PenyelesaianMr senyawa = 80massa = 300 g (0.3 kg)titik beku air = 0 °Ctitik beku larutan = −31 °Ckonstanta titik beku air [tex](K_f = 1.86\ ^\circ{C},~\text{kg mol}^{-1})[/tex][tex]\Delta T_f = 0 - (-31)[/tex][tex]= 31^\circ{C}[/tex][tex]m = \frac{\Delta T_f}{K_f}[/tex][tex]= \frac{31}{1.86}[/tex][tex]= 16.67\ \text{mol kg}^{-1}[/tex][tex]n = m \times \text{massa pelarut (kg)}[/tex][tex]= 16.67 \times 0.3[/tex][tex]= 5~\text{mol}[/tex][tex]m_s = n \times M_r[/tex][tex]= 5 \times 80[/tex][tex]= 400~\text g[/tex]massa senyawa = 400 g