Question

The freezing point of water is 0.00°C at 1 atmosphere.

If 13.81 grams of nickel(II) bromide, (218.5 g/mol), are dissolved in 290.9 grams of water ...

The molality of the solution is _______m

The freezing point of the solution is _______ °C.

Answer 1

1)

Lets calculate molality first

Molar mass of NiBr2,

MM = 1*MM(Ni) + 2*MM(Br)

= 1*58.69 + 2*79.9

= 218.49 g/mol

mass(NiBr2)= 13.81 g

number of mol of NiBr2,

n = mass of NiBr2/molar mass of NiBr2

=(13.81 g)/(218.49 g/mol)

= 6.321*10^-2 mol

m(solvent)= 290.9 g

= 0.2909 Kg

Molality,

m = number of mol / mass of solvent in Kg

=(6.321*10^-2 mol)/(0.2909 Kg)

= 0.2173 molal

Answer: 0.2173 molal

2)

lets now calculate ΔTf

ΔTf = Kf*m

= 1.86*0.2173

= 0.404 oC

This is decrease in freezing point

freezing point of pure liquid = 0.0 oC

So, new freezing point = 0 - 0.404

= -0.404 oC

Answer: -0.404 oC