Question

Calculate the boiling point of a solution when 2g of na2so4 is dissolved in 50g of water

Answer 1

Ionization of Na₂SO₄

2Na⁺ + SO₄²⁻ : The number of ions obtained per formula unit is 3 and is indicated by i.

Change in boiling point is :

ΔBP = K°mi

Where :

K° for water = 0.52

m = Molality of the solution

i = Number of ions obtained per formula unit.

Molality = moles of Na₂SO₄ / Kilograms of water.

Moles of Na₂SO₄:

2/142 = 0.01408 moles

Molality = (1000/50) × 0.01408 = 0.282 m

ΔBP = 0.52 × 0.28 × 3 = 0.44°C

Bp increases by 0.44 °C

The new BP is :

100+0.44 = 100.44°C

Answer 2

Hey !!

Na₂SO₄ an ionizing produces 3 ions

  i = 3 (Vant Hoff factor)

ΔTb = 3 × 0.52 × 2 × 1000 / 142 × 50

 ΔTb = 0.4394 K

ΔTb ≈ 0.44 K

ΔTb = Tb - Tb°                                         Tb° = Boiling point of pure water

   Tb = 0.44 + 373.15

   Tb = 373.59 K (or) Tb = 100.44°C

GOOD LUCK !!