Calculate the volume of glycol $(C_2H_6O_2,\;\rho = 1.11 \frac{g}{cm^3} )$ to be added to $15.0\;L$ of water to make a solution that freezes at $-30.0^ oC$
→ Cryoscopic and ebullioscopic constants
$g$=glycol and $e$= water: $\mu_g$ $=$ $\frac{1000\cdot n_g}{m_e}$ $=$ $\frac{1000\cdot m_g}{M_g\cdot m_e}$ $\Delta T$ $=$ $K_{fus}\mu_g$ so: $m_g$ $=$ $\frac{\Delta T\cdot M_g\cdot m_e}{1000\cdot K_{fus}}$ $=$ $\frac{30\cdot 62\cdot 1500}{1000\cdot 1.86}$ $=$ $1500\;g$ $V_g$ $=$ $\frac{m_g}{\rho_g}$ $=$ $1351 \;cm^3$
