\(m_{H_2O}=V.d=100.1=100gam\)
\(n_{H_2}=\dfrac{m}{M}=\dfrac{0,48}{2}=0,24mol\)
2M+2H2O\(\rightarrow\)2MOH+H2
\(n_M=2n_{H_2}=2.0,24=0,48mol\)
M=\(\dfrac{3,33}{0,48}=6,9375\approx7\)(Li)
\(n_{LiOH}=n_{Li}=0,48mol\rightarrow m_{LiOH}=0,48.24=11,52gam\)
\(m_{dd}=3,33+100-0,48=102,85gam\)
C%LiOH=\(\dfrac{11,52}{102,85}.100\approx11,2\%\)