Question

The normal freezing point of water, H2O is 0.00 °C and its Kfn value is 1.86 °C/m. Assuming complete dissociation of the electrolyte, if 12.39 grams of sodium nitrate (NaNO3, 85.00 g/mol) are dissolved in 245.8 grams of water what is the freezing point of the solution?

Answer 1

solving 2nd part first.

AT, = ikim

where, $\Delta T_{_{b}}$ is the elevation in boiling point. i is the van't hoff factor, K_b is the molal boiling point elevation constant

m is the molality of the solute.

molality is number of moles of solute per Kg of solvent.

i = 2 as KCl dissociates into two molecules in solution.

no of moles of KCL dissolved in 205.5 gm of water = 13.39 / 74.55 = 0.179 moles.

so no of moles of KCL dissolved in 1000 gm of solvent is 1000 x (0.179/205.5) moles = 0.874 moles

$\Delta T_{_{b}}$ = 2 x 0.512 ^oC / molal x 0.874 molal = + 0.895 ^oC is the elevation in boiling point

The boiling point of solution = 100.895 ^oC

1)  

AT, = iKfm

here i = 2 as NaNO₃ dissociates into 2 ions.

number of moles of NaNO₃ dissolved in 245.8 gm of water is 12.39/85 moles = 0.145 moles

number of moles of NaNO₃ dissolved in 1000 gm of water is (0.145/245.8) x 1000 moles = 0.593 moles

$\Delta T_{f}$ = 2 x 1.86 ^oC / molal x 0.593 molal = - 2.206 ^oC

Freezing point of the solution is - 2.206 ^oC