1 mol of each MgCl2, NaCl and glucose are dissolved in 3kg of water. NaCl is completely ionised and MgCl2 is ‘10x’% ionised. If the solution boils at 100.867°C. What is x ? Kb = 0.52 K kg mol–1

ΔTb=i×m×Kb⇒0.867=13(2+1+i)×0.52⇒0.867=0.523(3+i)(3+i)=3×0.8670.52=5i=2⇒i=(1+2α)⇒2=(1+2α)α=1/2=0.5=50%

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1 mol of each MgCl₂, NaCl and glucose are dissolved in 3kg of water. NaCl is completely ionised and MgCl₂ is ‘10x’% ionised. If the solution boils at 100.867°C. What is x ? K_b = 0.52 K kg mol^–1

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Detailed Solution

$\begin{array}{l} {ΔT}_{b} = i \times m \times K_{b} \\ \Rightarrow 0 .867 = \frac{1}{3} (2 + 1 + i) \times 0 .52 \\ \Rightarrow 0 .867 = \frac{0 .52}{3} (3 + i) \\ (3 + i) = \frac{3 \times 0 .867}{0 .52} = 5 \\ i = 2 \Rightarrow i = (1 + 2 α) \Rightarrow 2 = (1 + 2 α) \\ α = 1 / 2 = 0 .5 = 50 % \end{array}$