When 2.56 gram of sulphur was …
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Sia ? 6 years, 3 months ago
Given:
Kf = 3.83 K kg mol-1
Mass of solute = 2.56 gt
Mass of solvent = 100 g
Therefore,
Molality of the solution, {tex}m=\frac{2.56}{32} \times \frac{1000}{100}=0.8 m{/tex}
The depression in freezing point of a solution is given as
{tex}\Delta \mathrm{Tf}{/tex} = iKfm
0.383 = i {tex}\times{/tex}3.83 {tex}\times{/tex} 0.8
i = {tex}\frac {1} {8}{/tex}
Hence, 8 sulphur atoms are undergoing association, as shown below:
1Thank You