1. Tardigrade
  2. Question
  3. Chemistry
  4. The elevation in boiling point of a solution of 9.43 g of MgCl2 in 1 kg of water is (Kb = 0.52 K kg mol-1, Molar mass of MgCl2 = 94.3 g mol-1)

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Q. The elevation in boiling point of a solution of $9.43\, g$ of $MgCl_2$ in $1\, kg$ of water is $(K_b = 0.52\, K\, kg\, mol^{-1}$, Molar mass of $MgCl_2 = 94.3\, g\, mol^{-1})$

Solutions

Solution:

$ MgCl_{2} \rightleftharpoons Mg^{2+} + 2Cl^{-}$, $i = 3$
$\Delta T_{b} = iK_{b}m$
$ = 3 \times0.52 \times \frac{9.43}{94.3 \times 1} = 0.156$