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The volume of a couridal particle , $\;V_{c}\;$ as compared to the volume of a solute particle , $\;V_{s}\;$ in a true solution could be

$(a)\;\large\frac{V_{c}}{V_{s}} \approx 10^{-3}\qquad(b)\;\large\frac{V_{c}}{V_{s}} \approx 10^{3}\qquad(c)\;\large\frac{V_{c}}{V_{s}} \approx 1\qquad(d)\;\large\frac{V_{c}}{V_{s}} \approx 10^{23}$

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Answer : $\large\frac{V_{c}}{V_{s}} \approx 10^{3}$
Explanation :
For a true solution , the diameter range is 1 to $\;10A^{0}\;$ , and for colloidal solution , diameter range is $\;10 - 1000 A^{0}\;$ .
Taking lower limits ,
$\large\frac{V_{c}}{V_{s}}=\large\frac{\large\frac{4}{3} \pi r_{c}^{3}}{\large\frac{4}{3} \pi r_{s}^{3}}=(\large\frac{r_{c}}{r_{s}})^{3}$
We know ,
$r_{c}=\large\frac{10}{2}=5A^{0}\qquad \; , r_{s}=\large\frac{1}{2}=0.5A^{0}$
Therefore , $\;\large\frac{V_{c}}{V_{s}}=(\large\frac{5}{0.5})^{3}=10^{3}$
answered Mar 18, 2014 by yamini.v
edited Mar 18, 2014 by yamini.v
 

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