Answer to Question #71484 in General Chemistry for Jasmine

Question

What is the molar concentration of normal 0.90% saline (58.44 g/mol) solution? (Assume the density of the normal saline solution is 1.01 g/mL.)

Answer 1

C = n/V
n(saline) = m(saline) / M(saline) = 90 g / 58.44 g/mol = 1.54 mol
V(solution) = m(solution) / (solution) = 100 g / 1.01 g/mL = 99 mL = 0.099 L
C(saline) = n(saline) / V(solution) = 1.54 mol / 0.099 L = 15.5 M