$a,m_{H_2SO_4}=50.9,8\%=4,9g.$
$⇒m_{H_2O}=50-4,9=45,1g.$
$b,PTPƯ:Zn+H_2SO_4\xrightarrow{} ZnSO_4+H_2↑$
$n_{H_2SO_4}=\dfrac{4,9}{98}=0,05mol.$
$Theo$ $pt:$ $n_{H_2}=n_{H_2SO_4}=0,05mol.$
$⇒V_{H_2}=0,05.22,4=1,12l.$
$c,m_{H_2SO_4}=4,9+(150.8\%)=16,9g.$
$m_{ddH_2SO_4}=50+150=200g.$
$⇒C\%_{H_2SO_4}=\dfrac{16,9}{200}.100\%=8,45\%$
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