Answer :


Kb = 0.52 K kg/mol

Mass of solute = 4g

Mass of solvent = 100g

Molality =

Hence, Molality of the solution, m = = 0.28

Na2SO4undergoes complete ionization and yields 3 moles of constituent ions for every mole of Na2SO4. Therefore, i = 3

Elevation in boiling point, Tb = iKbm

= 3*1.86*0.28


The boiling point of the solution(Tb) = 373.15 + 1.562 = 374.71K


(i) Colligative properties of solutions are properties that depend upon the concentration of solute molecules or ions, but not upon the identity of the solute. Colligative properties include vapor pressure lowering, boiling point elevation, freezing point depression, and osmotic pressure.

(ii) Ideal solution, homogeneous mixture of substances that has physical properties linearly related to the properties of the pure components. The classic statement of this condition is Raoult’s law, which is valid for many highly dilute solutions and for a limited class of concentrated solutions, namely, those in which the interactions between the molecules of solute and solvent are the same as those between the molecules of each substance by itself.


A. Given:

Kf = 3.83 K kg/mol

Mass of solute = 2.56g

Mass of solvent = 100g

Therefore, Molality of the solution,

The depression in freezing point of a solution, ΔTf = iKfm

0.383 = i x 3.83 x 0.8

i = 1/8

Hence 8 small sulphur atoms undergo association. Hence formula of Sulphur is S8.

B. (i) Since 1.2% sodium chloride solution is hypertonic wrt 0.9% sodium chloride solution, the blood cells which do not have cell wall unlike plant, when placed in 1.2% sodium chloride solution, water flows out of the cells and the cells shrink.

(ii) Since 0.4% sodium chloride is hypotonic wrt 0.9% chloride solution, when the blood cells are placed in 0.4% sodium chloride solution, water flows in to the cells and the cells swell.

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Match the terms gChemistry - Exemplar

Note : A statemenChemistry - Exemplar

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Concentration terChemistry - Exemplar