When dT = Kf * molality * i
= Kf*m*i
and when molality = (no of moles of solute) / Kg of solvent
= 2.5g /250g x 1 mol /85 g x1000g/kg
=0.1176 molal
and Kf for water = - 1.86 and dT = -0.255
by substitution
0.255 = 1.86* 0.1176 * i
∴ i = 1.166
when the degree of dissociation formula is: when n=2 and i = 1.166
a= i-1/n-1 = (1.166-1)/(2-1) = 0.359 by substitution by a and c(molality) in K formula
∴K = Ca^2/(1-a)
= (0.1176 * 0.359)^2 / (1-0.359)
= 2.8x10^-3
298 g of calcium carbonate CaCO₃
Explanation:
We have the following chemical reaction:
CaCN₂ (s) + 3 H₂O (l) → CaCO₃ (s)+ 2 NH₃ (g)
number of moles = mass / molar weight
number of moles of H₂O = 161 / 18 = 8.94 moles
Knowing the chemical reaction we devise the following reasoning:
if 3 moles of H₂O produces 1 mole of CaCO₃
then 8.94 moles of H₂O produces X moles of CaCO₃
X = (8.94 × 1) / 3 = 2.98 moles of CaCO₃
mass = number of moles × molar weight
mass of CaCO₃ = 2.98 × 100 = 298 g
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