$M=\frac{10 \times \% w / w \times d}{M_{s d d d e}}$
Here $M=3.6, \% \mathrm{w} / \mathrm{w}=29, \mathrm{d}=$ density, Msolute $=98$
$\therefore 3.6=\frac{10 \times 29 \times d}{98}$
$\Rightarrow d=1.22$
$^{\mathrm{M}}\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}=267.46 \;\mathrm{g} / \mathrm{mol}$ $\mathrm{M}_{\mathrm{AgNO}_{3}}=169.87 \;\mathrm{g} / \mathrm{mol}$