10% (m/m) aqueous potassium iodide has a density of 1.202 g mL^-1 . The true statements about this solution are

10% solution of Kl contains, 10 g of KI in 90 g (or 0.09 kg) of water. 
Molar mass of KI = (39 +127) g mol^-1 = 166 g mol^-1
$\therefore \mathrm{molality} \mathrm{of} \mathrm{solution}=\frac{\mathrm{Number} \mathrm{of} \mathrm{moles} \mathrm{of} \mathrm{KI}}{\mathrm{Mass} \mathrm{of} \mathrm{water} (\mathrm{in} \mathrm{kg})}$
                       $\begin{array}{l} =\frac{\left(10\mathrm{g}/166\mathrm{g}{\mathrm{mol}}^{-1}\right)}{0.09\mathrm{kg}}\\ =0.67\mathrm{mol}{\mathrm{kg}}^{-1}\end{array}$
$\mathrm{Further},\mathrm{Molarity} \left(\mathrm{M}\right)=\frac{ \mathrm{Moles} \mathrm{of} \mathrm{KI} }{ \mathrm{Volume} \mathrm{of} \mathrm{solution}(\mathrm{in}\mathrm{L}) }$
                                 $\begin{array}{l}=\frac{\left(10\mathrm{g}/166\mathrm{g}{\mathrm{mol}}^{-1}\right)\times 1000}{100\mathrm{g}/1.202\mathrm{g}{\mathrm{mL}}^{-1}}\\ =\frac{10/166\times 1000}{83.19}\end{array}$
$[\because \text{ Volume }=\text{ Mass }/\text{ density }=100/1.202=83.19\mathrm{mL}]$
                                                      $=0.72{\mathrm{molL}}^{-1}=0.72\mathrm{M}$
$\text{ Number of moles of }\mathrm{KI}=\frac{10\mathrm{g}}{166\mathrm{g}{\mathrm{mol}}^{-1}}=0.06\mathrm{mol}$
$\text{ Number of moles of water }=\frac{90\mathrm{g}}{18\mathrm{g}{\mathrm{mol}}^{-1}}=5\mathrm{mol}$
$\text{ Mole fraction of }\mathrm{KI}=\frac{{\mathrm{n}}_{\mathrm{KI}}}{{\mathrm{n}}_{\mathrm{KI}}+{\mathrm{n}}_{{\mathrm{H}}_2\mathrm{O}}}$
                                $=\frac{0.06}{5+0.06}=\frac{0.06}{5.06}=0.01186$