Answer to Question #145615 in General Chemistry for Nikayla

Question

Concentration of MgCl2 in a solution is 4.89 w/w%.
What is the freezing temperature of the solution in °C?

M(MgCl2) = 95.210 g·mol-1, Kf = 1.86 K·kg·mol-1

Answer 1

concentration of MgCl₂ = 4.89% w/w

This means for every 100g of the solution, 4.89g of MgCl₂ is in it.

Therefore,

mass of MgCl₂ = 4.89g

mass of water = 100g - 4.89g = 95.11g

no. of moles of MgCl₂ = $\frac{4.89g}{95.210g/mol}$ = 0.0514moles

mass of water (kg) = 95.11g/1000kg = 0.09511

molality (m) = no. of moles/mass of water(kg) = 0.0514/0.09511 = 0.54moles/kg

$∆T_f = ik_fm$

where i = 3, because MgCl₂ dissociates completely to give 3 moles of ions in a solution.

$\therefore ∆T_f$ = 3 × 1.86 × 0.54 = 3.0132K

T_f = 273.15 (normal boiling point) - 3.0132 = 270.1368K

Therefore, the new freezing point is 270.14K.