a) What is the freezing point of a solution with a mass of= $100\; g $ containing $11.7\; g$ table salt?
$ NaCl$ is a strong electrolyte which completely dissociates in water.
b) In reality the freezing temperature is somewhat higher. Why?
a)
Let be $s$=salt et $e$= water:
$m_e=100-11.7=88.3\;g$
If $n_s$ is the number of moles of salt, there will be $i\cdot n_s$ with $i=2$ moles of particles (ions) in solution:
$\mu_s$ $=$ $\frac{1000\cdot 2\; n_s}{m_e}$ $=$ $\frac{1000\cdot 2\; m_s}{M_s\cdot m_e}$
$\Delta T$ $=$ $K_{fus}\mu_c $ $= $ $K_{fus}\frac{1000\cdot 2\; m_s}{M_s\cdot m_e}$ $=$ $1.86\frac{1000\cdot 2\; 11.7}{58.5\cdot 88.3}=8.4K $
$t_{fus}$ $=$ $-8.4^oC$
b)
There are some associations between $Na^+$ ions and $Cl^-$ in this solution so that the number of particles is a little less high $(i \lt 2)$.
