Question

calculate molality of NaCl solution whose elevation in boiling point is equal to the depression in freezing point 0.5 m sodium carbonate solution in water asuming complete dissociation of salt​

Answer 1

Molality of NaCl solution is 2.7 m

Explanation:

Depression in freezing point is given by:

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

$\Delta T_f$ = Depression in freezing point

$k_f$ = freezing point constant = 1.86

i= vant hoff factor = 3  (ions on complete dissociation)   $Na_2CO_3\rightarrow 2Na^++CO_3^{2-}$

m= molality  = 0.5

$=1.8\times m$

$\Delta T_f=3\times 1.86\times 0.5=2.79^0C$

Now elevation in boiling point is given by:

$\Delta T_b=k_b\times m$

$\Delta T_b$ = elevation in boiling point

$K_b$ = boiling point constant = $0.52K/kgmol$

m= molality  = ?

$2.79=2\times 0.52\times m$  

$m=2.7$

Molality of NaCl solution is 2.7 m

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