Given: Volume of solution = 200ml =0.2L, Mass of CH3COOH in gram =4.9g
Molecular Formula = H2SO4
Molecular mass of H2SO4 = 2(H)+1(S)+4(O) = 2 × 1 + 32 + 4 ×16 = 98m
Mass of H2SO4 in Mole (mol)
= Mass of H2SO4 in gram(g)/ Molecular mass(m)= 4.9/98 = 0.05 mol
Concentration of H2SO4 (g/L)
=Mass of solute (g)/Volume(L) = 4.9g/0.2L = 24.5 g/L
Concentration of H2SO4 (mol/L)
= Mass of H2SO4 (mol)/Volume of H2SO4 (L)
= 0.05mol/0.2L = 0.25 mol/L
Answer is 24.5 g/L & 0.25 mol/L