Question

1.33 dm3 9f H2O at 70°C are saturated by 2.25 moles of Pb(NO3)2 and 1.3 dm3 of H2O at 18°C are saturated by 0.53 moles of the same salt .if 4.50 dm3 of the saturated solution are cooled from 75- 18 °C , calculate the amount of solute deposited in moles and gram

Answer 1

Let us introduced a coefficient which will give

$extra \ mole\ of \ solute\ dissolved\ per\ unit \ increase\ in\ temprature\ per\ dm^3\ of solvent\\ volume.$

If volume of second solution is 1.33$\ dm^3$ then saturated solution would have been made at 0.54 mole

$\because \ \frac{1.33}{1.3}\times0.53=0.54\ mole$

$i.e.\\=\frac{amount\ of\ excess\ solute\ dissolved}{increase\ in \ temprature\ \times volume\ (in\ dm^3)}$

Let us denote it by $k_s$

$k_s=\frac{2.25-0.54}{(70-18)\times(1.33)}=0.024\ mol\ dm^{-3}\degree C^{-1}$

Amount of solute diposited(in moles) = $k_s\times (75-18)\times 4.5\ mole=0.024\times 57\times4.5=6.156mole$

Amount of solute dissolved (in gram) = molar mass\times\ no. of moles diposited$=331.2\times 6.156=2038.9\ g$