$m_{dd}=1.1,09=1,09g$
$\to n_{HCHO}=\dfrac{1,09.37\%}{30}=\dfrac{0,4033}{30}(mol)$
Trong $10ml$ $X$ có $\dfrac{0,4033}{30}$ mol $HCHO$
$\to 4ml$ $X$ có $\dfrac{0,16132}{30}(mol)$ $HCHO$
$\to n_{Ag}=4n_{HCHO}=\dfrac{0,64528}{30}(mol)$
$\to m=108n_{Ag}=2,323g$