The molal freezing point constant of water is -1.86. at what temperature will a solution containing 8.27 g of cacl2?

Temperature of fussion, for the solution is -0.41°C

Explanation:

To solve this we need to apply the colligative property, freezing point  the pression.

ΔT = Kf . m . i

ΔT = T° f pure solvent - T° f solution

Kf= cryoscopic constant (1.86° C / m for water)

m = molality (mol of solute/ 1kg of solvent)

i = Van't Hoff factor (number of ions dissolved in solution)

CaCl₂ → Ca²⁺ + 2Cl⁻   (for this salt, i = 3)

Molar mass calcium chloride = 110.98 g/m

8.27 g/ 110.98 g/m = 0.0745 m

0°C - T°f solution = 0.0745 m . 1.86°C /m . 3

0°C - T°f solution = 0.41°C

T°f solution = - 0.41°C

passing a magnet thought the

mixture , because iron gets attracted to magnet .