Question

Assuming 100% dissociation, calculate the freezing point ( T f ) and boiling point ( T b ) of 2.40 m Na 2 SO 4 ( aq ) . Colligative constants can be found in the chempendix.

Answer 1

Answer: The freezing point of the solution is $-13.4^0C$

The boiling point of the solution is $103.7^0C$

Explanation:

Depression in freezing point is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=T_f^0-T_f=(0-T_f)^0C$ = Depression in freezing point

i= vant hoff factor = 3 (for electrolyte undergoing complete dissociation, i is equal to the number of ions produced)

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

$K_f$ = freezing point constant = $1.86^0C/m$

m= molality = 2.40 m

$(0-T_f)^0C=3\times 1.86\times 2.40$

$T_f=-13.4^0C$

Thus the freezing point of the solution is $-13.4^0C$

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

$(T_b-100)^0C=3\times 0.512\times 2.40$

$T_b=103.7^0C$

Thus the boiling point of the solution is $103.7^0C$