Question

Calculate the boiling point elevation for a solution prepared by adding 10 g of MgCl₂ to 200 g of water assuming MgCl₂ is completely dissociated. 

(K_b for Water = 0.512 K kg mol^–1 , Molar mass MgCl₂ = 95 g mol^–1 )

Answer 1

Number of moles of \(MgCl_2 = \frac{10}{95}\)

Mass of water = 200 gm

Molality of MgCl2 solution \(\frac{10 \times 1000}{95\times 200} = \frac{10}{19}\)

i = 3 for MgCl₂

_{\(\Delta T_b = iK_bm\)}

_{\(= 3 \times 0.512 \times \frac{10}{19}\)} 

= 0.81 K

\(\therefore\) Elevation in boiling point of MgCl₂ solution = 0.81 K