Answer is: molarity of this solution is 0,55 mol/l.
If we use 100 ml of solution:
V(solution) = 100 ml ÷ 1000 ml/l = 0,1 l.
ω(NaClO) = 4,00% = 0,04.
d(solution) = 1,02 g/ml.
m(solution) = V(solution) · d(solution).
m(solution) = 100 ml · 1,02 g/ml.
m(solution) = 102 g.
m(NaClO) = ω(NaClO) · m(solution).
m(NaClO) = 0,04 · 102 g = 4,08 g.
n(NaClO) = m(NaClO) ÷ M(NaClO).
n(NaClO) = 4,08 g ÷ 74,44 g/mol = 0,055 mol.
c(NaClO) = 0,055 mol ÷ 0,1 l = 0,55 mol/l.