Organsko otapalo A često se koristi u industriji boja i lakova.

Volumen uzorka otapala A računamo iz definicije gustoće $\rho = \frac{m}{V}$$\rho = \frac{m}{V}$:
$V(\text{A}) = \frac{m(\text{A})}{\rho(\text{A})}$$V(\text{A}) = \frac{m(\text{A})}{\rho(\text{A})}$
Uvrštavanjem zadane mase $m(\text{A}) = 30{,}0\text{ g}$$m(\text{A}) = 30{,}0\text{ g}$ i gustoće $\rho(\text{A}) = 1{,}04\text{ g cm}^{-3}$$\rho(\text{A}) = 1{,}04\text{ g cm}^{-3}$ dobivamo:
$V(\text{A}) = \frac{30{,}0\text{ g}}{1{,}04\text{ g cm}^{-3}} = 28{,}85\text{ cm}^3 \approx 28{,}9\text{ cm}^3$$V(\text{A}) = \frac{30{,}0\text{ g}}{1{,}04\text{ g cm}^{-3}} = 28{,}85\text{ cm}^3 \approx 28{,}9\text{ cm}^3$
Odgovor: $28{,}9\text{ cm}^3$$28{,}9\text{ cm}^3$

Relativna molekulska masa jednaka je omjeru mase jedne molekule i unificirane jedinice atomske mase ($u = 1{,}66 \times 10^{-24}\text{ g}$$u = 1{,}66 \times 10^{-24}\text{ g}$):
$M_r(\text{A}) = \frac{m_f(\text{A})}{u}$$M_r(\text{A}) = \frac{m_f(\text{A})}{u}$
Uvrštavanjem mase jedne molekule dobivamo:
$M_r(\text{A}) = \frac{1{,}79 \times 10^{-22}\text{ g}}{1{,}66 \times 10^{-24}\text{ g}} = 107{,}83 \approx 108$$M_r(\text{A}) = \frac{1{,}79 \times 10^{-22}\text{ g}}{1{,}66 \times 10^{-24}\text{ g}} = 107{,}83 \approx 108$
Odgovor: 108