39.
$n_{Ag}=\dfrac{10,8}{108}=0,1(mol)$
$\Rightarrow n_{C_6H_{12}O_6}=\dfrac{1}{2}n_{Ag}=0,05(mol)$
$C_{M_{C_6H_{12}O_6}}=\dfrac{0,05}{0,5}=0,1M$
40.
$n_{C_6H_{12}O_6}=\dfrac{18}{180}=0,1(mol)$
$\Rightarrow n_{Ag}=0,1.2=0,2(mol)$
$m_{Ag}=0,2.108=21,6g$