Question

The normal boiling point of water, H₂O is 100.00 °C and its K_bp value is 0.512 °C/m.  

Assuming complete dissociation of the electrolyte, if 10.30 grams of magnesium sulfate (MgSO₄, 120.4 g/mol) are dissolved in 227.9 grams of water, what is the boiling point of the solution?

Answer 1

Answer:

We know that elevation in boiling point is

∆Tb=(Tb)soution - (Tb)solvent=i x Kb x m

Where (Tb)solu solu=boiling point of solution, (Tb) solvent=boiling point of solvent=100°C for water.

I=Van't Hoff factor=2 for MgSO4

Since MgSO4 is strong electrolyte and dissociates into

MgSO4 ------> Mg^+2 + SO4^2- (2 ions)

Kb=boiling point constant=0.512 °C/m

m=molality=moles of solute/mass of solvent

m= moles of MgSO4/mass of water.

Given mass of MgSO4=10.30 g and molar mass=120.4 g/mol

Therefore moles of MgSO4=mass/molar mass=10.30 g/120.4 g/mol=0.0855 mol

Mass of water=227.9 g=0.2279 Kg (since 1 Kg=1000 g)

Therefore molality=moles of MgSO4/mass of water=0.0855 mol/0.2279 Kg=0.3753 m.

Therefore (Tb) solution=(Tb)water + (ixKbxm)

(Tb) solution=100°C+(2 x 0.512 °C/m x 0.3753 m)

(Tb) solution=100.384°C.

Therefore boiling point of solution ~ 100.38°C.

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