$1.50\;g $ of a mixture of naphthalene $C_{10}H_{8}$ and of anthracene $C_{14}H_{10}$ are dissolved in $20.0 \; g$ benzene. This solution freezes at $2.81^oC.$ What is the composition of the mixture
→ Cryoscopic and ebullioscopic constants
$na$=naphtalène, $an$= anthracène, $mel$= their mixture, $b$= benzene: $\mu_{mel}$ $=$ $\frac{1000\cdot n_{mel}}{m_b}$ $\Delta T$ $=$ $K_{fus}\mu_{mel}$ so: $n_{mel}$ $=$ $\frac{\Delta T\cdot m_b}{K_{fus}\cdot 1000}$ $=$ $\frac{(5.50-2.81)20.0}{4.90\cdot 1000}$ $=$ $0.011\;mol$ Two equations: $n_{na}+n_{an}$ $=$ $0.011$ $1.50$ $=$ $n_{na}\cdot 128+n_{an}\cdot 178$ so: $n_{na}$ $=$ $0.0092\; mol$ $n_{an}$ $=$ $0.0019\; mol$
