A. When 2.56 g of sulphur was dissolved in 100 g of 
the freezing point lowered by 0.383 K. Calculate the formula of sulphur 
(
for 
Atomic mass of Sulphur 
]
B. Blood cells are isotonic with 0.9% sodium chloride solution. What happens if we place blood cells in a solution containing
(i) 1.2% sodium chloride solution?
(ii) 0.4% sodium chloride solution?
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, when the blood cells are 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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