Answer to Question #209803 in General Chemistry for Japles27

Question

The FP of the solution that results from 223 g of Al2(SO4) dissolved in 1450 g of solvent would be ____°C (round answer to whole number)

Answer 1

For a solution that has a solute dissolved, we know that

$\Delta T_f = T_{f(pure\,solvent)} - T_{f (solution)} \\ \Delta T_f = K_f \times m$

Then, if we substitute for water K_f=1.86 °C/m and T_{f(pure solvent)} = T_{f (water)} = 0 °C we find the new temperature as:

$T_{f(pure\,solvent)} - T_{f (solution)} = K_f \times m$

$\large{\implies 0°C - T_{f (solution)} = 1.86 \frac{°C}{m} \times m}$

$\large{\implies T_{f (solution)} = -1.86 \frac{°C}{m} \times m}$

Then we have to calculate the morality of the solution so we divide the mass of Al2(SO4)3 into the amount of water that was used as solvent and we proceed to calculate the rate in mol/kg, thus the result will be given in temperature scale or °C:

$\implies T_{f (solution)} = -(1.86 \frac{°C}{m} )[\frac{223g\,Al_2(SO_4)_3}{1450\,g\,water}][\frac{1\,mol\,Al_2(SO_4)_3}{342.15\,g\,Al_2(SO_4)_3}][\frac{1000\,g\,water}{1\,kg\,water}]$

$\implies T_{f (solution)} = -[\frac{(223)(1000)(1.86)}{(1450)(342.15)}]\,°C$

$\implies T_{f (solution)} = -0.836\,°C=-0.84\,°C$

In conclusion, the FP of the solution is -0.84 °C.

Reference:

• Atkins, P., & De Paula, J. (2011). Physical chemistry for the life sciences. Oxford University Press, USA.