Question

A cosmetic company manufactures hand sanitizers, and one of the main ingredients used is 70.0 %
isopropyl alcohol by volume. [Note: The density of isopropyl alcohol is 0.786 g/cm3 and Mr of
isopropyl alcohol (CH3CHOHCH3) is 60.0 g/mol]
a) What is the IUPAC name of isopropyl alcohol?
b) Calculate the molality of isopropyl alcohol.
c) Calculate the molarity of isopropyl alcohol.
d) Calculate the % by mass of isopropyl alcohol.
e) Calculate the ppm by mass of isopropyl alcohol.

Answer 1

a) 2-methylpropanol

b) Let the volume of solution to be 100 cm³. As there are 70% of isopropyl by volume, then the volume of isopropyl is 70 cm³, and that of water 30 cm³.

$molality = \frac{moles _{solute}}{mass_{solvent} ( kg)}$

$mass _{isopropyl} = V\times d = 70 cm^3\times 0.786 \frac{g}{cm^3}=55.02 g$

$moles _{isopropyl} = 55.02 g \times (\frac{1 mole}{60.0 g})=0.917 mol$

$mass_{water} = 30 cm^3 \times 1 \frac{g}{cm^3}\times \frac{1 kg}{1000 g}=0.03 kg$

$molality = \frac{0.917 mol}{0.03 kg} = 30.6 m$

c) $molarity =\frac{moles_{solute}}{volume _{solvent}(L)}=\frac{0.917 moles}{0.1 L}=9.17 M$

d) $\% mass = \frac{mass_{solute}}{mass _{solution}} = \frac{55.02}{55.02+30} \times 100\% =64.7 \%$

e) $ppm=\frac{mass_{solute}(g)}{mass _{solution)(g)}}\times 10^6= \frac{55.02}{55.02+30}\times 10^6=6.47\times 10^5$