Calculate the molarity of 90.0 ml of a solution that is 0.92 % by mass nacl. assume the density of the solution is the same as pure water.

Answer is: molarity is 0,155 M.
V(solution) = 90,0 mL = 0,09 L.
ω(NaCl) = 0,92% ÷ 100% = 0,0092.
d(solution) = 1 g/mL.
m(solution) = V(solution) · d(solution).
m(solution) = 90 mL · 1 g/mL = 90 g.
m(NaCl) = 90 g · 0,0092 = 0,828 g.
n(NaCl) = 0,828 g ÷ 58,4 g/mol.
n(NaCl) = 0,014 mol.
c(solution) = 0,014 mol ÷ 0,09 L.
c(solution) = 0,155 mol/L.

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