Question

a 50 ml sample of ( c_{6}h_{14}(l) ) is mixed with a 50 ml sample of ( h_{2}o(l) ), and the mixture is shaken vigorously. the two liquids do not stay mixed but instead form two separate layers. the density of hexane is ( 0.66 g/ml ), and the density of water is ( 1.00 g/ml ). a ( 1.0 g ) sample of ( i_{2}(s) ) is added to the mixture, which is shaken again. which of the following best predicts what happens to the ( i_{2}(s) )?

a ( i_{2} ) will be found mainly in the top layer because it will dissolve more in the ( h_{2}o(l) )

b ( i_{2} ) will be found mainly in the bottom layer because it will dissolve more in the ( h_{2}o(l) )

c ( i_{2} ) will be found mainly in the top layer because it will dissolve more in the ( c_{6}h_{14}(l) )

d ( i_{2} ) will be found mainly in the bottom layer because it will dissolve more in the ( c_{6}h_{14}(l) )

Explanation:

Step1: Analyze solubility principle

Like dissolves like. \(C_{6}H_{14}\) is non - polar (hydrocarbon), \(H_{2}O\) is polar. \(I_{2}\) is non - polar.

Step2: Determine layer density

Density of \(C_{6}H_{14}(0.66\ g/mL)\lt\) density of \(H_{2}O(1.00\ g/mL)\). So \(C_{6}H_{14}\) (top layer) is non - polar, \(H_{2}O\) (bottom layer) is polar.

Step3: Predict \(I_{2}\) distribution

Since \(I_{2}\) is non - polar, it will dissolve more in non - polar \(C_{6}H_{14}\). And \(C_{6}H_{14}\) is the top layer.

Answer:

C. \(I_{2}\) will be found mainly in the top layer because it will dissolve more in the \(C_{6}H_{14}(l)\)