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  • Calculate the boiling point of solution when 4g of  (Mg = 120 g mol-1) was dissolved in 100g of water, assuming <img alt="MgSO_{4}" src="https:/

Calculate the boiling point of solution when 4g of MgSO_{4} (Mg = 120 g mol-1) was dissolved in 100g of water, assuming MgSO_{4} undergoes complete ionization.

(K_{b} for water = 0.52 K kg mol-1)

 

 

 

 
 
 
 
 

Answers (1)

K_{b} for water = 0.52 K kg mol-1 

M = 120 g mol-1

Mass of solute = MgSO_{4} = 4g

Mass of solvent = H_{2}O = 100g

Complete ionization of MgSO_{4}, i=2

T_{b}= ?

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

T_{b}^{\circ}=373.15K

\Delta T_{b}=iK_{b}m

\Delta T_{b}=2 \times 0.52 \times \frac{4}{120}\times \frac{1000}{100}

\Delta T_{b}=2 \times 0.52 \times \frac{1}{3}

\Delta T_{b}=0.347K

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

T_{b}= 0.34+373.15

T_{b}= 373.49K

Posted by

Sumit Saini

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