Cryometry

Exercise 5

    

An aqueous solution $\mu = 0.0096 \frac{mol}{kg}$ thorium nitrate $(Th (NO_3) _4)$ gives rise to a cryoscopic lowering of $0.0703 ^ oC$ What is the number $i$ of dissolved ions produced by molecule (see discussion $i$ →       here )

→   Cryoscopic and ebullioscopic constants

If $i$ is this ion number, the molality is multiplied by $i$: With: $t$= thorium nitrate and $e$= water: $\mu$ $=$ $i\mu_t$ $=$ $i\cdot 0.0096$ $\Delta T$ $=$ $K_{fus}\mu$ so: $i$ $=$ $\frac{0.0703}{0.0096\cdot 1.86}\approx 4$