Question

1 mole of na2so4 is dissolved in 100 g of water. Calculate:



a. The depression in freezing point



b. The freezing point of the solution.

Answer 1

a. 10 Kf

b. 273- 10kf K

Answer 2

a) The depression in freezing point is $55.8^0C$

b) The freezing point of the solution is $-55.8^0C$

Explanation:

Molality of a solution is defined as the number of moles of solute dissolved per kg of the solvent.

$Molarity=\frac{n}{W_s}$

where,

n = moles of solute

 $W_s$ = weight of solvent in kg

$Molality=\frac{1mole}{0.1kg}=10mole/kg$

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

$\Delta T_f=3\times 1.86\times 10=55.8^0C$

$T_f^0-T_f=(0-T_f)^0C=55.8^0C\\T_f=-55.8^0C$

Thus the freezing point of the solution is $-55.8^0C$

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