17.4% (mass/Vol) K₂SO₄ solution at 27°C is isotonic to 5.85% (mass/Vol) NaCl solution at 27°C. If NaCl is 100% ionised, the % ionisation of K₂SO₄ in aqueous solution is x × 10 then the value of x is ____

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answer is 6.

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Detailed Solution

$\begin{array}{l} π_{K_{2} {SO}_{4}} = π_{NaCl}; i_{1} \times C_{1} RT = i_{2} C_{2} RT \\ \Rightarrow i_{1} C_{1} = i_{2} C_{2}; i_{1} \times \frac{17 .4}{174} = 2 \times \frac{5 .85}{58 .5} \\ i_{1} = 2; α = \frac{i_{1} - 1}{n - 1} = \frac{2 - 1}{3 - 1} = \frac{1}{2}; \\ 10 x % = \frac{1}{2} = 50 %; x = \frac{50}{10} = 5 \\ \Rightarrow \frac{100}{133 .5} = 0 .75 \Rightarrow P_{(new)}^{H} = 6 \end{array}$