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  • Calculate the boiling point of solution when 2g of  (M = 142 g mol-1) was dissolved in 50g of water, assuming <img alt="Na_{2}S

Calculate the boiling point of solution when 2g of Na_{2}SO_{4} (M = 142 g mol-1) was dissolved in 50g of water, assuming Na_{2}SO_{4} undergoes complete ionization.

(K_{b} for water = 0.52Kkgmol^{-1})

 

 

 

 
 
 
 
 

Answers (1)

K_{b} (water)=0.52Kkg mol^{-1}

M (Na_{2}SO_{4})=142 g mol^{-1}=0.142Kgmol^{-1}

Mass of solute \left (Na_{2}SO_{4} \right )=2g = 0.002kg

Mass of solvent (water)=50g = 0.05kg

T_{b}=?

Complete ionisation, i=3

\Delta T_{b}=T_{b}-T_{b}^{\circ}

\Delta T_{b}=i\times K_{b}\times m

\Delta T_{b}=\frac{3\times 0.52\times 0.002}{0.142\times 0.05}

\Delta T_{b}=0.439K

\Delta T_{b}=T_{b}-T_{b}^{\circ}

\Delta T_{b}=373.15+0.439

\Delta T_{b}=373.589K

Posted by

Sumit Saini

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