Question

Calculate the amount of CaCl2, which must be added to 2 kg water so
that the freezing point is depressed by 61 K (Kf of H2O =1.86 K kg mol^-1,
Atomic mass of Ca = 40, Cl = 35.5)

Answer 1

Explanation:

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Answer 2

2.43 kg of CaCl₂ must be added to 2kg so that freezing point depressed by 61K

given, mass of water = 2kg

molar mass of CaCl₂ = 40 + 35.5 × 2 = 40 + 71 = 111 g/mol

mass of solute (CaCl₂) = M

molality, m = no of mole of solute/mass of solvent in kg

= (M/111)/2

= M/222

using formula, ∆T_f = ik_f × m

dissociation of CaCl₂ ;

CaCl₂ ⇔Ca^+ + 2Cl^-

so, i = 1 - α + α + 2α = 1 + 2α

for complete dissociation, α = 1

and then, i = 1 + 2 = 3

⇒61K = 3 × 1.86 K kg/mol × M/222

⇒61 × 222/3 × 1.86 = M

⇒M = 2426.88 g = 2.42688 kg ≈ 2.43 kg