Answer:
0.104 M
Step-by-step explanation:
A saline solution contains 0.770 g of NaCl (molar mass = 58.55 g/mol) in 133 mL.
The molar mass of the solute (NaCl) is 58.55 g/mol. The moles corresponding to 0.770 g are:
0.770 g × (1 mol/55.85 g) = 0.0138 mol
The volume of solution is 133 mL. In liters,
133 mL × (1 L/1000 mL) = 0.133 L
The molarity of NaCl is:
M = moles of solute / liters of solution
M = 0.0138 mol / 0.133 L
M = 0.104 M