Otopina dobivena otapanjem 330,0 g natrijeva klorida u 1500 g vode pri 20 °C ima gustoću $1,130\text{ g cm}^{-3}$$1,130\text{ g cm}^{-3}$. Koliko iznosi masena koncentracija pripremljene otopine?

Ukupna masa otopine jednaka je zbroju masa otopljene tvari i otapala:
$m(\text{otopina}) = m(\text{NaCl}) + m(\text{H}_2\text{O}) = 330{,}0\text{ g} + 1500\text{ g} = 1830{,}0\text{ g}$$m(\text{otopina}) = m(\text{NaCl}) + m(\text{H}_2\text{O}) = 330{,}0\text{ g} + 1500\text{ g} = 1830{,}0\text{ g}$
Volumen otopine izračunavamo iz njezine mase i gustoće pomoću relacije $V = \frac{m}{\rho}$$V = \frac{m}{\rho}$:
$V(\text{otopina}) = \frac{1830{,}0\text{ g}}{1{,}130\text{ g cm}^{-3}} = 1619{,}47\text{ cm}^3 = 1{,}6195\text{ dm}^3$$V(\text{otopina}) = \frac{1830{,}0\text{ g}}{1{,}130\text{ g cm}^{-3}} = 1619{,}47\text{ cm}^3 = 1{,}6195\text{ dm}^3$
Masenu koncentraciju otopine dobivamo dijeljenjem mase otopljene tvari s volumenom otopine: $\gamma(\text{NaCl}) = \frac{m(\text{NaCl})}{V(\text{otopina})}$$\gamma(\text{NaCl}) = \frac{m(\text{NaCl})}{V(\text{otopina})}$.
$\gamma(\text{NaCl}) = \frac{330{,}0\text{ g}}{1{,}6195\text{ dm}^3} = 203{,}8\text{ g dm}^{-3}$$\gamma(\text{NaCl}) = \frac{330{,}0\text{ g}}{1{,}6195\text{ dm}^3} = 203{,}8\text{ g dm}^{-3}$
Odgovor: B