b} = 2.4 k g / m o l M = 32 n M = \frac{5 \times 1000}{32 n \times 100} Δ T_{b} = K_{b} / m 0.476 = 2.4 / \frac{5 \times 1000}{32 n \times 100} = \frac{24}{32} \times 5 = \frac{15}{4 n} n = \frac{15}{4 \times 0.476} = \frac{15}{1.9} = 8$

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Q. 45. The b.p. of a solution of 5 g of sulphur in 100 g of carbon disulphide is

{0.476}^{0}

above that of pure solvent. Determine the moles of sulphur in the solvent. The b.p. of pure carbon disulphide is

{46.30}^{0} C

and its heat of vaporisation is 84.1 calories per gram.

Q. The boiling point of a solution of

5 g

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of carbon disulphide is

{0.476}^{\circ} C

above that of pure solvent. Determine the molecular formula of sulphur in this solvent. The boiling point of pure carbon disulphide is

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Q. When a non-volatile solute of

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and its heat of vapourisation is

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Q. The standard heat of formation of carbon disulphide (l), given that the standard heat of combustion of carbon (s), sulphur (s) and carbon disulphide (l) are

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respectively is:

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