Question

A compound is known to be a potassium halide, KX. If 3.00 g of the salt is dissolved in exactly 100 g of water, the solution freezes at –0.938 °C. Identify the halide ion in this formula. Kfp for H2O is –1.86 C/m.

X- = ______?

Answer 1

Tf   = 0.938⁰C

KX(aq) -------------> K^+ (aq) + X^- (aq)

i   = 2

Tf    = i*Kf*m

0.938   = 2*1.86*m

m     = 0.938/(2*1.86)   = 0.252m

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

0.252      = 3*1000/G.M.Wt * 100

G.M.Wt   = 3*1000/(0.252*100)   = 119

gram molecular weight of KX   = 1119g/mole

39 + x = 119

x           = 119-39   = 80

The halide is Br

The formula of potassium halide is KBr