Gọi kim loại đã cho là R, ta có:
$2R+2H_2O→2ROH+H_2$
$n_{H_2}=\dfrac{0,22}{2}=0,11$
$⇒n_{ROH}=2n_{H_2}=0,22(mol)$
$⇒n_{OH}=n_{ROH}=0,22(mol)$
$⇒m_{ROH}=m_R+m_{OH}=5,06+0,22.17=8,8(g)$
Ta có: $m_{dd}=70,28.1,19=83,6332(g)$
$⇒$%$C_{ROH}=\dfrac{8,8.100}{83,6332}≈10,52$%