750 mili liter larutan NaOH 10 molal. Jika Mr NaOH = 40 dan massa jenis larutan = 1,12 g/ml. Tentukan molaritas larutan tersebut.

[tex] V_{larutan} = 750 ml \\ \rho= 1.12\, g/ml \\ molalitas = 10 \, molal \\ \\ p = 750ml \times 1.12\, g/ml = 840\ gram \\m = \frac{m}{Mr} \times \frac{1000}{p} \\ 10= mol \, NaOH \times \frac{1000}{840} \\ \\ mol \ NaOH = \frac{10 \times 840}{1000} \\ NaOH = 8,4\ mol \\ \\ molaritas = mol \times 1000/V \\ molaritas = 8,4 \ mol \times \frac{1000}{750} \\ molaritas = 11.2\ mol / liter \\ \\Persen\ massa \\ massa\ NaOH = 8.4\ mol * 40\ gr/mol = 336\ g \\ p = massa\ seluruh\ campuran = 840\ g[/tex]

Persen massa = 336 g / 840 g * 100 % = 0.4 * 100 % = 40 %