Question

The normal freezing point of water is 0oC and it's freezing point depression constant is 1.86oC/m. If a 2.00 molal solution of Na2SO4 is prepared, what is the freezing point of the mixture?

Answer 1

The answer is -11.16°C

Step-by-step explanation:

Answer 2

`\large \boxed{\text{-11.2 $^(\circ)$C}}`

Step-by-step explanation:

The formula for the change in freezing point ΔT_f caused by an electrolyte is

`\Delta T_(f) = -iK_(f)b`

1. Calculate i

Na₂SO₄(aq) ⟶ 2Na⁺(aq) + SO₄²⁻(aq)

1 mol Na₂SO₄ gives 3 mol ions.

i = 3

2. Calculate the freezing point

(a) Change in freezing point

`\Delta T_(f) = -iK_(f)b = -3 * 1.86 * 2.00 = -11.2 \,^(\circ)\text{C}`

(b) Freezing point

`\Delta T_(f) = T_(f) - T_(i)\\ T_(f) = \Delta T_(f) + T_(i) = 0.00 - 11.22 = \large \boxed{\textbf{-11.2 $^(\circ)$C }}`