Question

What mass of CaCl2 should be dissolved in 175g of water so that the solution freezes at -4.50 deg.C? Assume 100% dissociation. For water, Kf = 1.86 deg.C/m.

A. 15.7g

B. 47.0g

C. 23.6g

D. 5.23g

E. 10.5g

Please provide explanation.

Answer 1

Tf = 4.5⁰C

Kf   = 1.86⁰C/m

CaCl2 (aq) -----------------> Ca^2+ (aq) + 2Cl^-(aq)

   i = 3

Tf   = i*m*Kf

4.5   = 3*m*1.86

m    = 4.5/3*1.86   = 0.81m

molality   =   W*1000/G.M.Wt * weight of solvent in g

0.81      = W*1000/111*175

W          = 0.81*111*175/1000   = 15.7g

mass of CaCl2    = 15.7g >>>>answer