What is the molarity of K⁺ in aqueous solution that contains 17.4 ppm of K₂SO₄ (molar mass = 174 g mo^-1) ?

${\mathrm{K}}_2{\mathrm{SO}}_4\text{ is }17.4\mathrm{ppm}\text{, i.e. }{10}^6\mathrm{g}\text{ has }17.4{\mathrm{gK}}_2{\mathrm{SO}}_4$
$\begin{array}{r}1\mathrm{L}\left({10}^3\mathrm{mL}\right)\text{ has }{\mathrm{K}}_2{\mathrm{SO}}_4=\frac{17.4\times {10}^3}{{10}^6}=0.0174\mathrm{g}/\mathrm{L}\\ =\frac{0.0174}{174}\mathrm{mol}/\mathrm{L}\end{array}$
$\begin{array}{l}\therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\left[{\mathrm{K}}_2{\mathrm{SO}}_4\right]=1\times {10}^{-4}\mathrm{M}\\ \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{K}}_2{\mathrm{SO}}_4\rightleftharpoons 2{\mathrm{K}}^{+}+{\mathrm{SO}}_4^2\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\left[{\mathrm{K}}^{+}\right]=2\times {10}^{-4}\mathrm{M}\end{array}$