The relative volumes of chloroform and water that should be used is 9:10
Concentration of solution in chloroform = [tex]90[/tex] ( moles of chloroform )
Concentration of solution in water = [tex]10[/tex] ( moles of water )
Dissociation constant at [tex]25^oC[/tex]; [tex]K_D = 10[/tex]
[tex]K_D =[/tex] Concentration of solution in chloroform / Concentration of solution in water
Meaning;
[tex]K_D = \frac{\frac{mole\ of\ chloroform}{volume\ of\ chloroform} }{\frac{mole\ of\ water}{volume\ of\ water} }[/tex]
Since [tex]90[/tex] mole is present in chloroform and [tex]10[/tex] mole is present in water, Total mole of Caffeine present is [tex]100[/tex]
Now, we substitute our given values into the equation
[tex]10 = \frac{\frac{90}{volume\ of\ chloroform} }{\frac{10}{volume\ of\ water} }\\\\10 *\frac{10}{volume\ of\ water} = \frac{90}{volume\ of\ chloroform} \\\\\frac{100}{volume\ of\ water} = \frac{90}{volume\ of\ chloroform}\\\\\frac{volume\ of\ chloroform}{volume\ of\ water} = \frac{90}{100}\\\\ \frac{volume\ of\ chloroform}{volume\ of\ water} = \frac{9}{10}\\\\ \frac{volume\ of\ chloroform}{volume\ of\ water} = 9:10[/tex]
Therefore, the relative volumes of chloroform and water that should be used is 9:10
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