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Chapter 10: Problem 22

Which of the following is more soluble in \(\mathrm{CCl}_{4} ?\) In each case, explain your answer. (a) hexane \(\left(\mathrm{C}_{6} \mathrm{H}_{14}\right)\) or \(\mathrm{CaCl}_{2}\) (b) \(\mathrm{CBr}_{4}\) or \(\mathrm{HBr}\) (c) benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) or ethyl alcohol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) (d) \(\mathrm{I}_{2}\) or \(\mathrm{NaI}\)

Short Answer

Expert verified
a) Hexane (C6H14) or Calcium chloride (CaCl2) b) Carbon tetrabromide (CBr4) or Hydrogen bromide (HBr) c) Benzene (C6H6) or Ethyl alcohol (C2H5OH) d) Iodine (I2) or Sodium iodide (NaI) Answer: a) Hexane (C6H14) b) Carbon tetrabromide (CBr4) c) Benzene (C6H6) d) Iodine (I2)

Step by step solution

01

(a) Determine the polarity of hexane and CaCl2

Hexane (C6H14) is a non-polar hydrocarbon. Meanwhile, Calcium chloride (CaCl2) is an ionic compound and thus polar.
02

(a) Compare hexane and CaCl2 with the polarity of CCl4

Carbon tetrachloride (CCl4) is a non-polar compound. Since like dissolves like, hexane, a non-polar compound, is more soluble in CCl4 than the polar compound CaCl2.
03

(b) Determine the polarity of CBr4 and HBr

CBr4 is a non-polar compound due to its symmetric structure. Hydrogen bromide (HBr) is a polar compound because of the electronegativity difference between hydrogen and bromine.
04

(b) Compare CBr4 and HBr with the polarity of CCl4

Since CCl4 is a non-polar compound, it will dissolve the non-polar compound CBr4 better than the polar compound HBr.
05

(c) Determine the polarity of benzene and ethyl alcohol

Benzene (C6H6) is a non-polar hydrocarbon. Ethyl alcohol (C2H5OH) is a polar compound due to the polar hydroxyl (OH) group.
06

(c) Compare benzene and ethyl alcohol with the polarity of CCl4

As CCl4 is a non-polar compound, it will dissolve the non-polar compound benzene better than the polar compound ethyl alcohol.
07

(d) Determine the polarity of I2 and NaI

Iodine (I2) is a non-polar compound since it is composed of two iodine atoms with equal electronegativities. Sodium iodide (NaI) is an ionic compound and thus polar.
08

(d) Compare I2 and NaI with the polarity of CCl4

Because CCl4 is a non-polar compound, it will dissolve the non-polar compound I2 better than the polar compound NaI.

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