70 Решение задачки по химии №06 745

3. The resistivity of a saturated solution of sparingly soluble salt A at 298 K is equal to ρ. The specific resistance of water at the same temperature is ρ H2O = 1 • 104 Ohm • m (the values of the specific resistance of water at the same temperature are different, since the researchers took water of different purity).
Calculate the solubility of salt A in water and the solubility product. The solubility product can be approximately calculated using the equation l=c c-, where c and c are the concentration of ions, kion/m3
Molar electrical conductivities at this temperature T and infinite dilution for substances B, C and D are equal to µ∞B µ∞C µ∞D and are given in Table. 34.

Option T, K A r * 10-4. Ohm * m rH2O * 10-4. Ohm*m
3 281 CaF2 0.0271 0.466

Option B µ∞ B*103 C µ∞ C*103 D µ∞ D *103
3 Ca(ClO4)2 21.4 ZnF2 18.6 Zn(ClO4)2 20.2

If you do not open the file - install the RAR archiver. Inside the archive you will find a solution in Word format.