Bài giải :
$-D_{H_2O}=1(g/ml)$
a.
`-n_{SO_3}=\frac{160}{80}=2(mol)`
`-m_{H_2O}=240.1=240(g)`
`SO_3+H_2O→H_2SO_4`
2 → 2 (mol)
`⇒m_{chất..tan..H_2SO_4}=2.98=196(g)`
`⇒m_{dd..sau.pứ}=m_{SO_3}+m_{H_2O}=160+240=400(g)`
`⇒C%H_2SO_4=\frac{196}{400}.100%=49%`
b.
`-n_{Na}=\frac{46}{23}=2(mol)`
`-m_{H_2O}=250.1=250(g)`
`2Na+2H_2O→2NaOH+H_2↑`
2 → 2 2 1 (mol)
`⇒m_{chất..tan..NaOH}=2.40=80(g)`
`⇒m_{dd..sau..pứ}=m_{Na}+m_{H_2O}-m_{H_2}=46+250-1.2=294(g)`
`⇒C%NaOH=\frac{80}{294}.100%≈27,21%`