Question

Since CaCl2 dissociates into 3 ions , i = 3

Mass of 1 L water will be 1 kg as the density is 1.0 g/mL

kb = 0.515oC/m

Elevation in boiling point = 103.5 - 100 = 3.5oC

So putting all the values in the formula , we get :

3.5 = 3 x 0.515 x m

m = 1.359 m

Molality = moles of solute / mass of solvent in kg

1.359 = n / 1

n = 1.359

So the mass of CaCl2 added = 1.359 x 110.98

= 150.8 grams

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