Question

What mass percent of KBr in water would give you a -1.5^O freezing point? What would be the new boiling point?

Answer 1

Freezing point depression is one of the colligative properties of matter, which means it is affected by the number of particles, not the chemical identity of the particles or their mass. When a solute is added to a solvent, its freezing point is lowered from the original value of the pure solvent.

We know, K_f of water = 1.86 °C kg/mol

To find the temperature change elevation of a solvent by a solute, we use the freezing point depression equation:

ΔT = iK_fm

where
ΔT = Change in temperature in °C = 1.5⁰ C
i = van 't Hoff factor
K_f = molal freezing point depression constant in °C kg/mol
m = molality of the solute in mol solute/kg solvent.

For KBr, KBr ----> K⁺ +Br^-

So, the van't hoff factor (i) = 2

So, m= ΔT/(i*K_f)

m= 0.403 molal KBr

Basis: 1kg of solvent (water)

molality (m) of KBr = moles of KBr/kg water

0.403 molal KBr = Moles of KBr/1 Kg

So, moles of KBr = 0.403

moles= mass of KBr/Molecular weight of KBr

Molecular weight of KBr= 119 g/mol

So, mass of KBr = moles* Molecular weight of KBr = 0.403moles* 119g/moles = 47.957 g of KBr

So, mass percent of KBr in water = [47.957/(1000+47.957)]* 100 = 4.57%

Similarly for boiling point elevation,

ΔT = iK_bm

We have, i= 2

k_b for water = 0.512 °C kg/mol

m= 0.403 molal Kbr

So, ΔT = 0.412 °C

New Boiling point = 100°C (normal boiling point of water) + 0.412 °C = 100.412 °C