Question

What is the change in freezing point for a solution that contains 235 g Snly

Answer 1

Answer:

T° Freezing solution = - 2.91°C

Explanation:

To solve this, we apply this formula:

ΔT = Kf . m . i

Our solute is SnI₄

ΔT = Freezing T° of pure solvent - Freezing T° of solution

m is molality

i, numbers of ions dissolved.

SnI₄ →  Sn⁴⁺  +  4I⁻

5 moles of ions have been formed, 1 mol of Sn⁴⁺ and 4 moles of iodide.

Let's determine molality. Firstly we need the moles of salt.

235 g . 1mol / 626.3 g = 0.375 moles

molality = mol /kg of solvent → 0.375 mol /1.20kg = 0.313 m

Kf reffers to Cryoscopic constant, for water is 1.86°C/m. We replace data at formula →  0°C - T°F solution = 0.313 m . 1.86°C/m . 5

T°F solution = - 2.91°C