n Na2SO⁴ = 19.88 : 142 = 0.14 mol
500 ml = 0.5 l
CM Na2SO⁴ = n/V = 0.14/0.5 = 0.28 M
\(n_{Na_2SO_4}=\dfrac{19,88}{142}=0,14\left(mol\right)\)
\(C_{M_{ddNa_2SO_4}}=\dfrac{0,14}{0,5}=0,28M\)
\(n_{Na_2SO_4}=\dfrac{19,88}{142}=0,14\left(mol\right)\\ C_{MddA}=C_{MddNa_2SO_4}=\dfrac{0,14}{0,5}=0,28\left(M\right)\)