In each of the following pairs, which one has the lower viscosity?

Question

a. $"HCOOH"$ or $"C"_8"H"_18$
b. $"C"_2"H"_5"Cl"$ or $"CH"_3"OH"$

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Answer 1 · 2015-07-31T16:47:53

a. $"C"_8"H"_18"$ and b. $"C"_2"H"_5"Cl"$ have the lower viscosities.

The viscosity of a liquid is determined by its intermolecular forces.

The stronger the intermolecular forces, the more difficult it is for the molecules to slide past each other.

a. $"HCOOH"$ or $"C"_8"H"_18$

$"C"_8"H"_18$ has only weak London dispersion forces, while $"HCOOH"$ has strong intermolecular hydrogen bonds.

So we predict that $"C"_8"H"_18$ has a lower viscosity than $"HCOOH"$ .

The prediction is confirmed by the actual viscosities.

The viscosity of $"C"_8"H"_18$ is 0.542 mPa·s, while the viscosity of $"CH"_3"-OH"$ is 1.78 mPa·s.

b. $"C"_2"H"_5"Cl"$ or $"CH"_3"-OH"$

$"C"_2"H"_5"Cl"$ or $"CH"_3"-OH"$

$"C"_2"H"_5"Cl"$ has weak dipole-dipole forces from the $"C-Cl"$ bonds, but $"CH"_3"-OH"$ has strong hydrogen bonds.

Confirmation: The viscosity of $"C"_2"H"_5"Cl"$ is 0.279 mPa·s, while the viscosity of $"CH"_3"-OH"$ is 0.593 mPa·s.