Question

The boiling point of water/H₂O is 100.00 °C at 1 atmosphere.

If 12.84 grams of magnesium acetate, (142.4 g/mol), are dissolved in 233.0 grams of water ...

The molality of the solution is _____m.

The boiling point of the solution is ______°C.

Answer 1

1)

mass(solute)= 12.84 g

number of mol of solute,

n = mass of solute/molar mass of solute

=(12.84 g)/(142.4 g/mol)

= 9.017*10^-2 mol

m(solvent)= 233.0 g

= 0.233 Kg

Molality,

m = number of mol / mass of solvent in Kg

=(9.017*10^-2 mol)/(0.233 Kg)

= 0.387 molal

Answer: 0.387 molal

2)

Mg(CH3COO)2 dissociates to give 3 particles

So, i=3

lets now calculate ΔTb

ΔTb = i*Kb*m

= 3*0.512*0.387

= 0.60 oC

This is increase in boiling point

boiling point of pure liquid = 100.0 oC

So, new boiling point = 100 + 0.60

= 100.6 oC

Answer: 100.6 oC