Question

What is the boiling point of water when 175.0 g of Na2SO4, a strong electrolyte is dissolved in 1.000 Kg of water?

Answer 1

Answer: $101.9^0C$

Explanation:

Elevation in boiling point is given by:

$\Delta T_b=i\times K_b\times m$

$\Delta T_b=T_b-T_b^0=(T_b-100)^0C$ = Elevation in boiling point

i= vant hoff factor = 3 (number of ions an electrolyte produce on complete dissociation)

$Na_2SO_4\rightarrow 2Na^++SO_4^{2-}$

$K_f$ = freezing point constant = $0.512^0C/m$

m= molality

$\Delta T_b=i\times K_b\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}$

Weight of solvent (water)= 1.000 kg

Molar mass of solute $Na_2SO_4$ = 142 g/mol

Mass of solute $Na_2SO_4$  = 175.0 g

$(T_b-100)^0C=3\times 0.512\times \frac{175.0g}{142g/mol\times 1.000kg}$

$T_b=101.9^0C$

Thus the boiling point of water when 175.0 g of $Na_2SO_4$, a strong electrolyte is dissolved in 1.000 Kg of water is $101.9^0C$