\(\rm a)\)

\(\rm m_{H_2SO_4}={50.20}/{100}=10(g)\)

\(\rm\to C\%_{H_2SO_4\ sau}=\dfrac{10}{50}.100\%=20\%\)

\(\rm b)\)

\(\rm V_{dd\ H_2SO_4\ sau}=\dfrac{50}{1,1}\approx 45,5(mL)=0,0455(L)\)

\(\rm n_{H_2SO_4}=\dfrac{10}{98}=\dfrac{5}{49}(mol)\)

\(\rm\to C_{M\ H_2SO_4\ sau}=\dfrac{\frac{5}{49}}{0,0455}\approx 0,224M\)

a, m_{H_2SO_4}={C% xx m_(dd)}/100={50 xx 20}/100=10(g)

-> C%_{H_2SO_4\ s\ a\ u\ k\ h\ i\ p\ h\ a\ l\ o\ ã\ n\ g}=10/50 xx 100=20%

b,

V_{dd\ H_2SO_4\ s\ a\ u\ k\ h\ i\ p\ h\ a\ l\ o\ ã\ n\ g}={m_(dd)}/D=50/{1,1}~~ 45,5(ml)

n_{H_2SO_4}=m/M=10/98~~0,102(mol)

-> C_{M\ H_2SO_4\ s\ a\ u\ k\ h\ i\ p\ h\ a\ l\ o\ ã\ n\ g}=n/V={0,102}/{0,0455}~~2,24(M)

(45,5ml=0,0455l)